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 $n-1$.

  • $\begingroup$ 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

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 $0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.