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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!

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    $\begingroup$ 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. $\endgroup$
    – user1551
    Oct 6, 2021 at 10:36
  • $\begingroup$ @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? $\endgroup$
    – user5i
    Oct 6, 2021 at 13:05
  • $\begingroup$ This is unlikely to be true. Try a numerical experiment. $\endgroup$
    – user1551
    Oct 6, 2021 at 14:00
  • $\begingroup$ You are right, I found a counterexample. I will update my question as A being real symmetric. Thanks. $\endgroup$
    – user5i
    Oct 6, 2021 at 15:28

2 Answers 2

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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 $$ \lambda_i' = \lambda_i(BA) = \lim_{t \to 0^+}\lambda_i(BA_t) \geq \lim_{t \to 0^+} \lambda_i(A_t) = \lambda_i(A) = \lambda_i. $$

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  • $\begingroup$ This is a quite nice proof. It is really helpful. Thanks! $\endgroup$
    – user5i
    Oct 8, 2021 at 16:39
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    $\begingroup$ @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. $\endgroup$ Oct 8, 2021 at 16:52
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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 $(D^2A)=(BA)$

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  • $\begingroup$ This is a nice proof, thanks! $\endgroup$
    – user5i
    Oct 8, 2021 at 16:49

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