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How can we show that eigenvalues for $A^*A$ are real and positive without using the Singular Value Decomposition theorem (where $A$ is a complex square matrix and $A^*$ its Hermitian conjugate)?

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Do you know how to show that the eigenvalues of a self-adjoint matrix are real? –  Qiaochu Yuan May 2 '11 at 23:08
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Hi, have you tried to prove it by proving $A^* A$ is positive-semidefinite? a hint would be $(A^* Ax, y) = (Ax, Ay)$, where $(\cdot, \cdot)$ is the standard inner product –  Shuhao Cao May 2 '11 at 23:11

1 Answer 1

The eigenvalues of $A^*A$ can be zero, so what you want to show is that they are non-negative: Take any eigenvalue $\lambda$ of $A^*A$, Let $v$ be an associated non-zero eigenvector. $0\leq\langle Av,Av\rangle=\langle A^*Av,v \rangle=\lambda ||v||^2$, which implies that $\lambda\geq 0$. In particular, $\lambda$ is real and non-negative.

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Another way is to say that for every matrix $B$, $\lambda$ is an eigenvalue of $B$ if and only if $\bar{\lambda}$ is an eigenvalue of $B^*$, which implies (how?) that the eigenvalues of $B$, if $B$ is self-adjoint, are real ($B=B^*$). Since $A^*A$ is self-adjoint (check!) the result follows. –  Dennis Gulko May 2 '11 at 23:14
    
I was looking for the non-negative part and the above answer is a nice one. –  citrucel May 2 '11 at 23:18
    
@user10386: Then vote it up and accept... :) –  Dennis Gulko May 2 '11 at 23:27
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Please do not give complete answers to homework problems. –  Qiaochu Yuan May 3 '11 at 0:13

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