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.

- anonymous

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- anonymous

[v, (I + A)v] ?

- anonymous

Is that their inner product?

- anonymous

i think so. wording confused me.

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## More answers

- anonymous

should i just close the question?

- anonymous

nah

- anonymous

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

- dan815

time was always priceless

- anonymous

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

- anonymous

yep. got that.

- anonymous

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

- anonymous

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

- anonymous

but yea, i think thats all we are given.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yep. im with u

- anonymous

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

- anonymous

but...

- anonymous

I'm not sure

- anonymous

hold on

- anonymous

k

- experimentX

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

- anonymous

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

- anonymous

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!

- anonymous

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

- anonymous

mmm. any ideas experimentX

- experimentX

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

- anonymous

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

- anonymous

O.o

- anonymous

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

- anonymous

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

- anonymous

I will figure this out

- anonymous

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

- anonymous

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 \(Ay=A(I+A)x=(A+A^2)x=(1+k)Ax\).

- anonymous

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 \(k=\langle v,Av\rangle/\langle v,v\rangle\):$$x=y-\frac1{1+\langle v,Av\rangle/\langle v,v\rangle}Ay=y-\frac{\langle v,v\rangle}{\langle v,v\rangle+\langle v,Av\rangle}Ay$$

- anonymous

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

- anonymous

I TOLD YOU I WOULD FIGURE IT OUT!!!

- anonymous

hahaha pellet. that is crazy.

- anonymous

you are a lad mate.

- anonymous

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

- anonymous

haha I wish!

- dan815

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

- anonymous

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

- dan815

haha that comment is jokes doe

- anonymous

hahahaha oath!

- dan815

oldrin knows way too much

- anonymous

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

- anonymous

yeah he's a full blown genius

- dan815

hahahaha true

- anonymous

i need ur 99 status

- anonymous

looks so dope

- dan815

lol pls SS doesnt mean anything

- anonymous

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

- anonymous

that would be sick for jobs i reckon

- dan815

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

- anonymous

is that the ceo? noiceee

- dan815

ye

- anonymous

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

- anonymous

that would be so dope especially this is considered as volunteering

- anonymous

when did they start doing the owl bucks tho?

- anonymous

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

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