anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
[v, (I + A)v] ?
anonymous
  • anonymous
Is that their inner product?
anonymous
  • anonymous
i think so. wording confused me.

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anonymous
  • anonymous
should i just close the question?
anonymous
  • anonymous
nah
anonymous
  • anonymous
gotta do some geology though. time is priceless right now for me.
dan815
  • dan815
time was always priceless
anonymous
  • anonymous
$$\langle v, (I + A)v\rangle=\langle v,v+Av\rangle=\langle v,v\rangle+\langle v,Av\rangle\ne0$$
anonymous
  • anonymous
yep. got that.
anonymous
  • 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
  • anonymous
this question is kinda a generalisation of a question i did before it. not good a generalisations.
anonymous
  • anonymous
but yea, i think thats all we are given.
anonymous
  • 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
  • anonymous
yep. applying eigen-knowledge. i like the sound of this.
anonymous
  • anonymous
We know \((k+1)\langle v,v\rangle\ne0\) and \((I+A)v=(k+1)v\)
anonymous
  • anonymous
yep. im with u
anonymous
  • 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
  • anonymous
but...
anonymous
  • anonymous
I'm not sure
anonymous
  • anonymous
hold on
anonymous
  • anonymous
k
experimentX
  • experimentX
this looks like quadratic forms. what info we have on rank of A and dimension of W?
anonymous
  • anonymous
rank of A is \(1\), dimension of \(W\) is at least \(1\)
anonymous
  • 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
  • anonymous
The denominator is just \(\langle v,(I+A)v\rangle\) hence why we're given it's non-zero...
anonymous
  • anonymous
mmm. any ideas experimentX
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
  • 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
  • anonymous
O.o
anonymous
  • anonymous
i've gotta go now. ur welcome to keep posting ideas. i'll give u a medal for persisting though:)
anonymous
  • anonymous
argh this is the only problem today I haven't been able to figure out
anonymous
  • anonymous
I will figure this out
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
I TOLD YOU I WOULD FIGURE IT OUT!!!
anonymous
  • anonymous
hahaha pellet. that is crazy.
anonymous
  • anonymous
you are a lad mate.
anonymous
  • anonymous
have u finished like a 7 maths degrees at 7 universities or something. nice work bro. respect.
anonymous
  • anonymous
haha I wish!
dan815
  • dan815
LOL "have u finished like a 7 maths degrees at 7 universities or something. nice work bro. respect."
anonymous
  • 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
  • dan815
haha that comment is jokes doe
anonymous
  • anonymous
hahahaha oath!
dan815
  • dan815
oldrin knows way too much
anonymous
  • 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
  • anonymous
yeah he's a full blown genius
dan815
  • dan815
hahahaha true
anonymous
  • anonymous
i need ur 99 status
anonymous
  • anonymous
looks so dope
dan815
  • dan815
lol pls SS doesnt mean anything
anonymous
  • anonymous
i wonder if its an idea to buy that emporium thing where the ceo writes somethign about you for ur CV
anonymous
  • anonymous
that would be sick for jobs i reckon
dan815
  • dan815
yeah sure preetha's got PHDs its legit, maybe ill get one soon xd
anonymous
  • anonymous
is that the ceo? noiceee
dan815
  • dan815
ye
anonymous
  • anonymous
id do it when i'm applying for jobs i reckon
anonymous
  • anonymous
that would be so dope especially this is considered as volunteering
anonymous
  • anonymous
when did they start doing the owl bucks tho?
anonymous
  • anonymous
i'm guessing thats an idea for them to generate an income some how?

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