# Relation between eigenvalues of $BA$ and eigenvalues of $A$, where $B=\operatorname{diag}([b_1\:b_2 \cdots b_m])$ and $b_i>1$ for all $i$.

I have a real symmetric matrix $$A\in\Bbb R^{m×m}$$, which has one eigenvalue $$\lambda_1=0$$ and the rest of the eigenvalues are all positive and assume $$\lambda_2<\lambda_3<\cdots<\lambda_m$$. I have another matrix $$B \in \Bbb R^{m×m}$$, which has elements only on the diagonal, i.e. $$B=\operatorname{diag}([b_1\:b_2 \cdots b_m])$$ and $$b_i>1\;$$ for all $$i$$. Let $$\lambda_1'<\lambda_2'<\cdots<\lambda_m'$$ be the eigenvalues of $$BA$$.

Is $$\lambda_i'>\lambda_i\;$$ for $$i=2,\cdots,m$$ ?

Intuitively, I know it is true but I need a proof of this. I have been looking for a result from the literature but I am not able to find it yet.

Any help would be appreciated. Thanks!

• It is true if $A$ is real symmetric, but false otherwise. The eigenvalues of $BA$ can also be non-real when $A$ is not symmetric. It should be easy to generate a random counterexample by computer. Oct 6, 2021 at 10:36
• @user1551 Thanks for the comment, I have edited my question. Yes, A is real but not necessarily symmetric and with the update (now, only considering the real part of the eigenvalues) I think the statement is true. What do you think? Oct 6, 2021 at 13:05
• This is unlikely to be true. Try a numerical experiment. Oct 6, 2021 at 14:00
• You are right, I found a counterexample. I will update my question as A being real symmetric. Thanks. Oct 6, 2021 at 15:28

We can at least show that $$\lambda_i' \geq \lambda_i$$.

One way to show this is as follows. For $$t > 0$$, define $$A_t = A + tI$$. Because $$A_t$$ is positive definite, it follows that it has an (invertible) square root $$A_t^{1/2}$$. Now, note that $$BA_t$$ is similar to $$A_t^{1/2}(BA_t)A_t^{-1/2} = A_t^{1/2}BA_t^{1/2}.$$ We note that the matrix $$M = A_t^{1/2}BA_t^{1/2} - A_t = A_t^{1/2}(B - I)A_t^{1/2}$$ is positive definite. It follows that $$\lambda_i(BA_t) = \lambda_i(A_t^{1/2}BA_t^{1/2}) = \lambda_i(A_t + M) \geq \lambda_i(A_t) + \lambda_1(M) > \lambda_i(A_t).$$ Now, because eigenvalues depend continuously on matrix entries, we can conclude that $$\lambda_i' = \lambda_i(BA) = \lim_{t \to 0^+}\lambda_i(BA_t) \geq \lim_{t \to 0^+} \lambda_i(A_t) = \lambda_i(A) = \lambda_i.$$

• This is a quite nice proof. It is really helpful. Thanks! Oct 8, 2021 at 16:39
• @user5i Glad to help! If you're satisfied with either of these answers, please accept the answer that you prefer by clicking the checkmark ($\checkmark$) on the left of the question. Oct 8, 2021 at 16:52

another way of doing this, for $$k\in \big\{1,2,...,m\big\}$$, place the eigenvalue of interest, $$\lambda_k$$, in the kernel of $$A$$:

$$A^{(k)}:= A-\lambda_kI$$
This translates all eigenvalues by $$\lambda_k$$ so the resulting eigenvalues are $$\eta_j = \lambda_j -\lambda_k$$

$$D:=B^\frac{1}{2}$$ and do a congruence transform:
$$D^TA^{(k)}D = DA^{(k)}D = \big(DAD -\lambda_k B\big)$$
has the same signature as $$A^{(k)}$$, in particular the kth eigenvalues must agree so $$\gamma_k = \eta_k = 0$$
so we add a well chosen symmetric PSD matrix to our real symmetric matrix and get
$$M:=\big(DAD -\lambda_k B\big)+\lambda_k\big(B-I\big)=DAD -\lambda_k I=\big(DAD -\lambda_k B\big)+\lambda_k\sum_{j=1}^n \sigma_j \cdot \mathbf q_j\mathbf q_j^T$$
where the RHS writes out PSD matrix as a sum of outer products, via spectral theorem, each being PSD

This creates an interlacing of eigenvalues, e.g.
$$M_{1}:=\big(DAD -\lambda_k B\big)+\lambda_k\cdot \sigma_1\cdot\mathbf q_1\mathbf q_1^T$$
$$\gamma_m\leq \eta_{m_1}' \leq \gamma_{m-1}\leq \eta_{{m-1}_1}'\leq \dots\leq 0=\gamma_k\leq \eta_{k_1}'\leq \dots \leq \gamma_1\leq \eta_{1_1}$$
and the eigenvalues of $$M_2 =M_1+ \lambda_k\cdot\sigma_2 \cdot \mathbf q_2\mathbf q_2^T$$ interlace with those of $$M_1$$ and and $$M_3$$ interlaces those of $$M_2$$ and so on, giving us the monotone sequence

$$0=\gamma_k\leq \eta_{k_1}'\leq \eta_{k_2}'\leq\eta_{k_3}'\leq \dots \leq \eta_{k_{m-1}}'\leq \eta_{k_m}'=\eta_{k}'$$
where we write $$\eta_{k}':=\eta_{k_m}'$$ to cleanup notation, so

$$0=\eta_k =\gamma_k\leq \eta_k'\implies \lambda_k =\eta_k + \lambda_k\leq \eta_k' +\lambda_k = \lambda_k'$$
finally recall that $$(DAD)$$ has the same eigenvalues as $$(D^2A)=(BA)$$

• This is a nice proof, thanks! Oct 8, 2021 at 16:49