Eigenvalues of $A^HA$ I was wondering if we know something about the eigenvalues of $A^HA$ if we know the eigenvalues of $A$. Although I do know that the expression $A^HA$ is diagonalizable and positive definite, I want to know more about the relation between the eigenvalues of $A$ and $A^HA$ or is there no relationship between them?
 A: There are some connections between eigenvalues and singular values. For instance, the largest singular value of $A$ is always bounded below by the spectral radius of $A$. Also, we have $|\det(A)|=|\prod_i\lambda_i(A)|=\prod_i\sigma_i(A)$. So, given the eigenvalues of $A$, the singular values are not entirely arbitrary.
Yet, in general, given the eigenvalues (resp. singular values) of $A$, the singular values (resp. eigenvalues) are not uniquely determined. For example, the eigenvalues of $A=\pmatrix{1&p\\ 0&-1}$ are $-1$ and $1$ regardless of $p$, but the singular values of $A$ are the square roots of the roots of the quadratic polynomial $x^2-(p^2+2)x+1$, which are dependent on $p$.
A: Let the eigenvalues of $A$ be $\lambda_k$.
For the normal matrix $A$, you have a Schur decomposition $A = U \Lambda U^*$, where $U$ is unitary and $\Lambda$ is complex diagonal. Then $A^*A = U |\Lambda|^2 U^*$, i.e., the eigenvalues of $A^*A$ are $|\lambda_k|^2$ (up to possibly different number of zero eigenvalues if $A$ is not square).
Now, for a general case, consider
$$A := \mathcal{J}_2(0) = \begin{bmatrix} 0 & 1 \\ & 0 \end{bmatrix},$$
i.e., $A$ is a Jordan block of order $2$ associated with an eigenvalue $0$. Then
$$A^*A = \begin{bmatrix} 0 \\ & 1 \end{bmatrix}.$$
So, we had one eigenvalue $0$ of partial multiplicity $2$ in $A$, but two different eigenvalues, $0$ and $1$ (both of partial multiplicities $1$), in $A^*A$.
Now, take
$$A := X \mathcal{J}_2(0) X^{-1} = \begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix}, \quad X := \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}.$$
So, this $A$ is similar to the previous one and has the same eigenvalue $0$ with the partial multiplicity $2$. However,
$$A^*A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix},$$
with the eigenvalues $0$ and $4$, so I conclude that there is no connection (at least not without considering some additional properties as well) beyond the well-known relation between the zero eigenvalues (due to $A$ and $A^*A$ having the same rank).
