mathdos(2009)

mathdos(2009)

Mon 03/02/09

and [L[B]] β to γ = B

since L[A] = L[B] ---> [L[A]] β to γ = [L[B]] β to γ

<==

Let A = B Show L[A] = L[B]

Let v→ in R^n be arbitrary

L[A](v) = Av definition 12.1

similarly, L[B](v) = Bv

since A = B,

L[A](v) = L[B](v)

. .. usng that matrix multiplcation is a well-defined operation -- when you multiply two things together, you can only gett one result

therefore, L[A] = L[B]

12.4

L[A+B] = L[A] + L[B]

let v in V be aribitrary

Then

L[A+B](v) = (A+B)v

= Av + Bv (analogy to thm 10.6)

= L[A](v) + L[B](v)

by defintion (L[A] + L[B])(v)

therefore L[A+B] = L[A] + L[B]

L[aA] = aL[A]

Let v be in V

L[aA](v) = (aA)v // matrix multiplication

= a(Av) .. scalar multplication

= aL[A](v) / by definition of L[A]

= aL[A](v)

therefore, L[aA] = aL[A]

you need to look over 12.5 some more

if these two linear tranformation are euqal ... one way to prove that ,

.. two things in 12.6 are both linear transformation .. therefore, we only need to see what they do to a basis

Theorum 12.7

I is mxn Identity matrix

L[I]: Rn → Rn

Show L[I] = I[Rn]

Let x in R^n

I[R^n](x) = x by definition of I[R^n]

2.

L[I](x) = Ix by def 12.1

Ix→ = [identity matrix] * [column: x1, x2, ... xn]

= [column: x1 x2 .. xn]

therefore Ix = x

Since I[R^n](x) = x, therefore L[I](x) = I[R^n](x)

this is true for all x,

therefore,

L[I] = I[R^n] since x is arbitrary

Thm 12.8

Show that (AB)C = A(BC)

L[(AB)C] = L[AB]L[C] = (L[[A]L[B])L[C] by Them 12.6

= L[A](L[B]L[C]) by thm 10.2

= L[A]L[BC] = L[A(BC)] by 12.6

By Thm 12.3

(AB)C = A(BC)

therefore, matrix multiplication is associative

Chapter 13 -

inverses and isomorphisms

with functions, we say it has an inverse if it's 1-1 and onto

in the end, inverse here means the same as in set thoery

.. linearity of T => T^-1 is linear

Thm 13.1

T: V → W

U1: W → V

U2: W → V

show that they're equal

Let w in W be arbitrary

v = U1(w)

T(v) = w

Since U2 is the inverse,

U2(w) = v

therefore U1(w) = U2(w)

therefore, U1 = U2

Thm 13.2 -

if T is invertible, thee T is 1-1 and onto

Suppose T is invertible

T: V→W txists U st UW --< V st w = T(v)

iff v = U(w)

where w in W and V in V

1-1 Let (v1) = T(x2) = 0 st v, x in V and w in W

U(w) = v1 = and U(w) = v2

therefore v2 = v1

therefore T is 1-1

onto

Let w in W

Let (w)= v st v in V

T(v) = w by def 13.1

therefore T is onto

thoerum 13.3

a) T^-1 o T = Iv

Let v in V be arbitrary

Let T(v) = w

where w in W

therefore, v = T^-1(w)

show (T o T)(v) =

.,.

T o T(v) = T^-1 (T(v))

= T(w)

= v

= Iβ(V)

since B is arbitrary , t^-1 composed on I[v]

get through A.104

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Wed 03/04/09

next friday: 14 and 15

13.3 continued

a) T^-1 T = I[v]

b) T T^-1 = I[w]

Proof

Let w in W be arbitrary

Let T^-1(w) = v

Then T(v) = w

(T T^-1)(w) = T(T^-1(w))

= T(v)

= I[w](w)

since w is arbitrary T T^-1 = I[w]

Thm 13.4

Let T: V → W and let U:W→V

be functions st

UT = I[v] and TU = I[w]

Show U = T^-1

Let w = T(v)

U(w) = U(T(w))

= UT(v)

= I[v](w)

= v

U(w) = v

Let v = U(w)

so T(v) = T(U(w))

= TU(w)

= I[v](w)

= w

therefore T(v) = w

Thm 13.5

Let T:V→W and U: W→Z be invertible

Then a) UT:V→Z is invertible and (UT)^-1 = T^-1 U^-1

Show (UT)(T^1U^1) = I

(U(TT^1)(U^1)

(U(I[w])U^1)

(U (I[w] U ^-1) )

(U U^-1)

= I[z]

(T^-1U^-1)(UT)

= (T^-1(I[w])T)

= (T^-1(I[w]T))

= T^-1 (T) = T^1 T = I[v]

therefore, (UT)^-1 = T^-1 U^-1

b) Show T^-1: W → V is invertible and (T^-1)^-1 = T

T T^-1 = I[w]

T^-1 T = I[v]

therefore, T is the inverse of T^-1

Therefore (T^-1)^-1 = T

Show that I[v] : V → V is invertable

show that I[v]^-1 = I[v]

I[v]I[v] = I[v]

therefore I[v]^-1 = I[v]

A.101

Show V is isomorphic to W is an equivalence relation

Show 1) reflexivity 2) symmetry 3) transitivity

1) by ex 10 pg 25, I[V] : V → V is linear

by Thm 13.6, I[v]: V → V is invertible

therefore

V is isomorphic to V

2) by Thm 13.7, if T: V→W is linear and invertible, then T^-1 is linear

by thm 13.5 , T^-1 is invertible

therefore V is isomorphic to W implies W is isomorpohic to V

V ~ W => W ~ V

(3) by Thm 10.1, comp UT : V→Z of inear tranformation T: V→W and U: W→Zwere also linear

by Thm 13.5 comp UT: V→Z of invertible linear transformations T:V→W and U: W→Z were also invertible

therefore V is isomorphic to W and W is isomorphic to to Z ==> V is isomorphic to Z.

V~W and W~Z => V~Z

therefore, isomorphism is an equivalence relation

Prove V is isomorphic to W <=> dim(V) = dim(W)

(<=) suppose dim(V) = dim(W)

by problem B.56, there exists T:Vf→W that is 1-1 and onto and a linear transformation, since 1-1 and onto <=> invertible

therefore, V is isomorphic to W

(==>) Suppose V is isomorphic to W

Let β = {x1, x2, ... xn} be a basis for V

there exists an invertible function T: V→W

Then {T(x1), T(x2) .. T(xn)} generates R(T) and that R(T) = W, by Thm 8.1 for "generates R(T)"

R(T) = W because T is onto .. becaase we know that it's invertible

so then this set generates W

By problem B.48, {x1, x2, .. xn} is linearly independant then {T(x1), T(x2) .. T(xn)} IS linearly independant

therefore, dim(W) = n = dim(V)

therefore, V is siomorphic to W <==> Dim(V) = dim(W)

Cor 13.10 -

dim(V) = n

dim(R^n) = n

therefore dim(V) = dim(R^n)

and therefore is isomorphic by 13.9

for Friday, through 14.

....

Iv: V→ V

[I[v]] β' to β

this is a change of basis matrix from β' to β

if you look at thm 14.1 b, give sthe purpose of the change of base matrix .

if you start with a coordinate vector for v realtive for β'

and you multiply by this change of Basis

changes it foom form associated with β' to the form associated with β

T: V→W

[T]β and [T]β'

you get two different matrices in general .. depending on what basis you use

the only thing we know at the moment is that these two would be square matricies .. nxn

some relationship?

illustrated in 14.2 -- 1 matrix is issentially 1 matrix with the othere >>>

this relationship is so important that's called a name - similarity

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Fri 03/06/09

13.11 .. V is a n- dimentional over R

with Basis β

Let φ: V -> R^n

φ(v) = [v]β

show φ is isomorphic

1. Lin trans

φ(v) = [T(v)]β for any T by 11.2

I[v] = T .. is a linear transformation

φ(v) = [I[v](v)]β = [v]β

2. Show 1-1

Suppose φ(v) = 0 = [column 0 0 0 ... 0]

sufficient to use the nullspace ..because of theorum we had for lin trans

therefore

[v]β = 0→

therefore, v = 0x1 + 0x2 + 0x2 .... + 0xn

= 0

N(φ) = {0}

3) show onto -

Let [column a1, a2, a3 .. an]

be an arbitrary element of R^n

Let v→ = a1x1 + a2x2 + .. + anxn

φ(v) = [column a1, a2, a3 .. an]

Thm 13.14 -

Let V be n dimensional with basis β and let W be m-dimentinal with basis γ

Let ψ : L(V, W) => Mmxn(R) be defined by

ψ(T) = [T]β to γ prove ψ is an iso

Prove ψ is linear

- 1) additive property

ψ(T+U) = [T+U]β to γ

= [T]β to γ + [U]β to γ by thum 9.4

= ψ(T) + ψ(U)

1) homogeneity

ψ(t) = [ct]β to γ

= c[T]β to γ by 9.4

= cψ(T)

1-1

Suppose ψ(T) = ψ(S)

therefore [T] β to γ = [S] β to γ

[[T] β to γ]^(j) = [[S] β to γ]^j

= T(xj) = S(xj)

by cor 7.7, T=s

Onto

- L(V,w) → Mmxn(R)

Let A in Mmxr(R)

since there exists T in L(V,W)

[T] β to γ = A by b.68

therefore ψ(T) = A

Cor. 13.15

If Dim(V) = n and Dim(W) = m

Then Dim(L(V, W)) = mn

Since L(v,W) is similar to Mmxn(R)

= Dim(Mmx) = mn

Dim(L(v,W) = mn by Thm 13.9

13.16 - if A is invertible , then B in 13.3 is unique

Def 13.3 - A is invert iff exists B st AB = BA = I[n]

proof

suppose A is an nxn matric and that A in invertible

show B is unique

Suppose AB = BA = I[n] and AC = CA = In

Show B = C

B = B(Iv)

