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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?

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See singular value decomposition. –  1015 Jul 4 '13 at 21:26
If $A$ is not symmetric, then the eigenvector of $A$ and $A^H$ are not the same and I can't see any relation between the eigenvalues. The eigenvalues of $A$ and $A^H$ are always the same, and you can check that when $A$ is symmetric, the eigenvectors are also the same, therefore the eigenvalues will be $\lambda^2$ being $\lambda$ the eigenvalues of $A$, although this case is trivial because when $A$ is symmetric, that product is $A^2$ –  MyUserIsThis Jul 4 '13 at 21:27
Precisely. The eigenvalues of $A^*A$ are the squares of the singular values of $A$. So you are asking about relations between eigenvalues and singular values of $A$. –  1015 Jul 4 '13 at 21:32
If $A$ is symmetric, the singular values are the absolute values of its eigenvalues. Not that bad. –  1015 Jul 4 '13 at 21:36
@MyUserIsThis No, the eigenvalues of $A$ and $A^*$ are not the same, but are complex conjugates. You are correct about the symmetric case. Even if $A$ is "only" normal with the eigenvalues $\lambda_k$, the eigenvalues of $A^*A$ are $|\lambda_k|^2$ (comes from the Schur/eigenvalue decomposition). I don't think there is a connection in the general case. –  Vedran Šego Jul 4 '13 at 21:40

2 Answers 2

up vote 2 down vote accepted

There are some connections between eigenvalues and singular values. For instances, the maximum 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 $p$.

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how did you see this square root $x^2-(p^2+2)x+1$ thing? I do not now how one can see the singular values? –  Xin Wang Jul 5 '13 at 12:25
@Lipschitz The singular values of $A$ are the square roots of the eigenvalues of $AA^H=\pmatrix{p^2+1&-p\\ -p&1}$ and the characteristic equation of $AA^H$ is $x^2-(p^2+2)x+1=0$. Hence the claim. –  user1551 Jul 5 '13 at 12:33

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).

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