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I need help to show given that $\mathbf A$ is a complex nxn matrix that
$\mathbf {AA}$* is a Hermitian matrix and the eigenvalues > 0

Where * is taking the transpose of the complex conjugate of the matrix. I got the first part. However for the second part, I tried checking if $\mathbf {AA}$* was a positive definite matrix but I don't think that works. Is there any information I can use from the fact that $\mathbf {AA}$* is Hermitian?

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Let $\lambda$ be an eigenvalue of $AA^*$, and let $\mu$ be an eigenvalue of $A^*$. Then $$\lambda x^*x = x^*\lambda x = x^*AA^*x = (A^*x)^*(A^*x) = (\mu x)^*(\mu x) = \overline{\mu}\mu x^*x = |\mu|^2 x^*x$$ So we get $\lambda x^*x = |\mu|^2 x^*x$. Note that $x^*x$ is a scalar. Argue why we can cancel it, and why the resulting equation shows that $\lambda$ must be non-negative.

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  • $\begingroup$ Thanks! I appreciate your help $\endgroup$
    – lllll
    Nov 26 '14 at 2:51

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