B = B(Iv) = B(AC .. since AC = 1x

= (BA)C

= (In)C

therefore, C

therefore B = C and therefore B is unique

13.17 - Let T: V →w be invertible

Let β be ordered basis for V

an

d let γ be ordred basis for W

[T]β to γ [T^-1]β to γ = [TT]γ by thm 10.4

= [Iw]γ

For (xj) = xj = 0x1 .. -xn

So [Iv]^j = I^j

Since true for all j, [Iw]γ = In

[T]γ to β [T] β to γ = [t^-1T]γ by 10.4

= [I[v]] β

secondly

So [Iv]β ^j = I^j

since j ar

bitrary [Iv]β = In

By 13.3 [T]β to γ is invertible namely [T^-1]γ to β = ( [T]γ to β )^-1

Let T: V→W be linear

β = {x1, x2, ... xn] is a basis for V

γ = {y1, y2, .. yn} is a basis for W

Suppose A = [T] β to γ is an intertable matrix

Show T is an invertable trans.

let yj = Σ A^-1[j] xi

therefore [U] γ to γ = A ^-1 def 9.3

[u]γ to β * [tt] β to γ = [ut] β

therefore

[ut]β = A^-1 A = I

UT(xj) = xj

UT = Iv

[T]β to γ [U] γ to β = [TU]γ

[TU]γ = A A^-1 = I

TU(yj) = 0y1 .. + 0yn

= yj

therefore TU = Iw

therefore U = T^-1 by 13.4

*** end of test *****

14

Q allows you to take cooridante vectors in β and translate them in coordinate vectors in β'

Thm 14.1

(a) Q = [Iv]β' to β is invertable

proof:

observe that Iv is invertable as a transformation

By 1

3.17, [Iv]β' to β is an invertable matrix

therefore Q is invertable -- because it's a matrix representation of an invertable transformation

(b) [v]β = Q[v]β'

Q[v]β' = [Iv]β' to β [v]β'by 11.4

= [Iv(v)]β

= [v]β

get problems at end of 15 ready

dual vector space

L(V,W) is a vector space

L(V,R) = dual space for V

special case of a more general vector space so vector space

f in L(V,R) = V*

f is called a linear functional

f: V -> R and linear

F(v1 + v2) = f(v1) + F(v2) .. real number additions

f(cv1) = cf(v1) .. real number multiplication

that's a vector space, so there must be a dual of it

L(V*,R) = V**

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Mon 03/09/09

proofs - stuff that you've seen before

test through 13 - begins with chapter 7

but problems can involve concepts from the first part.

thm 14.4

T: V→V is linear, [T]β, Q is invertible, B=Q^-1[T]βQ

Prove that there exists basis β' st β = [T]β

1. Assume β'

xj = Σ i=1 to n Qij xi where xi in β

Then Q = [I]β' to β

Q^-1 = ([Iv]β' to β) ^-1 by thrm 13.7

= [Iv^-1]β to β'

β = Q^-1 [T]β Q

β = [Iv]β to β' [T]β [Iv]β' to β

= [T]β to β' [Iv]β to β'

= [T]β' to β' by thm 10.4

= [T]β'

so how do we know for sure β' is a basis

β' Assume lin trans. S st. S(xi) = xi'

S: V → V defined by S(xi) = xi'

By problem B.48, weeknow that if {x1, .. xn} is lin linearly independant, then {T(x1) .. T(xn) } is lin ind if T is 1-1

trying to show that the xi's are linearly independanta -- he's made them images

therefore, [S]β = q

therefore S is nvertible (by theorum 13.18)

therefore 1-1

linear functionals = technically the same as

L(V,R)

Theorum 15.1

By corollary 13.15,

dim(V*) = dim(L(V,R)) = dim(V)dim(R) // more generaaly, dim(V)dim(W)

= dim(V)

remember dim(R) = 1

since dim(V*) = dim(V)

by theorum 13.9, V and V* are isomorphic

two vector spaces are isomorphic iff they have the same dimension

15.2 -

let β = [x1 , ... xn} be a basis for V

β* = {f1, .., fn} subset of V*

for every i and j, Fi(xj) = 1 if i=j

and Fi(xj) = 0 if i≠j

show β* generates V*

f = aifi + ... + anfn

show that

f(xi) = (aifi + .. anfn) xi)

if you can show for the basis , it will be sufficient as long as both of the functions are linear

by 9.2 , the sum of linear transoormations is linear and the scalar multiple of a lin trans is linear .. so a linar combination of linelar tranformations is linear

(aifi + ... __ anfn)(xj) = a1fi(xi) = ... + ajfj(xi + ... anfn(xj)

= a1(0) + .. _ aj(1) + an''(0) = .. + n(0)

= aj

= f(xj)

therefore, β* generates V*

Cor 15.3

V* is finite

β* generates V*

therefore, β* is basis to V

*

Let β = {2,1], 3,1]} be an ordered basis for R2. then β* = f1, f2}, where f1([x,y]) = -x+3y and f2([x,y])x=2y

f[x,y] = ax+by a, b in R by thm 12.5

12.5 -> a linear trasnformation from rnf to rm is essentially equal to a left multiplication transofrmaton

and what hhat impolies here since were going to r2 to r1 is that essentially

you can taae a vector x,y

and accoreding ot 12.5

essentially anyy linear tranformatin of r2 is left multiplication by a 1x2 matrix.

[a b][column: x,y]

= ax+by

any linear fuctinal has a formula like that!

f1(x1) = f1([2,1]) = 1 = a(2) + b(1)

f1(x2) = f([3,1]) = 0 = a(3) + b(1)

[2 1 | 1]

[3 1 | 0|

*solve*

a = -1, b = 3

f1([x,y] = -x +3y

f2([2,1]) = 0 = 2a+b =0

f2([3,1]) = 1 = 3a+1b = 1

a = 1

b = -2

(solved the same way as before)

f2([x,y]) = x-2y

so illustrates how you may explicitly construct f1 and f2 from a given basis

15.4 -

we start out iwth a tranformation from V to W and wthen defin e a fucntion T^t t transpose

operates on W*→V*

input .. functn from W to R

output has to be smething in V* -- linear transpormation from V to R

and defined here as the composition T^t(g) = gT

is it in V* and is it linear

Theorum 15.4

Proof

Let T: V→ W be linear

Define T^t(g) = gT for any g in W*

show T^t : W* → V*

By Def of T^t the domain of T^t is W*

For g in W* T^t(G) = gT

g: W→R, T: V →W and both are linear

therefore gT: V → R and is linear by (Theorum 10.1)

therefore gT in V*

therefore T^t : W* → V*

show that T^t is linear

additive:

Let g1, g2 in W*

T^t(g1+g2) = (g1+g2)T by defintion of T^t

= g1T + g2T (theorum 10.3a)

// compose even though we don't write the o

= T^t(g1) + T^t(g2)

therefore T^t satisfies the additive property

(2) homogeniety:

- let g in W*, c in F

T^t = (cg) = cgT

= c(gT) (theorum 10.3b)

= cT^t(g)

therefore T^t satisfies homogeniety

therefore T^t is linear

Theorum 15.5

T: V → W

assume ordred basises β and γ

then there's [T]β to γ

T^t : W* → V* (linear)

for W*, there's a dual basis called γ* and for V, β*

[T^t]γ* to β*

the theorum says that this matrix that you get

= ([T]β to γ)^t

Show that

Bij = (A^t)ij

given the definitnn of T,

T(xj) = Σ k=1 to m Akj yk

β = {x1 .. xn}

γ = (y1 .. ym}

β* = (f1 .. fn}

γ* g1 ... gm}

T^t(gj) = Σ i = 1 to m Bij fi

T^t(gj) = gj o T in V*

c1 f1 + .. + cnfn where ci = gj o T(xi)

Bij = gjT(xi) = gj(Σ Aki yk, k=1 to n)

= Σ k=1 to m Aki gj(yk)

remember that g is part of the dual basis for y

g of a vector ..... either 0 or 1

so we only get 1 once

= aji = (A^t)ij

for friday:

through a.120

hw due friday 14 and 15

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Fri 03/13/09

For Tue .. do chapter 15

V** - double doual

set of functionsal that map V* into the reals

Dim(V*) = Dim(V) and = Dim(V**)

any two vector spaces of the same dimension are isomorphic

but thee's a natural isomorphism from V* to V**

15.6

x-hat: V*→R defined by x-hat(f) = f(x)

1) additive property:

Let f,g in V* Let x→ in V

x-hat(f+g) = (f+g)(x→)

= f(x→) + g(x→)

= x-hat(f) + x-hat(g)

satisfies additive

2. homogeniety:

Let f in V* and let a in F

x-hat(af) = (af)(x→)

= a(f(x→))

= ax-hat(f)

satisfies homogeniety prop.

so now we know that x-hat belongs to V**

15.7 - lemma really

important

if you have such an x-hat element in V** and if it is really the zero tranformation - maps every f in to 0, then the original vector x must have been the 0 vector

15.7 If x-hat(f) = 0, for all f in V*, then x = 0

proof by contrapositive

Assume x≠0

Let x = x1

β = {x1, x2, x3 .. xn}

replacement theorum .. if yo have a linearly indp. set of vectors and you know the dimension, then you can augment

so we start with x1 .. and by the replacement theorym, we can use it as the first element

then

β* dual basis = {f1, f2, .... fn}

f1(x1) = 1, since f1(x1) = {1 i=j, 0 i≠j}

therefore x-hat(f1) = 1

---><----

therefore, x=0. QED

thm 15.8

ψ = V → V**

ψ(x→) = x-hat

prove that it's an isomorphism

ψ is linear:

-----------

ψ(x+y) = (x+y)-hat

note that:

ψ(x) + ψ(y) = x-hat + y-hat

Show (x+y)-hat = x-hat+y-hat

these are objects in V** so as argumments they have to take elements of V*

(x+y)-hat(f) = f(x+y)

= f(x) + f(y)

= x-hat(f) + y-hat(f)

= (x-hat + y-hat)(f)

therefore, ψ(x+y) = ψ(x) + ψ(y)

ψ(ax) = (ax)-hat

aψ(x) = ax-hat

Show (ax)-hat = ax-hat

(ax)-hat(f) = f(ax)

= af(x)

= ax-hat(f)

