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If $\mathbf{A}_{n\times n}$ is a positive semi-definite matrix with eigenvalues $\{\alpha_k\},\ k\in\{1,...,n\}$, and $\mathbf{B}_{m\times n}$ is an arbitrary matrix with singular values $\{\beta_k\},\ k\in\{1,...,\min(m,n)\}$, can anything be said about the singular values $\{\gamma_k\},\ k\in\{1,...,\min(m,n)\}$ of the matrix $\mathbf{\Gamma}=\mathbf{BA}$? Is there a way I can relate $\gamma_k$ to $\alpha_k$ and $\beta_k$?

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Are these real-valued matrices? Is $\mathbf{A}$ symmetric? – cardinal Mar 27 '11 at 22:01
yes, $A$ is hermitian. The matrices are not necessarily real valued, but if it helps to consider only all real or all complex, by all means, do so. Such a solution is acceptable too. – user7815 Mar 27 '11 at 22:09
What do we know about singular values? I've never worked much with them? How close to the eigenvalues are they actually? – David Kohler Mar 30 '11 at 22:07
up vote 9 down vote accepted

The singular values of $B$ are the square roots of the eigenvalues of $B^* B = C$, and those of $BA$ are the square roots of the eigenvalues of $(BA)^* BA = A C A$. So the question is whether for positive semidefinite matrices $A$ and $C$ there is any relation between eigenvalues of $A C A$ and those of $A$ and $C$. The answer is, in general, not much. You might consider some simple $2 \times 2$ examples, with different $C$ having the same eigenvalues, but the corresponding $A C A$ having different eigenvalues.

Of course the product of the eigenvalues is the determinant, and $\det(ACA) = \det(C) \det(A)^2$. Also, the sum of the eigenvalues is the trace, and using Cauchy-Schwarz you have a bound ${\rm tr}(ACA) = {\rm tr}(CA^2) \le ({\rm tr}( C^2))^{1/2} ({\rm tr}(A^4))^{1/2}$.

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I don't think there is a simple answer. Well, I'm not familiar with singular values. My understanding is that they are some sort of generalization of eigenvalues, when the matrices are not square. But even in the case when both matrices $\mathbf{A}$ and $\mathbf{B}$ are square and positive definite, the eigenvalues of $\mathbf{A}$ and $\mathbf{B}$ do not pin down eigenvalues of $\mathbf{BA}$. One needs more structural information about the matrices. For instance: Let $$\mathbf{A}= \left( \begin{array}{cc} 1 & 0 \\ 0 & 2 \\ \end{array} \right), \mathbf{B}_1= \left( \begin{array}{cc} 3 & 0 \\ 0 & 4 \\ \end{array} \right), \mathbf{B}_2= \left( \begin{array}{cc} 4 & 0 \\ 0 & 3 \\ \end{array} \right).$$ Obviously $\mathbf{B}_1$ and $\mathbf{B}_2$ have the same set of eigenvalues, but $\mathbf{B}_1\mathbf{A}$ and $\mathbf{B}_2\mathbf{A}$ have different sets of eigenvalues $\{3,8\}$ and $\{4,6\}$ respectively.

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for symmetric square matrices the singular values are the squares of eigenvalues, so your example is spot on. – mpiktas Mar 31 '11 at 20:53

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