Question

Let W be a linear space with an inner product and
A : W → W be a linear map whose image is one dimensional (so in the case of matrices,
it has rank one). Let a vector v not equal to 0 be in the image of A, so it is a basis for the image. If
[v, (I + A)v] doesn't = 0, show that I + A is invertible by finding a formula for the inverse.

Answer 1

[v, (I + A)v] ?

Answer 2

Is that their inner product?

Answer 3

i think so. wording confused me.

Answer 4

should i just close the question?

Answer 5

nah

Answer 6

gotta do some geology though. time is priceless right now for me.

Answer 7

time was always priceless

Answer 8

$$\langle v, (I + A)v\rangle=\langle v,v+Av\rangle=\langle v,v\rangle+\langle v,Av\rangle\ne0$$

Answer 9

yep. got that.

Answer 10

Do we know anything about \(\langle v,v\rangle\)? does it have any definition or do we just know its an inner product?

Answer 11

this question is kinda a generalisation of a question i did before it. not good a generalisations.

Answer 12

but yea, i think thats all we are given.

Answer 13

Well, we know \(Av=kv\) for some scalar \(k\) (since \(v\) is the basis of \(A\)). This means \(\langle v,v\rangle+\langle v,kv\rangle=\langle v,v\rangle+k\langle v,v\rangle=(k+1)\langle v,v\rangle\)

Answer 14

yep. applying eigen-knowledge. i like the sound of this.

Answer 15

We know \((k+1)\langle v,v\rangle\ne0\) and \((I+A)v=(k+1)v\)

Answer 16

yep. im with u

Answer 17

It follows then that to recover \(v\) we merely multiply by \(1/(k+1)\) which is safe since we know \(k+1\ne0\).

Answer 18

but...

Answer 19

I'm not sure

Answer 20

hold on

Answer 21

k

Answer 22

this looks like quadratic forms.
what info we have on rank of A and dimension of W?

Answer 23

rank of A is \(1\), dimension of \(W\) is at least \(1\)

Answer 24

I found $$(I+A)^{-1}=I-\frac{\langle v,v\rangle}{\langle v,v\rangle +\langle v,Av\rangle}A$$but I'm not sure how to arrive at that conclusion!

Answer 25

The denominator is just \(\langle v,(I+A)v\rangle\) hence why we're given it's non-zero...

Answer 26

mmm. any ideas experimentX

Answer 27

ideas is what i lack ... but i remember somewhere if quadratic form is >0, then matrix is positive definite or something like that.

Answer 28

Well for some \(u\) we have \((I+A)u=u+Au=u+kv\). Passing through \(A\) again we find:$$A(u+kv)=Au+kAv=k(v+Av)=k(I+A)v=(I+A)kv$$... now I'm lost again

Answer 29

O.o

Answer 30

i've gotta go now. ur welcome to keep posting ideas. i'll give u a medal for persisting though:)

Answer 31

argh this is the only problem today I haven't been able to figure out

Answer 32

I will figure this out

Answer 33

Okay. We know that \(Av=kv\) and \(\langle v,Av\rangle=\langle v,kv\rangle=k\langle v,v\rangle\implies k=\langle v,Av\rangle/\langle v,v\rangle\)

Answer 34

Observe that for all \(u\) we have \(Au=tv\) so \(A(Au)=A(tv)=tAv=tkv=kAu\), so \(A^2=kA\). Consider \((I+A)x=y\). It follows that \(Ay=A(I+A)x=(A+A^2)x=(1+k)Ax\).

Answer 35

From the given we know \(\langle v,(I+A)v\rangle=\langle v,v+kv\rangle=\langle v,v\rangle+k\langle v,v\rangle=(1+k)\langle v,v\rangle\ne 0\) so \(1+k\ne 0\). We then can safely multiply by \(1/(1+k)\):$$\frac1{1+k}Ay=Ax$$Recall that \(y=(I+A)x=x+Ax\). It follows then that:$$x=y-Ax=y-\frac1{1+k}Ay$$Lastly, substitute \(k=\langle v,Av\rangle/\langle v,v\rangle\):$$x=y-\frac1{1+\langle v,Av\rangle/\langle v,v\rangle}Ay=y-\frac{\langle v,v\rangle}{\langle v,v\rangle+\langle v,Av\rangle}Ay$$

Answer 36

I guess your prof supposes we are dealing with a normed vector space where \(\|v\|^2=\langle v,v\rangle\) so:$$x=y-\frac{\|v\|^2}{\|v\|^2+\langle v,Av\rangle}Ay$$

Answer 37

I TOLD YOU I WOULD FIGURE IT OUT!!!

Answer 38

hahaha pellet. that is crazy.

Answer 39

you are a lad mate.

Answer 40

have u finished like a 7 maths degrees at 7 universities or something. nice work bro. respect.

Answer 41

haha I wish!

Answer 42

LOL "have u finished like a 7 maths degrees at 7 universities or something. nice work bro. respect."

Answer 43

hahaha, i can't belive this is 2yrs. i like to look at past questions that i've done. i dunno why hahah

Answer 44

haha that comment is jokes doe

Answer 45

hahahaha oath!

Answer 46

oldrin knows way too much

Answer 47

i can't believe i'm stil mid 60's. i need to spend alot more time on this to get my points up. i swear i was 67 2yrs ago. i went missing for a bit on open study haah

Answer 48

yeah he's a full blown genius

Answer 49

hahahaha true

Answer 50

i need ur 99 status

Answer 51

looks so dope

Answer 52

lol pls SS doesnt mean anything

Answer 53

i wonder if its an idea to buy that emporium thing where the ceo writes somethign about you for ur CV

Answer 54

that would be sick for jobs i reckon

Answer 55

yeah sure preetha's got PHDs its legit, maybe ill get one soon xd

Answer 56

is that the ceo? noiceee

Answer 57

ye

Answer 58

id do it when i'm applying for jobs i reckon

Answer 59

that would be so dope especially this is considered as volunteering

Answer 60

when did they start doing the owl bucks tho?

Answer 61

i'm guessing thats an idea for them to generate an income some how?