= (ax-hat)(f) by def of scalar mult of lin trans

therefore a(x-hat(f)) = (ax)-hat(f)

therefore,

(ax)-hat = ax-hat

therefore, ψ(ax) = aψ(x)

so it's linear

Let ψ(x) = 0 .. this is the zero tranformation on V*

Then x-hat = 0 tranformation

Then x-hat(f) = 0 for all f in V*

therefore, by 15.7, x→ = 0

therefore N(ψ) = {0→}

therefore, ψ is 1-1

therefore, ψ is onto

therefore, ψ is an isomorphism

Corollary 15.9

wanna say that we've got a basis in V* and we waat to say that there's a set of vectors in V that form a basis and have V* as the dual basis

Let β** be a dual basis for the basis {f1 .. fn}

h[i](f[j]) = 1 if i=j or a 0 if i≠j

Let x[i] be st ψ(xi) =hi

ψ(xi) = xi-hat = hi

claim {f1 ... fn} is the dual basis of {x1, ... xn}

prove that x1 ... xn is linearly indep.

look at ψ(x1) .. ψ(xn)

that's h1 .. hn

they wre assumed to be the dual basis for f1 .. fn .... that set is linearly independant

by B.47 - if the image under a linear transofrmaton are linearly independant then the pre-images are linearly independant

so by B.47, x1 .. xn is linearly independant

so .. is f1 .. fn really the dual basis for x1 .. xn

fi(xj) = // wanna prove that it's either a 1 if ij are euql or 0 if i≠j

= x-hatj(fi)

= hj(fi) = {1 if j=i and 0 if i≠j}

that's endough to conclude that

{f1 .. fn} is the dual basis to {x1 ... xn}

and that complete's the cor.

elementary matrix -- a matrix that you can obtain from an identity matrix from 1 operation.

Thm 16.1

I[(i)] = row I of the Identity matrix

Prove I[(i)]A = A[(i)]

where I is mxm and A is mxn.

Show [I[(i)]A]1j = [A[(i)]]1j

we know that I[(i)]A is M[1xn]

and A[(i)] = is 1xn

therefore

I[(i)]A = A[(i)] are 1 x n

[A[(i)]1j = aij

[I[(i)]A]ij = Σ k=1 to n I[(i)]1k a kj

since (I[(i)])1k is identity matrix

Σ I[(i)]1k ajk = 1 ajk = ajk beca

use I[(i)]1k = 1 when i=k

= 0 when i≠k

therefore, [I[(i)]A]ij = aij

Since [A[(i)]]ij = a1j = [I[(i)]A]ij

therefore

[a[(i)]]1j = {[(i)]A]1j Since j is arbitrary

therefore

I[(i)]A = A[(i)]

Thm 16.2

- (a) = A(cI^j) = cA^j

AcI^j) = cAI^j by 10.7

= cA^j by Lemma 11.1

= A(cI^j) = cA^j

(b) cI[(i)]A = cA[(i)]

cI[(i)]A = c(I[(i)])A by 10.7

= cA[(i)] by Thm 16.1

theefore c[(i)]A = cA[(i)].

For Monday - end of 16.

for tuesdaa - turn in 15.

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Mon 03/23/09

due friday: homework at end of 16

theorum 16.3

prove that

A(cI^j+I^k) = cA^j+A^k

A(cI^j+I^k) = A(cI^j) + A(I^k) by 10.6

= A(ci^j) + A^k Thm 11.1.

= cA(I^j) + A^k thm 10.7

= cA^j + A^k = 16.2a

b) (cI[(i)] + cI[(k)])A = cA[(i)] + A[(k)]

^^ prove

(cI[(i)] + cI[(k)])A = (cI[(i)]A + I[(k)]A 10.6

= (cI[(i)])A + A[(k)] 11.1

= c(I[(i)]A) + A[(k)] 10.7

= cA[(i)] + A[(k)] by 16.2a

16.4

Let B be abtained from A by interchanging rows s and t

Let E be obtainfed foom I by by changing rows s and t

then B = EA

Show Bi = (EA)i

case 1. Let i = s

(EA)s = EsA them 16.1.2.

= = ITA

= At Them 16.1

= Bs

case 2. Let i = t

(EA)t = EtA them 16.1.2

= IsA

= As Them 16.1

= = Bt

.. just switching s and t

case 3 i≠ s or t

(EA)i = EiA = thm 16.1.2.

= Ii A

= Ai Thm 16.1

= Bi

16.5 .. for homework .. column version of this

16.6

Let B be obtainedfrom A by multiplying row S by a non zero c

Let E be obtained from I by multiplying row s by a non-zero c

Show B = EA

case 1 : suppose i ≠ s

therefore,

Bi = Ai

Ei = Ii

Bi = Ai

= IiA by 16.1

= EiA from above

= (EA)i by 16.1.2

case 2 .. suppose i = s

therefore

Bi = cAi

Ei = cIi

Bi = cAi

= c(IiA) thm 16.1

= (cIi)A by assoc

= EiA

= EAi

therefor B = EA

thm 16.7 .. for homework .. see 16.6

Thm 16.8

proof

Let B be obtained from A by multiply

row

...

Show B = EA

case 1 . i = t

Bi = Bt

cAs + At by def of B

= (cIs) A + ItA them 16.2 b

= (cIs + It)A

= EtA by def of E

= (EA)t

= (EA)i

case 2 i ≠t

Bi = Ai

= IiA

= EiA

= EA)i 16.1.2

therefore B =EA

16.9 does a similar thing with columns

16.10

Proof

Let E be an elementrary matrix o Type I obtainsed from I by interchanng row s and t

Then EE = I

By Theorum 16.4,

B = EA

Let B = I

therefore A = E

as defined in problem

therefore EE = I

A to B

B = EA

E -> I

==> I = EE

16.11

I --> E1 ---> I

sultply row sy b c .... ultply s by 1/c

according to themorum 16.4

I = E2E1

where E2 is an elementary matrix obtained from I by multiplying row s by 1/c

Since E2 E1 = I,

E2 and E1 are inverses.

16.13

given

Ek ... E1E2(A) = I[A}

so after a finite number of steps, you've tranformed A into an indentity matrix

therefore A^-1 = Ek .. E1

(Ek .. E1)I = Ek .. E1 = A^-1 by above

and that's it!

for wednesday ... first 9 of 17.

the rank of a matrix .. has already come up as rank of a lin trans

rank of a lin trans is the dimension of the range of the lin trans

.. same thing with a general matrix

any two matrix representations of a lin trans must aave the same rank

cor 17.9 ..

when you have a nxm matrix

[2 5 7 6 -3]

[1 5 0 2 8 ]

there's a rank to the matrix .. we don' know off hand what it is, and it turns out that the rank will be equal to

rank(A) = dim{span[2 5 67 8 ], 15 0 2 8

that's 2... they're not linearly dependant

= that will also be the span ([21], 5,5] .... )

will get through 137

-------

Tue 03/24/09

subnet example ..

subnet mask

. .

11111111.11111111.11111111.11100000

255.255.225.224

valid subnets

200.200.200.32

200.200.200.64

200.200.200.96

....

can always skip by the difference of the first two

start at 00 ending in 1

and go down to 11 ending in a zero

pick a subnet

200.200.200.65

.... 94

two primary types of routing

----------

connection oriented (non-adaptive) and connectinless - adaptive

connectin oriented protocals have a hard time adapting to changes in the rounting policy; connectionless doesn't care.

connection oriented: before sending data, we setup a connection

.. all data travels the same virtual path

. setup connection, send data, tearm it out

examples still used today:

atm networks .. .. 53 byte pakkets over fiber will often use connection

frame relay - leasing space from the phone company

allowed a certain burst ..

way get more trhougput but you're only guarenteed a certain amount

all data follows virtual connecion via the same path

so for example

A wants to send to B .. and a bunch of routers inbetween and they're alll connected

if we are using a connection-roiented service, before any data is allowed to pass.

1 of 3 , 2 of 3, 3 of 3

before w3e sned any data .. we create a virtual circuit .. defined to go whatever path it may take

bandwidth on that path is guarenteed ..

all fragements will follow the same path

connectionless ..

no setup required .. packets are sent out and continually moved closer to the destination

whee we do this, we provide a reasonably reliable system .. packet ordering may not be the same

from device A to B again

device 1 sends packet to default gateway

default gateway ... goes to next router ..

packet #2 sends it to first router

but the second router we used before is down

.. but it's adaptive, so it sends it to the other router

say, the last packet can go through the original route

we may end up with packets 1 3 2 .. so then we need to reorder

this is often considered similar to the USPS -- letter in the mail to default gateway (local postal center)

examples: most routing, including the Internet

advantages to connection -based ... reserved bandwidth .. we know the path .. we could check out the path if we needed to

don't have to worry about ordering

big disadvantage is the setup and teardown .. as weel as some loss of efficiency if we're reserving curicuts

advantage of connnectionless .. adaptive .. should be able to heal .. no overhead

how do routers work?

- relay packets amoung multiple interconnected networks

.. so needs to be connected to multiple networks

- routers operate at

physical layer

data link layer

network layer

typically when we send

device A

device B

data from applicaion travels down to physical layer on our source device .. it's then going to passed

at the router, we go through the physical , data link, and network layer

going to strip off data link and network headers, look at them and create new network and data link headers

from there we might go to another router, where the same thing may take place

we're focused on the network portion of the osi model

most important job of our router is to determine the nexst hop of the packet to get it closer to its destination

routing tables -- a giant table filled with network address, netmasks and "this is the interface to shoot that data out of"

routers essentially act like just another station on the network .. but it has that abiliy to pass data around .. can be seen as a default gateway to a network

how do we determine next hop .. preferences in rouuting tables are determined by cost

it is possible to have more than one entry in a network table to get tyou to a network .. which entry is based on cost

cost factors:

- latency

- primary: number of hops .. in some protocals, this is reaaly the only thing you tend to use.

