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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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    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 product2011-05-02

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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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    Please do not give complete answers to $h$omework problems.2011-05-03