Proof: Eigenvalues of Matrix A*A

Bazzman · **Joined:** Tue, 16 Dec 2008 20:15:02 UTC **Posts:** 7

Hey all,

I have to prove that all eigenvalues of the matrix A*A are non-negative real numbers (where A* denotes the conjugate transpose matrix of A). How do you show this? Thank you for any help!

Baz

Opalg · **Posted:** Tue, 16 Dec 2008 23:17:14 UTC

Bazzman wrote:

I have to prove that all eigenvalues of the matrix A*A are non-negative real numbers (where A* denotes the conjugate transpose matrix of A). How do you show this?

If A*Ax = λx then $\langle Ax,Ax\rangle = \langle A^*Ax,x\rangle = \lambda\langle x,x\rangle$ . So $\lambda = \frac{\langle Ax,Ax\rangle}{\langle x,x\rangle}\geqslant0$ .

Bazzman · **Joined:** Tue, 16 Dec 2008 20:15:02 UTC **Posts:** 7

Thank Opalg.

Sorry I didnt reply earlier but I fell sick. Yes that makes sense - and they are real b/c of inner product <Ax,Ax> which would square all the complex terms in Ax.

I had two questions however if you dont mind. Im a bit of a noob with using the <,> notations and tend not to use it. So if you were to write <Ax,Ax> out how would it look (as in x*A*Ax,...)??

This would also probably answer my second question but why did you begin to work with <Ax,Ax>.

Thanks if you can answer.

Baz

Opalg · **Posted:** Thu, 18 Dec 2008 16:47:09 UTC

Bazzman wrote:

Im a bit of a noob with using the <,> notations and tend not to use it. So if you were to write <Ax,Ax> out how would it look (as in x*A*Ax,...)??

This would also probably answer my second question but why did you begin to work with <Ax,Ax>.

These are alternative notations for the same thing. Analysts usually think in terms of inner products and write $\langle Ax,y\rangle$ . Algebraists think in terms of matrices and write $x^*\!Ay$ . But these notations both mean exactly the same thing.

Choose whichever one you feel more at home with. As you can see, I'm an analyst.

Bazzman · **Joined:** Tue, 16 Dec 2008 20:15:02 UTC **Posts:** 7

Hi Opalg,

I realized they were alternative notations, yet I did not know who uses which type. My question however was how would you express your "analyst" equations in terms of "algebraic" equations?

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

(lambda)<x,x> = (lambda)x*x

How would <Ax,Ax> look/fit in?

Baz

Opalg · **Posted:** Thu, 18 Dec 2008 20:15:10 UTC

Okay, if A*Ax = λx then $(Ax)^*(Ax) = x^*A^*Ax = x^*(\lambda x) = \lambda (x^*x)$ . So $\lambda = \frac{(Ax)^*(Ax)}{ x^*x}\geqslant0$ .

Bazzman · **Joined:** Tue, 16 Dec 2008 20:15:02 UTC **Posts:** 7

Thank you very much for your time and help!

Baz

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Proof: Eigenvalues of Matrix A*A

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