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Let $M \in \mathbb{R}^{d\times d}$ be an invertible real matrix (not necessarily symmetric), and assume $M$ is positive semi-definite in the sense that $$ v^T M v \geq 0 $$ for all $v \in \mathbb{R}^d$. I have noticed experimentally that the matrix $$N = (I+a(M-M^T))^{-1}M$$ always has eigenvalues with strictly positive real part, for any $a \in \mathbb{R}^+$. However, I haven't been able to prove it. The matrix is not necessarily positive definite (which would imply eigenvalues with positive real part), and I've tried finding a similarity transformation which would yield positive definiteness in another basis, but without success. Any ideas?

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If $a=0$, then $N=M$ and the result is trivial. Suppose $a>0$. Write $M=S+K$ where $S$ is symmetric positive semidefinite and $K$ is skew-symmetric. For any eigenvalue $\lambda$ of $N$, let $u$ be an associated unit eigenvector. Then $u^\ast Su=s$ for some $s\ge0$ and $u^\ast Ku=ih$ for some $h\in\mathbb R$. Now we have \begin{align} \lambda u&=(I+2aK)^{-1}(S+K)u,\\ \lambda(I+2aK)u&=(S+K)u,\\ \lambda(1+2iah)&=s+ih,\\ \lambda&=\frac{(1-2iah)(s+ih)}{|1+2iah|^2} =\frac{(s+2ah^2)+ih(1-2as)}{|1+2iah|^2}. \end{align} If $\Re(\lambda)=0$, we must have $s=h=0$. But then $\lambda=0$ and in turn $N$ and $M$ are singular, which is a contradiction. Therefore $\Re(\lambda)=\frac{(s+2ah^2)}{|1+2iah|^2}>0$.

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  • $\begingroup$ +1, this is an interesting proof. You may wish to add a word or two about real versus complex scalars since your argument sort of straddles the two cases. For instance, we shouldn't replace with the assumption $u^TMu \geq 0$ for all real vectors with the analogous condition $u^*Mu \geq 0$ for all complex vectors since this stronger assumption would imply that $M=M^*$ narrowing the scope of the result. I just raise this as a possible point of confusion, I can see that your argument doesn't need the stronger assumption. $\endgroup$
    – Mike F
    Commented Aug 11, 2020 at 4:28
  • $\begingroup$ Thanks for the answer. As Mike pointed out, my only doubt is whether $u^* Su \geq 0$ holds... The assumption is that $u^T S u = u^T M u \geq 0$ for all real $u$, does this imply $u^* Su \geq 0$ for complex $u$? $\endgroup$
    – smalldog
    Commented Aug 11, 2020 at 7:09
  • $\begingroup$ @chaos Yes. Every real symmetric matrix is orthogonally diagonalisable. It follows that if $u^TSu\ge0$ for all real $u$, the eigenvalues of $S$ are nonnegative and hence $u^\ast Su\ge0$ for all complex $u$. $\endgroup$
    – user1551
    Commented Aug 11, 2020 at 7:20
  • $\begingroup$ Thanks, perfect answer :) $\endgroup$
    – smalldog
    Commented Aug 11, 2020 at 7:27
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Added disclaimer: This answer is wrong. It mistakenly proves the result for $N=(I+a(M+M^T))^{-1}M$ instead.

Just an initial comment that I (and others) would usually require a positive-(semi)definite matrix to be symmetric by definition, but I will work with your definition here. In any event, every matrix $M$ can be expressed uniquely as the sum of a symmetric matrix and an antisymmetric matrix, i.e. we can write $$M =S+A$$ where $S^T=S$ and $A^T=-A$. For any $v$, we then have $v^TMv=v^TSv$, so your assumption that $M$ is positive-semidefinite is equivalent to the assumption that its symmetric part $S$ is positive-semidefinite.

Next, we calculate that $$N =(I+2aS)^{-1}M.$$

Since $2aS$ is positive-semidefinite, symmetric, we have that $I+2aS$ is positive-definite, symmetric. It is, in particular, invertible and its inverse $P = (I+2aS)^{-1}$ is also positive-definite, symmetric. It follows that $P$ has a positive-definite, symmetric square root $P^{1/2}$.

It is an easy exercise to check that, whenever $X$ and $Y$ are matrices one of which is invertible, $XY$ is similar to $YX$. Thus we have that $N = PM$ is similar to $P^{1/2} M P^{1/2}$.

The symmetric part of $P^{1/2} M P^{1/2}$ is equal to $P^{1/2} S P^{1/2}$, which is positive-semidefinite symmetric, and so $P^{1/2} M P^{1/2}$ has nonnegative eigenvalues, which therefore implies the same of the similar matrix $N$.

I don't see any reason for $N$ to have strictly positive eigenvalues, but perhaps I missed something.


Added: To see that $N$ need not have strictly positive eigenvalues, just consider any example with $M$ not invertible. For example, taking $M=0$, we get $N=0$ also.

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    $\begingroup$ Since $M$ is assumed to be invertible, non-negative eigenvalues implies strictly positive eigenvalues. $\endgroup$
    – Jason
    Commented Aug 11, 2020 at 0:47
  • $\begingroup$ @Jason: Thanks, I missed the invertibility assumption on $M$. With your correction, everything works out fine. $\endgroup$
    – Mike F
    Commented Aug 11, 2020 at 1:51
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    $\begingroup$ Hi Mike, thanks for your answer. I think you made a mistake when writing $N = (I+2aS)^{-1}M$, it should be $(I+2aA)^{-1}M$. $\endgroup$
    – smalldog
    Commented Aug 11, 2020 at 7:04
  • $\begingroup$ @chaos: Good catch, I'll leave this up anyway and flag the error. $\endgroup$
    – Mike F
    Commented Aug 11, 2020 at 13:01

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