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.


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. $\lambda_1 = a_1, \mu_1 = a^2_1$.

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