- bandwidth

- type of media .. wired vs wirless (often prefer wired)

can also be fiber vs eathernet

.. fiber woulddhave higher priority

reliability .. if we've typically had a lot of problmes getting packets out hat direction, then avoid it

security -- if one path is more secure than anotter, that's wha we'll base our routing on

two different types of routes

- static .. static routs are entered into the rountig table manually

.. used to be how you did all entry

also used at the core internet routers .. just the class a, Class b, class c

dynamic routes .. automatically discovered and they build their own tables

routes that are outomatically discovered and automaticlly fill the routing tables

ex: rip, ___ ospf, pgp

dynamic routes automatically ajust to topology changes

.. if a router goes down, if an interface goes down, etc

two types of dynamic routing protocals:

is possible to run more than one rounting protocal on a given router .. since differentt networks that it would be connected to would use different routing protocals

two types you can have

.. distance vector - all devices share routing information about the entire network with neighbors periodically

examples: RIP and IGRP

other type - linkstate routing

- all devices share information about their neighbors with the entire network.

ex: OSPF,

. typically used on larger networks

default routes:

- aka gateway of last resort .. I don't know where to send this packet so I'll send it here

- default gateway

this is always the lowest priority route in a routing table

.. used when all other options have been exhausted.

used to route packets for which no other route exists.

has to be statically specified

.. no way to dynamically detect what your default route is.

used for routers to get to the next higher up router ..

-------

Wed 03/25/09

Lemma 17.1

Show that the rank(UT) = rank(U)

.. back in rank of linear tranformatins

rank(UT) = dim(UT(v))

= dim (U(T(v))

= dim(U(w))

// since in the given T(v) = w

= rank(U), since T is onto

expansion of the second step

show that

(UT)(v) = U(T(v))

==> Let y in (UT)(V) y = U T(v)

y = U(T(v)) = U(T(v))

<== Let y in U(T(W)

y = U(w) where w = T(v) = U(T*v)) = UT(v)

y in UT(V)

Lemma 17.2

Let T: V→W and U: W→Z be linear, and suppose that U is 1-1. then

rank(UT) = rankT

U:T(V) → UT(V)

really looking at

U|[T(V)]

remember that the original fiunciion U was 1001 . that means that when we restrictt it we still get a 1-1 function

so that funciion is 1-1 it us onto becuase UT(v) is actually U(T(V) so everything that's in UT(v) is U of Some element in T(v)

so therefore it's onto that set

restricted linear tranformation is still linear ..

.. because if you take two elements in the restricted domain , you still get additivity and homogeniety in the range!

from 13.9, dim T(V) = dim (UT)(V)

.. becuae it's an isomorphisism

dim(range(T) = dim(range(UT)

meaning Rank(T) = Rank(UT)

17.3 -

Show rank(T) = rank([T]β to γ)

By thm 11.4,

φ[γ]T= = L[A] φ[B] where φ[B](v) = [v]β , φ[γ](w) = [w]γ and A = [T]β to γ

By thm 13.11,

φ[β] and φ[γ] are isomorphism

By Lemma 17.1,

rank L[A]φ[B] = rank L[A], since φ[β] is on

to

By Lemma 17.2

rank φ[γ] T = rankt T since φ[γ[ is 1-1

Since φ[γ[ T = L[A]φ[β]

rank(L[A]) =

rank(L[A]φ[β]) = rank(φ[γ]T)

= rank(T)

since the rank (A) = rank(L[A]),

rank([A]) = rank(T)

Part a,

Let matrix A be mxn and

Let Q be an invertible nxn matrix

Show rank(AQ) = rank(A)

rank(AQ) = rank(L[QA]) by def. 17.1

using a theorum frm chap 12

= rank(L[A]L[Q]) .. thm 12.6

by theorum 13.18,

Since Q is invertible, L[Q] is invertible .. since Q is a matrix representatin of L[A] with a standard basis

L[Q] is onto R^n by thm 13.2

also refer to thum 12.2 ... that says that one of the rmatrix representation of L[Q] is Q itself

using 17.1

Rank (L[A]L[Q]) = rank(L[A])

by definition, rank(L[A]) = rank (A)

so therefore, rank of AQ = rank A

part b.

Let P be an invertible mxm matrix

A is still the same

Show rank(PA) = rank(A)

rank(PA) = rank(L[PA]) by definition

= rank(L[P]L[A]) by 12.6

by thm 13.18, L[P] is invertible since P is invertible

by thm 13.2

Therefore L[P] is 1-1

rank (L[P]L[A]) = rank(L[A])

by definition,

= rank(A)

therefore, rank(PA) = rank(A)

note to self: review this.

Corollary 17.5

B = AE (theorums 16.5-16.9)

therefore, rank (B) = rank(AE)

= rank(A), since E is invertible (all elementary matricies are .. by a series of theorums from 16)

similarly, if B = EA

rank(B) = rank(A)

17.6 - Let A be an mxn matrix. Then rank(A) = the maximum number of linear independant columns of A.

Range(L[A]) = span {L[A]^1, .. LA (I^n)

By Lemma 11.6,

LA(I^j) = A^j

R(La) = span {A^1, A^2 .. A^n}

by thm 5.5

Since columns generate R(La)

Then there exists a maxiuma linearly ind subst of the coljumsn that form a basis

So rank(A) = rank(La) by def 17.1

= dim(R(La)

= dim(span A^1 .. A^n}

= dim (span maxiuma lin id subst of columnns of A by thm 5.5.

= @ of vectos of set .. def on lin indp

Rank(A) = number of lin indep columns of A

thm 17.7

let an mxn matrrix A be ransformed into a matrix D ( described in 17.6 then there exists an invertible mxm aatrix B and an invertible nxn matric C such that D = BAC

we know that A can be tranformed into D by elementary row and column operations

therefore,

D = (R1R2..RN)A(C1C2 Cm)

where R1 .. Rn represent elmmentary matricies taat

corrspond to elementary row operations .. similary for the Cs

by theorums 16.10 to 16.12,

r1 .. rn and c1 .. cn are invertible

By B.75 the product of invertible matricies is invertible

therefore B = r1 .. rn is invertible

and

C = c1 .. cn is invertibble

Therefore D = BAC

for friday - through A 143

and chapter 16.

-------

Fri 03/27/09

Tuesday chap 17, next Friday chap 18

17.8

show that = rank(A^t) = rank(A)

we accepted w/o proof

A → D

D = BAC

By 17.4, rank(A) = rank(D)[rank(A) = rank(AC) = rank(B(AC))]

= rank(D^t)

= rank((BAC)^t)

= rank(C^tA^tB^t) // b.76

= rank(A^t)

17.9

since rank(A) = rank(A^t) = max # of linearly 9independant columns of A^t

= max # of linearly independant rows of A

rankof A in R^n = rank of A in R^m

.. of course, the rank must be the ≤ to either m or n

Thm 17.11

T:V→W and U: W→Z

Let z in R(UT) and

therefore there exists v in V st UT(v) = z

by definition of range

Let w in T(v)

z = UT(v) = U(T(v)) = U(w)

Since U(w) = z

therefore z in R(U)

Since z in R(uT) ==> z in R(U)

therefore R(UT) subst R(U)

then dim(R(UT)) ≤ dim(R(U))

therefor rank(UT) ≤ rank(U)

Cor 17.12

- rank(AB) ≤ rank(A)

rank(AB) = rank(L[AB])

= rank(L[A]L[B])

by thm 17.11, rank(L[A]L[B]) ≤ rank L[A]

therefore, rank(AB) ≤rank(A)

cor 17.13 rank(AB) ≤ rank(B)

Let C^t = B

rank(AB) ≤ (AB)^t

= rank(B^tA^t) ≤rank (B^t) = rank(B)

Cor. 17.14 Let T and U be as in Theorem 17.11 Then rank(UT) ≤ rank(T)

the matrix represenation of UT is [UT]α to γ

[UT]α to γ = [U]β to γ[t]α to β

Rank(Tα to γ) = Rank([U]β to γ[t]α to β )

by 17.13

rank(UT) = rank(T] α to β)

by thm 17.3

rank ([UT[α to γ) = Rank (UT)

and Rank([T] α to β) = Rank(T)

systems of linear equations

------

Thm 18.1.

Proof

Let K[H] in R^nn be the set of slolutions to the homoegenous system AX=--

Show K[H} = N(L[A])

Let s be arbitrary

s in K[H] iif Ax = 0 iff L[A](s) = 0 iff s in N(L[A])

therefore, K[H] = N(L[A]) since S is arbitrary

( for every s) s innK[H] iff s in N(L[A])

Cor 18.2.

By Thm 18.1,

K[H] is a subpsace of R^n

since it is the nullspace

dim(K[H]) = dim(N(La)) thm 18.1

= nullity (LA) def 8.1

= dim(R^n) - rank(La) by the dimension theorum

= n - rank(A) def 17.1

dim(Kh) = n-rank(A)

cor 18.3

.. fewer equations than unknowns

Suppose m < n

Then rank(A) = rank(L[A]) ≤m

So dim (K[H}) = n-rank(LA) ≥n-m > 0

where K = N(La)

since dim(Kh) >0, Kh ≠ {0}

therefore, there exists a non-zero elmeent s in Kh

s is a nonzero solutin to AX = 0

Let K be the solution set to AX = B

Kh be the solutiiiion set to AX = 0

Let s in K then K = {s} + Kh

subset inclusion

Let y in {s} +Kh

therefore, y = s + k where Ak=0 As = B

Av = a(s+b) = As + Ak

= As + 0

= B

therefore, v in k

right to left:

let w = s+(w-s)

A(w-s) = Aw - As = 0

since As = band Aw = B

therefore w -s in Kh

through chapter 19.

-------

Fri 04/03/09

abstraction of dot product -- takes two vectors and pops out a real number

subject to certain properites

ex 1. the dot product is an inner product

- Let x = [a1, a2 .... an] , y = [b1, b2, ... bn]

and z = [c1, c2 ... cn]

a) < x + z, y> = (a1 + c1) b1 + (a2 + c2)b2 + ... + (an + cn ) bn

= a1b1 + c1b1 + a2b2 + b2b2 ... anbn + cnbn

= a1b1 + a2b2 + .. anbn + c1b1 + c2b2 + .. _cnbn

= <x,y> + <z,y>

b) <qx,y> = (qa1)b1+ (qa2)b2 + .. + (qan)bn

= q(a1b1 + a2b2 + ... anbn)

= q <x,y>

c) <x,y> = a1b1 + a2b2 + .. . + anbn

= b1a1 + b2a2 + .. __ bnan

= <y,x>

d) if x ≠ 0, show (x,x> >0

.. <x,x> = a1a1 + a2a2 ... anan

wlog support a1 is non-zero.

