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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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.