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We have two real-valued Positive Semi-definite matrices, $A$ and $B$, both $n$ by $n$.

Can the minimum eigenvalue $\lambda_\mathrm{min}(AB)$ be lower-bounded using the following eigenvalue sequences?

$\lambda_1(A)\ge \lambda_2(A) \dots \lambda_\mathrm{min}(A)$

$\lambda_1(B)\ge \lambda_2(B) \dots \lambda_\mathrm{min}(B)$

Not much are known other than PSDness of two matrices, except rank of $B$ is $n-1$.

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  • $\begingroup$ The minimum eigenvalue is necessarily $0$. $\endgroup$ – Omnomnomnom Jun 20 '15 at 4:08
  • $\begingroup$ If the rank is not full, it must have a non trivial kernel. $\endgroup$ – copper.hat Jun 20 '15 at 4:08
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The minimum eigenvalue is necessarily $0$. We can prove that this is the case as follows:

We begin by noting that the product $AB$ has non-negative eigenvalues. We can see this by noting that if $A$ is invertible, then $AB$ is similar to the positive semidefinite matrix $$ A^{1/2}BA^{1/2} = A^{-1/2}(AB)A^{1/2} $$ If $A$ fails to be invertible, then it suffices to note that $AB = \lim_{\epsilon \to 0^+}(A+\epsilon I)B$.

Because $B$ is singular, the product $AB$ is singular.

So, $AB$ is a singular matrix with non-negative eigenvalues. So, its minimum eigenvalue must be $0$.

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  • $\begingroup$ Thanks for answering. simple and succinct. However that makes me wondering if the following is also true: $B$ is singular, then $UBU^T$ is also singular, where $U$ is a full rank matrix. $\endgroup$ – Sng L Jun 20 '15 at 4:22
  • $\begingroup$ Yep, that's also true, assuming all the matrices are square. $\endgroup$ – Omnomnomnom Jun 20 '15 at 4:45
  • $\begingroup$ If $U$ is not square (in particular: if $U$ has more columns than rows), then the product could be non-singular. $\endgroup$ – Omnomnomnom Jun 20 '15 at 4:48
  • $\begingroup$ Thanks again, however, before looking at your comment, I have just posted another question, with a little more details, just in case if you are interested in answering: link $\endgroup$ – Sng L Jun 20 '15 at 4:53
  • $\begingroup$ I see your deleted your question. Anyway, the thing to remember is that the rank of a product of matrices is limited to the lowest rank in the product, but can be smaller. $\endgroup$ – Omnomnomnom Jun 20 '15 at 5:03

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