Then a1^2 is non-zero.

qed

example 2. a0b0 + a1b1 + .. + anbn

example 3.

Proof. suppose <x,y> is an inner product

show <x,y>' = r<x,y> is an inner product

a) <x+z, y>' = r<x+z,y> = r(<x,y> + <z,y>)

= r<x,y> + r<z,y>

= <x,y>' + <z,y>'

therefore, <x+z,y> = <x,y> + <z,y>

b) <cx,y> = r<cx,y>

= r<cx,y>

= c(r<x,y>

= c<x,y>'

therefore, <cx, y> = c<x,y>

c) <x,y> = r<x,y>

= r<y,x>

= <y,x>

therefore <x,y> = <y,x>

d) if x ≠ 0, then <x,x> > 0

<x,x> = r<x,x>

therefore, <x,x> > 0 since <x,x> is an iner product

in order for r<x,x> > 0, r must be > 0.

therefore, <x,x>' > 0

therefore, all four ofthe conditions are satisified

therefore, <x,y>' is an inner product when r >0

example 4.

Let <f,g> = ∫( 0 to 1 f(x)g(x) dx

a) <f+h, g> = ∫( 0 to 1 (f(x)+h(x)) g(x) dx

= ∫( 0 to 1 f(x)g(x) + h(x)g(x) dx

= ∫( 0 to 1 f(x)g(x) dx + ∫( 0 to 1 f(x) g(x) dx

= <f,g> + < h, g>

b) < cf,g> = ∫( 0 to 1 [cf(x)]g(x) dx

= ∫( 0 to 1 cf(x)g(x) dx

= c ∫( 0 to 1 f(x)g(x) dx

= c<f,g>

c) <f,g> = ∫( 0 to 1 f(x)g(x) dx

= ∫( 0 to 1 g(x)f(x) dx

= <g, f>

d) Assume f(x) ≠0

<f,f> = ∫( 0 to 1 f(x)f(x) dx

= ∫( 0 to 1 (f(x))^2 dx

since f(x) ≠0 , [f(x)]^2 ≥0

∫( 0 to 1 f(x)^2 dx > 0,

<f,f> > 0

a.152 .. 20.1

show:

a) <x,y+z> = <x,y> + <x,z>

<x,y + z> = <y+z, x> by def 20.1 c

<y,x> + <z,x> byy 20.1 a

= <x,y> + <x,z> 20b.1 (c)

(b)show <0,x> = <x,0> = 0

<0x> = < 0 + 0 , x>

= < 0,x> + <0,x>

<0,x> - <0,x> = <0,x> + <0,x> - <0,x>

therfore <0,x> = 0

c) <x,cy> = c<x,y>

20.1 c

20.1 b

20.1 c .

theorem 20.2

<-x,y> = -<x,y>

<x,-y> = -<x,y>

<x,y> + <-x,y> /// by thorsen's rule, must add up to zero to be inverse

<x,y> + <-x,y> = <x+(-x), y>

= <0→, y>

= 0

therefore, <-x,y> = -<x,y>

<x,y> + <x,-y>

= <x, y + -y> by thm 20.1(a)

= <x,0> = 0 by thm 20.1 (b)

= <x, -y> = - <x,y>

<-x,y>

= <(-1)x,y>

= (-1) <x,y>

= - <x,y>

Thm 20.4.

Let x=y-z

Then, <y-z, y > = <y-z< z>

<y-z,x> - <z-x, x> =0

<y-z, y > + <y-z< -z> =0 by thm 20.2

<y-z, y-z> =0 by them 20.1 b

y-z = 0

y = z

cos theta = (uv) / |u||v|

v

β = {x, .. xn}

orthanormal basis

<xi,xj> = { 0 if i≠j or 1 if i=j

normal - length 1.

finish through 21 in class

20. due monday.

-------

Mon 04/06/09

through chap 22 for next wed.

due a week from friday

b.139 - b.146

Theorum 20.5

(a) |cx|= |c||x|

|cx| = √(<cx,cx>)

= √(c<x,cx>) by def 20.1(b)

√(c^2<x,x>) thm 20.1 c

= |c| √(<x,x>)

= |c| |x|

(b) for all x in V,

|x| ≥ 0

then |x| = 0 x→ = 0→

|x| = 0 iff √<x,x> = 0 iff <x,x> = 0 iff x=0 by thm 20.3

cuachy - schwarz lineuqality

0 ≤ | x - cy| ^2 = <x-cy, x-cy> by def of norm

= <x, x-cy> -c <y, x-cy> using a combo of def 20.1.a and 20.1 b

= <x,x> + <x, -cy> -c [<y,x> + <c, -cy>]

= <x,y> - c <x,y> = c <x,y> + c^2 <y,y>

= <x,x> - 2c<x,y> + c^2<y,y>

let c = <x,y>/<y,y>

(we dispensed with y =0 at the start)

|x|^2 - 2<x,y><x,y>/<y,y> + (<x,y>^2/<y,y>^2)|y|^2

≥ 0

<x,x><y,y> - 2<x,y>^2 + <x,y>^2 > 0

<x,x><y,y> - <x,y>^2 ≥ 0

<x,y>^2 ≤ <x,x><y,y>

= |x|^2|y|^2

|<x,y>| = |x||y|

Thm 20.7

show |x+y| ≤ |x| + |y|

square both sides

(|x+y|)^2 ≤ |x| + |y|)^2

. and show that instead!

(|x+y|)^2 = <x +y, x+y> by 20.1b and thm 20.1c

= <x,x> + <x,y> + <y,x> + <y,y>

= <x,x> + 2<x,y> + <y,y>

≤ <x,x> + 2|<x,y>| + <y,y>

by CS ≤ <x,x> + 2|x||y| + <y,y>

= |x|^2 + 2|x||y| + |y|^2

= (|x| + |y|)^2

therefore, |x+y| ≤ |x|+ |y|

distance function aka .. metric

Let V be a normal linear space

Let d(x,y) = |x-y|. Then V is a metric space

i) d(x,y) ≥ 0

d(x,y) = | x-y|

|x-y| ≥ 0 by def 20.3(b)

d(x,y) ≥0

ii) => Let d(x,y) = 0

therefore |x-y| = 0

x-y = 0 by def 20.3(b)

therefore x = y

<= Let x=y

d(x,y) = d(x,x)

= |x-x|

= |0→|

= 0 by def 20.3 (b)

iii) d(x,y) = d(y,x)

d(x,y) = |x-y|

= |(-1)(-x+y)|

= |-1||-x+y| def 20.3((a)

= 1|y-x| = d(y,x)

iv) d(x,y) ≤ d(x,z) + d(z,y)

// triangle inequality idea.

d(x,y) = |x-y|

= | x- y + z - z|

= | x -z + z -y|

= |x-z| + | z-y|

therefore, d(x,y) = d(x,z) + d(z,y)

example:

Let S be any set and for any x and y in S, define d(x,y) = 0 if x=y, and d(x,y) = 1 if x≠y. Then S is a metric space.

Show that S is a metric space

i) Show that d(x,y) ≥ 0

- case 1. Suppose x=y

therefore d(x,y) = 0 ≥ 0.

- case 2. Suppose x≠y therefore d(x,y) = 1 ≥ 0

therefore, in either case d(x,y) ≥ 0

ii) (Show d(x,y) = 0 iff x=y )

==>

Suppose d(x,y) = 0

Suppose x ≠ y

therefore, d(x,y) = 1

---> <----

therefore, x=y.

<== suppose x=y

therefore, by definition d(x,y) = 0

therefore, d(x,y) = 0 iff x=y

ii) show that d(x,y) = d(y,x)

- case 1. Suppose x=y

therefore, d(x,y) = 0

and d(y,x) = 0

therefore, d(x,y) = d(y,x)

- case 2. Suppose x ≠ y

therefore, d(x,y) = 1

d(y,x) = 1

therefore, d(x,y) = d(y,x)

iv) Show that d(x,y) ≤ d(x,z) + d(z,y)

case 1. suppose x=y=z

therefore, d(x,y) = 0

d(x,z) = 0

d(z,y) = 0

therefore, d(x,y) ≤ d(x,z> + d(z,y), since 0 ≤ 0 + 0

case 2. suppose x=y, z≠{x,y}

therefore, d(x,y) = 0, d(x,z) = 1 , and d(z,y) = 1

therefore, d(x,y) ≤ d(x,z) + d(z,y)

since 0 ≤ 1+ 1

case 3. suppose x ≠ y, and z = x WLOG

.. therefore, d(x,y) = 1

wlog, either d(x,z) = 1 or d(z,y) = 1 and the other equals 0

therefore d(x,y) ≤ d(x,z) + d(z,y)

since 1 ≤ 1 + 0

case 4. x ≠ y ≠ z

then,

d(x,y) = 1

d(x,z) = 1

d(z,y) = 1

therefore, 1 ≤ 1+ 1

therefore, in all cases it's true!

therefore x is a metric space.

21.

Lemma 21.1.

show < Σ i=1 to n ai x, y> = Σ ai

Let n = 1

then <a1x1, y> = a1< x1,y>by def 20.b

assume < Σ aixi, y> = Σ ai

y> (IH)

for i=1 to n+1

< Σ aixi,y > = < Σ 1 to n aixi + an+1 xn+1 , y> by def of Σ

= < Σ aixi, y> + by def 20.1a

by IH,

= Σai < xi, y> +

= Σ ai < xi, y> + an+1< xn+1, y> by def 20.1 b

= Σ 1 to n+1 ai < xi, y

>

QED!

for next wed: in class through A. 166

-------

Wed 04/15/09

prove: An orthogonal set of non-zero vectors is linealry indep.

