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#1
Nov 25 2014 07:20pm
Hey so I'm having trouble with 4b)



I think I would be able to do it if they asked about A*A instead of AA*.

If it was A*A I would just let x be an eigenvector corresponding to an eigenvalue lambda of A*A and then do

<A*Ax,x>=lambda<x,x>

And then using the complex inner product reach ||Ax||^2 = lambda||x||^2 proving lambda is >= 0. I'm not sure how to do it if I have AA* instead though...
From part a) we see that AA* is Hermitian but does that say anything about A? Can we then say that A is Hermitian and if it is then A is also normal, that way I can get AA*=A*A?

Thanks!
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#2
Nov 25 2014 08:51pm
* is the transpose of the complex conjugate?
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#3
Nov 25 2014 09:00pm
a) (AA*)* = (A*)*A* = AA* -> hermitian
b) x eigenvector of AA*, (AA*x|x) = (A*x|A*x) = ||A*x||² = lambda*||x||² -> lambda >=0
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#4
Nov 25 2014 09:19pm
Quote (rwarth @ Nov 25 2014 09:51pm)
* is the transpose of the complex conjugate?


Yes sorry should've said that earlier

Quote (HbSoe @ Nov 25 2014 10:00pm)
a) (AA*)* = (A*)*A* = AA* -> hermitian
b) x eigenvector of AA*, (AA*x|x) = (A*x|A*x) = ||A*x||²  =  lambda*||x||² -> lambda >=0


Can you explain how you go from (AA*x|x) to (A*x|A*x) ?

Nevermind, figured it out!

<AA*x,x> = x*AA*x = (A*x)*(A*x) = <A*x,A*x>

thank you!!!

This post was edited by Bloo_Guardian on Nov 25 2014 09:38pm
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#5
Nov 25 2014 10:05pm
Quote (Bloo_Guardian @ Nov 25 2014 09:20pm)
Hey so I'm having trouble with 4b)

http://i.imgur.com/C2sm2sg.png

I think I would be able to do it if they asked about A*A instead of AA*.

If it was A*A I would just let x be an eigenvector corresponding to an eigenvalue lambda of A*A and then do

<A*Ax,x>=lambda<x,x>

And then using the complex inner product reach ||Ax||^2 = lambda||x||^2 proving lambda is >= 0. I'm not sure how to do it if I have AA* instead though...
From part a) we see that AA* is Hermitian but does that say anything about A? Can we then say that A is Hermitian and if it is then A is also normal, that way I can get AA*=A*A?

Thanks!


how do you get ||Ax||² to lambda*||x||². isnt it ||AA|| ||x||^2
and AA isnt equal to lambda? I could be wrong i havent taken lin alg in a while

This post was edited by rwarth on Nov 25 2014 10:07pm
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#6
Nov 26 2014 01:44pm
Use the fact that x is an eigenvector corresponding to lambda of A*A and you have (A*A)x=lambda x

Then take inner product (use complex inner product) with A*Ax with x

<A*Ax,x>=lambda<x,x> can pull out lambda since it is a scalar
x*A*Ax = lambda<x,x>
(Ax)*(Ax) = lambda<x,x>
<Ax,Ax> = lambda<x,x>
||Ax||^2 = lambda ||x||^2

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