# Relating the eigenvalues of two positive semi-definite matrices

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