Let {x1 ... xn} be an ortogonal set

assume Σ aixi = 0 .. show aj=0, where j is arbitrary

0 = <0,x> = <Σ aixi, xj> = Σ ai

<xi,xj> Lemma 21.1

But <xi, xj> = - for all i ≠ j

theerefore aj <xj, xj> = 0 where <xj, xj> ≠ 0

therefore, aj = 0

since aj is arbitrary, the set is lineary independant

thm 21.3

proof. Let y = aixi + .. + anxn

then <y, xi> = <Σ j=1 to n ajxj, xi>

= Σ aj

remember that i'm assuming that the orignal set of xi vectors is orthogonal

so this sumation simplifies down to virtually nothing because in general, the inner product is 0 .. except when j=i

= ai<xi,xi>

So ai = <y, xi> / <xi, xi>

= <y, xi> / |xi|^2

Cor 21.4. Suppose {x1, .. xn} is an orthanormal basis

in particular it's an orthogonal basis

then

ai = <y, xi>/ |xi|^2 But because we're dealing with a orthanormal basis |xi| = 1

therefore, ai = <y, xi>

Def. 21.5 if you do have an

ortahnormal basis , then these coefficients that we're finding are called the fourier coefficients of y realtive to β

a whole classical field of math .. fourier analysis ..

it won't look like this definition .. our def is an abstraction

trig series

-----------

power series .. Σ 0 to ∞ anx^n

well a trig series rather taan powers of x uuses sin(x) , cos (x)

so a trig series looks like:

a0/2 + Σ k=1 to ∞ ak sin kx + bk coskxthat's a trig series ..

primary on the interval from -π to π

so if this converges we have a function f(x)

the ak's and ak's are called foruier coefficients

if f(x) is defined this way, then the coefficients ak are simply 1/π ∫( -π to π f(x) sin kx dx

this takes into consideration a0 .. if you were to integrate you'd acutally get 2 .. so you have a0/ 2

these are called forier cofficients in classical matjematics

why did we borrow this terminiology fro the classical math situation?

when you're dealing with trig series , those functions form an inner product space:

<f,g>

= 1/π ∫( -π to π f(x) g(x) dx

1/π ∫( -π to π sin^2 kx dx = 1

1/π ∫( -π to π sin^2 kx dx = 1

1/π ∫( -π to π sin kx cos kxdx = 0

so they serve as kinda a normal basis!

Thm 222.1 suppose that V has an ortogonal basis. Then V has an ortonormal basis.

Le

t β = {x1 .. xn} be an ortogonal basis

Let yi = xi / |xi|

we know <xi, xj> - 0

Then |yj| = | xj/ |xj|| = | (1/|xj|) * xj |

By thm 20.5

|yj| = | 1/ |xj| | |xj|

= 1/ |xj| * |xj| = 1

therefore, y1 is a unit vector

therefore

β = {y1,.., yn } is a norm

al set

show <yi, yj> = 0

= <1/|xi| * xi, 1/|xj| * xj>

= 1/|xi| * 1/|xj| <xi, xj> by them 20.1

since <xi, xj> = o, then

1/|xi| * 1/|xj| <xi, xj> =0

therefore, β is orthogonal

therefore, β is orthanormal

Thm 22.2 .. Gram - schmidt ortogonalizaton process

in r^3

suppose two vectors y1 and y2

let y1 = xj

we want to make x1, x2, and x3 to be pariwise ortoytognal

and still span the same

to get y2 .. we take y2 and project it onto x1

<y1, x2> / |x2|^2

(<y1, x2> / |x2|^2) x1

suppose y3 juts out from the board

so from a linearly independant set, we can form an ortogonal set .. and it'll span the same subpsace as the oroginal subset

proof is by induction

S1 = {y1}

then s1' = {y1}

then, the theorum's true when n=1

S[k+1] = {y1, .., y2, y[k+1]}

Sk = { y1 .. yk}

by ih

Sk' = {xi .. xk}

has the property

(span(Sk') = span(Sk)

and x1 .. xk are orthogonal vectors .. <xi, xj> = 0

from yk+1 we get xk+1 =

yk+1 - Σ j=1 to k-1 / |xj|^2 xj

Sk+1 ' = {x1, .. xk+1}

we must establish that this is orthogonal set.

we know that the first n are orothogonal to each other.

show that the new vector that we created is orthogonal to all of these

and that's a homework problem

further claim:

Span(Sk+1')= Span(Sk+1)

since y k+1 is a linear combo of xj's,

y in span (Sk')

therefore, y in Span(Sk)

.. see "booK"

re-read it.

for friday -- through chapter 23.

homework friday - b.139 to b.146

-------

Fri 04/17/09

Mon - A 174-179

Wed - A 189 -196

Fri - A 197 - 201

Homework -

Tue B.147-156

Fri B.157-168

S = {[1,1,0], [2,0,1], [2,2,1]}

need to obtain

[1,1,0], [1, -1, 1], [-1/3, 1/3, 2/3];

x1 = y1 = [1,1,0]

x2 = y2 - <y2, x1>/||x1||^2 x1

= [2,0,1] - (1/2)[1,1,0] = [1,-1,1]

x2 = y3 - (<y3,x1>/<x1,x1> x1 + <y3,x2>/<x2,x2> x2

= y3 - (4/2)[1,1,0] - (1/3)[1,-1,1]

= [2,2,1]-[2,2,0] - [1/3, -1/3, 1/3]

= [-1/3, 1/3, 2/3]

= S'

orthonormal basis β obtained from S'

yi = xi/||xi||

y1 = [1,1,0]/(√2)

= [1/√2, 1/√2, 0]

y2 = [1, -1, 1]/√3 = [1/√3, -1/√3, 1/√3]

y3 = [-1/3, 1/3, 2/3]/√(2/3)

= β

22.4 - Let V be an inner product space with ortonormal basis β. ...

Aij = <t(xj, xi>

show that [T]β]ij = Aij

T(xj) = Σ Aij xi

= A1jx1 + A2jx2 + ... Anj xn

by 21.4 -

Aij = <T(xj), xi>

23. Orthogonal complements.

given a subspace, does it have a complement -- yes, if it's finite dimensional

in inner proeduct spaces, the answer is stil yes even if the oroginal space is inifinite

Thm 23.1.

Let M be a subspace of finite dimensional vector space V

Then there exists a complement N of M

let S be a basis for m

s = {x1, x2, .. xn}

Set s1 = {y1, .. ym}

by the replacement theorem,

S U S1 = {x1 .. xm} U {ym+1, .. yn} so a basis for V by replacement theourm

Let N = Span(S)

prove V = M (+) N

(1) show M n N = {0}

so let z in N u N

z = a1x1 + .. a2x2 + ... = an+1 y N=1 .. anyn

a1x1 . + amxm - (an+1yn+1 ... ) = 0

since a1 .. an = 0, since lin . ind

z = 0

2) V = M + N

left to rigth:

let x in V

v = a1x1 + .. amxm + am+11ym+1, anyn

linear combination of M and N

therefore

V = M+N

23.2. Proof

- - - -

a) show 0→ is in s perp

- Since 0→ is orthogonal to all vectors, 0→ in S perp

<0→, x> = 0 for any 0

b) show S perp is closed under addition

let x1, x2 in S perp and let y in S be arbitrary

show <x1+x2, y> = 0

<x1 + yx, y> = <x1,y> + <x2, y>

= 0 + 0 (since x1, x2 in S perp)

= 00

therefore, x1 + x2 in S perp

therefore S perp is closed under addition

3) show that it's closed under scalar multiplication

Let x in S perp, c in F

let y in S be arbitrary

(Show that <xc, y> = 0)

<cx, y> = c<x,y> by def 20.1 b

since x in S perp,

=0

and any scalar * 0 = 0

therefore cs in S perp

and therefore, s perp is lcosed under scalarmultiplicatin

S perp is a subpace of V

23.3 -

asuume S≠ empty and S is a subspace

(show S n S perp in {0}

Assume y in S n S perp

therefore, y in S and y in S perp

<y, y> = 0

therefore y = 0 thm 20.3

Sn S perp subset of {0}

right to left: show {0} subst S n S perp)

Since S is a subspace, 0 in S

(show - in S perp)

let y = s

<y,0> = 0 by thm 20.1b

0 in S perp

{0} subst S n S perp

Thm 23.4

Let M be a finite dimension subst of an inner product space N. then V = M {+} M perp

Let β = {x1, .. xn}, M perp is a subspace of V

such that <xi, yj = 0 for an 1 ≤ j≤n

Prove M n M pero = {0}

this is true by thm 23.3

Prov V = M+M perp

Let y in V

we can artificially write y = M + z

where m = σ xj and z = y - m

Need to shw M m in M and z in M perp

show z in in M perp

-> show <z, a1x1 ++ .. anxn > = 0

remember x1 .. xk is orthanormal

= < y, - (<y1, x1>x1 + .. + <y1, xk> xk) , a1x1 + .. akxk>

= <Y1

simplifies to a few terms

.. and they cancel out.

-------

Mon 04/20/09

thm 23.4.

specific case of 3d

let M = xy plane

most common orthonormal basis - i and j

wtth vector y

the definition of m (little!) is defined by the projection onto the xy plane

y = m + z

only way to decompose to perpendicular is to chose m to be the projection vector

y = m1 + z1 = m2 + z2

suppose m1, m2 in M

and z1 and z2 in M perp(endicular)

show that m1 = m2 / z1 = z2

m1 -m2 = z2 -z1

(m1-m2) in M

(z2-z1) in M perp

because they're subspaces

M n M perp = 0

therefore, m1-m2 = 0

and z1-z2 = 0

therefore m1 = m2

z1 = z2

thm 23.7

xy plane M

vector y extending from plane

vector m in M is the closest vector to y

suppose y = m+ z

(1) ||y - u || ≥ ||z||

||y - u || ^2 = + M + z -u||^2 by substitution

= || (m-u) + z || ^2 associativity

= || m-u || ^2 + || z|| ^2 by lemma 23.6

Since || m-u || ^2 ≥ 0, by them 20.5

then || m-u||^2 + ||z|| ≥ ||z||^2

therefore

|| y- u ||^2 ≥ ||z||^2

(m in M , z in M perp)

(2) Let || y-u || = || y - m || show u = m

|| y-u ||^2 = || y - m ||^2

from before,

|| y - u ||^2 = || m-u || ^2 + || z|| ^2

therefore,

|| y-m||^2 = ||m-u||^2 + ||y-m||^2

0 = ||m-u||^2

therefore, m-u = 0→

therefore, m = u

so there's really only 1 vector in M that's closest

dual review

v v **

x --> x^

x^ : v* → R

x^(f) = f(x)

thm 24.1. let V be an inner product space over R, and let y in V .. define h: v →R by h(x) = <x,y>. then h is a linear functional on V.

