2
$\begingroup$

Let $A$ be an $n \times n$ positive semi-definite matrix, and let $B$ be an $m \times n$ matrix.

Can one say anything relating the eigenvalues of $M_1 := BAB^\top$ and $M_2 := BA^2 B^\top$? In particular, I am interested if a bound on the maximum eigenvalue of one matrix implies an analogous bound for the other matrix.

My elementary observations:

  • Both matrices are positive semi-definite.
  • If $B^\top B = I$, then $M_2 = M_1^2$, so the eigenvalues are directly related. But I am interested in what happens when $B$ does not have orthonormal columns.
  • By diagonalizing $A$, it suffices to prove the claim for diagonal $A$.

My hunch is that if $B^\top B$ were badly behaved in some way, the eigenvalues for $M_2$ can be made to be arbitrarily large/small, but I am not able to formalize this guess.

$\endgroup$
1
$\begingroup$

Let the eigenvalues of $M_1$ be $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_m \ge 0$, and the eigenvalues of $M_2$ be $\mu_1 \ge \mu_2 \ge \dots \ge \mu_m \ge 0$. Meanwhile, let $A = diag(a_i)$ where without loss we also assume $a_1 \ge a_2 \ge \dots \ge a_n \ge 0$.

If I understand you correctly, you would like a bound on $\mu_1$ as a function of $\lambda_1$. Presumably you also know some info about $a_i$, or else there seems little hope. Here I will prove that:

Claim: $a_1 \lambda_1 \ge \mu_1$, and this bound is tight in some cases (e.g. $B=I$)

Is this the kind of bound you're looking for?


Let $v$ be a unit eigenvector of $M_2$ corresponding to $\mu_1$, so we have:

$$M_2 v = \mu_1 v, ~~~~~~~~~~ v^\top M_2 v = \mu_1 v^\top v = \mu_1$$

Writing $y = B^\top v = (y_1, y_2, \dots, y_n)^\top$, we have:

$$\mu_1 = v^\top M_2 v = y^\top A^2 y = \sum_{i=1}^n a^2_i y^2_i$$

Meanwhile $v$ is not necessarily an eigenvector of $M_1$, but we can still say:

$$\lambda_1 \ge v^\top M_1 v = y^\top A y = \sum_{i=1}^n a_i y^2_i$$

where the inequality comes from either a geometric argument, or this algebraic argument:

  • Diagonalize $M_1 = P^\top D P$ where $D = diag(\lambda_i)$ and $P$ is orthonormal

  • Write $Pv = x = (x_1, x_2, \dots x_m)^\top$ and note that since $P$ is orthonormal, $|x| = 1$

  • $v^\top M_1 v = x^\top D x = \sum \lambda_i x^2_i \le \lambda_1 \sum x^2_i = \lambda_1$

Now we can put everything together:

$$a_1 \lambda_1 \ge a_1 \sum a_i y^2_i \ge \sum a^2_i y^2_i = \mu_1$$

This bound is tight when "things line up perfectly", e.g. when $B=I$ s.t. $\lambda_1 = a_1, \mu_1 = a^2_1$.

| cite | improve this answer | |
$\endgroup$

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.