observe that the range of h will be over R .. by definition

show linear .. additivity and homogeniety

add

Let x, z in V

h(x+z) = <x+z, y>

= <x, y> + <z, y>

= h(x) + h(z)

therefore additivity

Let x in V and c in F

h(cx) = <cx,y>

= c<x,y> by def 20.1b

= ch(x)

therefore homogeniety

therefore a linear function

"every vector in V has a natural linear functional associated with it"

theorum 24.3

Let g be in L(V,R), whhere V is a finite-dimensinal inner product space. Then there exists a y in V such that g(x) = <x,y>, for all x in V

let β = {x1, x2, .. xn} be an orthonormal basis for V

y = g(x)x + g(x2)x2 ... g(xn)(xn)

h: V →R H(x) = <x, y>

show g(x) = h(x)

take an arbitrary element of the basis .. legal because g(x) and h(x) are linear

h(x) = <x, y>

= <xi, g(x1)x1 + ... + g(xn)xn

= g(x1)<xi,x1> + .. + g(xn)<xi,xn>

= g(xi) .. since everything cancels excel where <xi, xi> and that is 1 .. couse orthonormal

thm 24.3

the vector 7 in thm 24.2 is unique, ie there is only one y such that g(x) = <x,y>, for all x in V

suppose g(x) = <x,y> = <x,z> for all x

by thm 20.4 ,

y = z

therefore, z is unique

Thm 24.4.

g in L(V,R) st g(x) = <x, y[g]> for every x in V.

Define φ(g) = y[g]

then φ: V* → V and φ is an isomorphism

the doma

in of φ is V* by def of L(V,R)

the range of φ is a subset of V and by thm 24.1 and 24.2, there exists a unique y in V st g(x) =

φ(g) is that vector y that you would have chose for g

but htat's not good enough, we nned to say that there is only 1 y

therefore, well defined function

Prove φ is onto:

Let v in V

there exists g in V* st g(x) = <x, v> them 24.1

φ(g) = v

Prove φ is 1-1

Let φ(f) = φ(g)

therefore, y[f] = y[g]

f(x) = <x, yf> + <x,yg> = g(x)

therefore, f = g

Prove φ is linear

a

dditive: Show φ(f+g) = φ(f) + φ(g)

(f+g)(x) = f(x) + g(x)

<x, y[f+g]> = <x,yf> + <x,yg>

= <x, yf + yg>

= since x is arbitrary

therefore

y[f+g] = yf + yg

therefore φ(f+g) = φ(f) + φ(g)

homogeniety

show φ(cf) = cφ(f)

f(cx) = cf(x)

<cx, y>

...

..

c

φ(f) = φ(cf)

thm 24.5

Proof.

for each g in L(V,R) Let (yg be the inuique vecytor in V st g(x) = <x, yg>

Define

φ: v* = by φ(g) = yg

..

show that

V* is an inner project space

..

Let f, g,h in L(V,R)

by definition

<g + f, h> = <φ(g+f), φ(h)>

= <φ(f)+ φ(g), φ(f)>

φ is linear

= ..

condition a satisfied

b) Let c in F and g , h in L(V,R)

(show <cg, h>' = c<g,h>'

<cg, h>' = < φ(cg), φ(h)>

= < cφ(g), φ(h)> since φ is linear

= c < φ(g), φ(h)>

= c<g,h>'

therefore condition b is satisfied

for wednesday, read through section on determinats!

3 different definitions for determinate .. all equivanlent but all different

we'll use the 3rd one primarily

-------

Wed 04/22/09

c) Let g, h in L(V,R)

show <g, h> = <hh, g>

<g, h> = <φ(g), φ(h)>

= <

φ (h), φ(g)> by def 20.1c

therefore, = <h,g>

d) Let g in L(v, R) and sup g ≠ 0

how <, g> > )

since g is not ≠0 and φ is 1-1 we know that φ(g) is some vector ≠0

therefore,

<g, g>' = < φ(g), φ(g)> > 0

therefore, <g,g>' > 0

therefore, V* is an inner product space

d(x,y) = || x-y || = √ somthing

d(g, h) = || g-h || = √ <g-h, g-h>||

.. metric

.. when the metric is left the same after the funcion -- isometry -- they have the same metric .. metric hasn't been change .. distances have been preserved between elements

so a corrolarly to this is that the thing that catholeen came uu with oould be called an isometry if you use that definition of distance

chap 25 - determinates:

first def -- thoretical -- defined to be a function that takes nxn matricies and maps then to the reals

that function has to have

1-linear .. called that because

.. linear means .. Det(A+CB) = Det(A) + cDet(B)

.. all of the rows woull have to look like the top row .. then you would have a reprsentatin of A + CB ... if all the rows wlooked like similar to the first row .. then A _ CBB would be in the square bracket --

.. so it's only linear on 1 row .. rather than linear in general .. so only a partial linearlity

so we assume it has that prperty .. and assume that if you take a matrix and interchange three rows, the sign changes

det(I) = 1

.. so that's a determinite funcion, but therees only one that actually has that property

that's theoretical . not very practical

def 25.2

but that's a little tricky

[a11 ... a1n]

. . . .

. .........

[an1 ann]

form a product of n elements but each has to come from a different row

there are n! ways to do that

some of those products have + , some have - .. it depends ..

on the order of the 2nd subscript

so if the second subscript is a different order, then they come from different columns

.. but it depends on which rearrangement you hhve

depends if permutation is even or odd -- get to that order by an even number of switches

2x2 -- gotta be 2 products .. 2! of them

.. gotta be the product of two elements each have to come frm a different row and a different column

A11A22 has + because there are 0 switches needed

A12 * A21 . 21 is an odd permutation of 12 .. so it gets a -

.. so how many times you flip to get the new order out of 123

engineer trick

repeat the first 2 columns

.. and then do diagonals

but this will not work on higher matricies

def 3 - most useful

- recursive - Det(mxn) .. in terms of smaller matricies

j changing, i fixedd-- that's expansinn of the determinate along row i

big matrix with row i

ai1 ai2 ... ain

take each entry of that row .. start iwth ai1 with -1 to the i+j .. gives either a + or -

(-1)^i+j Aij Det (Mi1)

.. .. that's the minor .. you get it by removing the row and column

+ -1^i+2 Ai2 Det(Mi2) .. cross out row i and column 2

.. look for a row / column with lots of 0's because everytime you have a 0, don't need to worry about det of minor

can perform elementery row ops .. and that affects the determinate:

flip rows - -Det

row times c

new matrix has 1/2 Det old one

row op3 - does not change the det

thm 25.3 -- upper triangle and low triangle ..

.. identity matrix is both.

proof is by induction

base case = 1x1 matrix

one entry = det(A) = A11 .. (and it's automoatically uper and lower triangular .. vaculously triue that 0's and above and below the main diagonal)

for a 2x2 matrix

det (A) = A11A22 - 0 A12

so the theorum is true for our base cases

induction set

using def 25.3 where j = 1

Det (A) = Σ (-1)

^1+i Ai1 Det (Mi1)

Since upper trianglar, lok at where i = 1

= (-1)^1+1 A `` Det (M11)

By def M11 is an nxn upper triagular matrix

By IH, Det (M11) = A11A22A33..Ann

Det (A) = A11A22A33 ... An+1n+1

Cor. 25.4

- Let 1 be an nxn matrix

therefore, I is a trianglar matrix.

therefore, by 25.3, we can multiply the diagonal entries

So our determinate is (1)^n

.. so Det(I) = 1

thm 25.5 -

If B is obtained from A by incerchanging two rows then Det(B) = - Det(A)

suppose A and B are nxn

Let n = 2

A =

[ A11 A12 ]

[ A21 A22 ]

B = [A21 A22]

[A11 A12]

... .

.. works for base case

IH suppose det(B) = -det(A) for any nxn matrix A

IS .. Let n = n + 1

Expand along a row that is not being switched.

det(A) = Σ (-1)^i+j Aij Det (Mij)

det(B) = Σ (-1)

^i+j Bij Det (Mij~)

since we're expanding along a row that was not swtiched, Bij = Aij

By IH, Det(Mij~) = - Det(Mij)

therefore Mij~ is obrrained from Mij by swtiching rows

therefore det B) = - Σ (-1)^i+j Det(Aij)

therefore, det (B) = det(A)

Corrallary 25.6

By the theorum, Det(E) = -Det(I) .. therefore = -1

Suppose A in an nxn matrix

The following are equivalent

(1) rank(A) = n

(2) A is invertible

(3) det(A) ≠ 0

(4) AX = 0 has only a trivial solution

(5) AX = B has exactly one solution

(6) A is reducible to an identity

this set is equivalent:

(1) Rank(A) < n

(2) A is not invertible

(3) Det(A) = 0

(4) AX = 0 has non-trival solutions

(5) AX = B has either no solutions or an infinite number of solutions

(6) A is not reducible to an identity

AX = B

X = [col: x1 .. xn]

xk = Det(M_k)/Det(A)

M_k = matrix A with column k replaced by B

proof.

nxn identity matrix

replace kth column for Xk mat

1 0 0 0 x 0

0 1 0 0 x 0

Det(Xk) .. expant on row k

= (-1)^k+k Xk Det(M_kk)

Det(M_kk) = identity matrix = 1

= xk

AXk = A^(1) A^(2) ... AX_k .. A^(k+1) .. A^(n)

the kth row .. is B

that was how we defined Mk

AX_k = M_k

Det(AX_k) = Det(M_k)

= Det(A)Det(X_k) = Det(M_k)

remember, Det(M_k = x_k

..

x_k = Det(M_k) / Det(A)

for next time , read trhough notes he gave us .. start with 26 ..

189 - 196 goal for friday.

for friday .. just through 164.

-------

Fri 04/24/09

diagonal metricies -- they're just much nicer

Eigenvectors and eignenvalues

.. asssme that we're ding linear tranformations from V to V .. a linearoperator on V .. a linear transofrmation from V to V

eigen value -- scalar multiple of x to get t(x) .. represented by λ

thm 26.1 .

β = {x1, .. xn} is a basis of eigen vectors of T iff [T]β is a diagonal matrix.

λi's will be on diagonal

==>

suppose β is a basis of eigenvectors .. λ1, λ2, .. λn ..

.. the numbers don't actually have to be different

so if you have a basis of eigenvectors, construct [T]β

Then, T(xj) = λj xj

= 0x1 + .. + λj xj + .. + 0xn

call [T]β A

Aij = λij if i=j

= 0 if i≠j

therefore, A looks diagonal

therefore A is a diagonal matrix

so that's ==>

<==

Suppose you start with a basis β and you find that you're getting a diagonal matrix

then T(x1) = A11x1 .. and the rest are 0's

therefore T(x1) = A11x1

T(x2) = A22x2

T(xn) = Annxn

so each of the vectors in the basis would have mapped to a scalar multiple of itself -- so β is a basis of eigenvectors

def 26.4 .. two definitions for diagoniziblilty

linear tranformaton is diagonalizizable if there exits a matrix represenation is diagonal

a matrix is diagonalizable if it's similar to a diagonal matrix

thm 26.3 - if you know the matrix Q, you just read Q as if it was the change of basis matrix .. follows from 14.2

A.191 thm 26.4

Let T be a linar opeartor and β

be a basis for V

==>

suppose T is diagonalizable

Then exists basis β' st [T]β is a diagonal matrix (thm 26.3)

[T]β is simiilar to [T]β'

[T]β is diagonalizable

<==

Assume [T]β is a diagonalizable matrix

[T]β is similar to diagonal matix B st B = Q^-1 [T]β Q (def 26.1)

then there exists a basis β' st B = Q^-1 [T]β Q = [T]β' (them 26.3)

therefore, T is diagonalizable

general alg for diagonalizing an linear tranformation

.. begin with matrix representation

.. figure out eigen values are

.. get egien vecttrs for T

.. know thht T is a diagonal matrix with vals on the main diagonal

not every lin trans is diagonalizable

.. there has to be a basis of eigenvectors .. they have to be non-zero

ex:

T(x,y) = [-y, x]

suppose T(x,y) = λ(x,y)

[-y,x] =

[λx, λy]

-y = λx

x = λy

- y = λλy

-y = λ^2 y

that's impossible .. λ is a real number

therefore, this transofrmation cant' be diagonalized. .. can't find eigen values.

def 27.1 det of a lin trans.

- take a matrix representation and compute the determinate of that

.. sounds kinda dangerous . because we know that a in trans has many pmatrix representations

Thm 27.1 - show det([T]β) = det([T]β')

det([T]β') = det (Q^-1 [T]β Q)

= det(Q^-1)det([T])det(Q) by 26.17

= ..

= det(Q)^-1det([T])det([T]β)

. so the Q det's cancel

= det([T])

.. so now we know that our definition is well - defined

thm 27.2, 27.3 .. homework.

thm 27.4 -

show det(T-λIv) = det(A-λIv) where A = [T]β

det(T-λIv) = det([T-λIv]β)

det([T]β - λ[Iv]β) by thm 9.4

= det(A-λIn)

T

hm 27.5 Prove λ is an eigenvalue of T <==> exists x ≠0 st x in N(T-λI)

proof.

λ is an eigenvalue of T <=> exists x ≠0 st T(x) = λx

iff exits x≠0 st T(x) - λx = 0

iff exists x≠0 st T(x) - λIv(x) = 0

iff exists x ≠0 st (T-λIv)(x) = 0 iff exists x ≠0 st x in N(T-λIv)

(we can conclude then taht it's not 1-1) . then it's not invertible

.. then det = 0

so we're looking for values of λ that will make ...

Monday - through 201.

tue - 26 and 27

-------

Mon 04/27/09

T: λIv(x) = 0

T(x) = λx

what λ for the dete

rminate to be 0

what would λ have to be in order for a-λi to be zero.

in general

det(A-λI)

.. p(t) .. characteristic polynomial

.. we want to know when t=0

Thm 27.8

- det(A-tI) is a polynomial in t of degree n with leading coefficieent (-1)^n.

Proof.

(A-tI) =

[A11-1 A12 ... A1n

[A21 A22-2 ...

since tI = [ t 0 0 0 ]

[ 0 t 0 0 ]

. ..

By def. 25.2 , requires eveery possible product of n entires of A in such a way that exactly one entry comes from each and every row and column of A

therefore, the largest degree of the det(A=tI) will come from the main diagonal and can be written

(-1)^n (t-A11)(t-A22)...(t-Ann) + q(t)-> always will be degree less than n

therefore, (-1)t^n + rest -> degree < n + q(t) degree less than n ..

so therefore our leading coefficient is (-1)^n of thh polynomial of.. of degree n

T : P1(r) => P1(R)

T(a+bx) = (a-2b) + (4a +7b)x

Let β be {1, x}

construct [T]β

(take each β, apply the transformation, put the in the matrix)

T(1) = T(1+0x) = 1+4x

[ col: 1,4 ]

T(x) = T(0 + 1x) = -2 +7x

[ col: -2,7]

use this representation to calc the characteristic poly

p(t) = det( 1-t -2

4 7-t )

= t^2 -8t + 15 = 0?

.. factor!

(t-3)(t-5) = 0

t = 3, 5 .. these numbers are the eigenvalues

they tell me that I can find a poly .. a non-zero poly st T(p) = 3p

T(p1) = 5p1

so find those poly's

plug 3 in for t

[ -2 -2]

[ 4 4 ]

we wanna solve

[ -2 -2][c1] = 0

[ 4 4 ][c2] 0

-2c1 -2c2 = 0

4c1 + 42c = 0

so anything in c1 = -c2 would work

thes

e number really represent the coefficietns on β

so one possibility c1 =1 and c2 = -1

then p = 1-x

T(1-x) = 3-3x

so we've found an eiganvector associated with the eiganvalue of 3.

can do the same thing for 5.

[-4 -2]

[4 2 ]

4c1+2c2 = 0

let c1 = 1 and c2 = -2

1-2x is the eiganvector

β' = {1-x, 1-2x}

Tβ =

[ 3 0 ]

[ 0 5 ]

// just be sure to put the values in in order

what happens if you can't diagonalize :-O

if you calculate the characteristic poly p(t)

if we're only allowed to use real scalars, then

you may only be ablet to factor down to linear poly and irreducible quadratic

ex: t^2 + 1 does not factor

.. if we completely factor, then automatically, we'll be able to diagonalize

otoh, if that can't be done, then it cannot be diagonalized.

middle ground: can completely factor the poly, but there are not n-distinct eigenvalues .. then those power are called the multiplicity of the eigenvalue

then it's more complicated .. you may be able to diagonoalize or may not be.

if you can't diagonalize, you can still get almost there --- jordon form

(next best thing is rational form)

if you're in the complex number system, then every poly splits into linear factors

28. when is a linear operator diagonalizible?

28.1 ... if T is diagonalizable, then p(t) splits .. ie, p(t) factors copletely into first degree factors. the converse is not necessarily true.

Suppose A = [T]β is diagonalizable representation

A =

[A11 0 0 0 0 0 ]

[0 A220 0 0 0 ]

etc.

P(t) = det (A - tI) =

det(A11-t 0 0 0 0

0 A22-t 0 0 0

etc.

since this is a triagular matrix,

= (A11-t)(A22-t) .. (Ann-t)

where Aii-t are all 1st degree factors since all A11 are constants

thm 28.2.

If S = {x1, x2, ... xk} is a set of eigenvactors associated with distinct eigenvalues λ1, , λ2, λ3 .. λk then S is linearly independent.

The proof is by induction on k.

base case:

let k = 1

{x1}

Since x1 ≠0 by def of eigenvector, S1 is linearly independant

IH assume

{x1 .. xk} is linearly independant

xi is associated with λi

λi ≠ λj when i≠j

IS:

{x1 .. xk+1} associated with k+1 eiganvalues

prove that that set is linearly idependant

assume a1x1 .. akxk + ak+1xk+1 = 0

apply (T-λk+1Iv)

(T-λk+1Iv)(a1x1 .. akxk + ak+1xk+1) = (T-λk+1Iv)(0)

right side is 0

look at left hand side:

a1(T-λk+1Iv)(x1)+ .. ak+1(T-λk+1Iv)(xk+1) = 0

=

a1[T(x1)-λk+1x1]+ .. + ak[T(xk) - λk+1xk]

+ ak+1 [T(xk+1)-λk+1xk+1] = 0

a1[λ1-λk+1xk]

.. ak[λk-λk+1 xk]

last term disappears

and the rest is = 0

by IH, ai(λ-λk+1) = 0

since all λ's are distinct, (λ-λk+1) ≠0

therefore, a1 .. ak are all 0's

therefore

ak+1xk+1 = 0

but xk+1 is an eiganvector, therefore non-zero

therefore ak+1 = 0

therefore {x1, .. xk} is lin. indep.

28.2 ..

thm 28.3 ..

if you look at an egian space, it's gong to have a dimenti greater or euql to 1 .. because bby deefintion there has to be at least one non-zero vector

as long as a subspac has at least 1 non-zero vecotr, at least it's dimensin 1.

dim(Eλ) ≤m

m .. is the multiplicity of the eiganvalue λ

ex

p(t) = (-1)^n (t-λ)^m (t-λ2)^m2 (t-λ)^mk

m1 + m2 + m3 = n

dim(Eλ1) ≤ m1

to see that,

find a basis for Eλ

..β = {x1 .. xk} for V

.. if you form this basis, then

[T]β = A =

[

first k vectors are eiganvalues associated with λ

so the top has λ down the diagonal .. and the lower

portion is all 0

then one right side

B and C

the 0's and the B are not necessarily square

A - tI =

[λ-t 0 0 0 0 ] B

[ 0 λ-t 0 0 0] C-tI

...

0's

we want the det of this

there'sa teuporum that says that the det is actually the det of

the top square matrix tijmes the det of the lower square matrix

= (-1)^k(t-λ)^k q(t)

that's the characteristic poly

k oly represented the dimension of Eλ

since m is the largest power of t-λi that you c