# Eigenvalues with positive real part for a matrix product

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?

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 $$\Re(\lambda)=\frac{(s+2ah^2)}{|1+2iah|^2}>0$$.

• +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. Commented Aug 11, 2020 at 4:28
• 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 $u$? Commented Aug 11, 2020 at 7:09
• @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 $u$. Commented Aug 11, 2020 at 7:20
• Thanks, perfect answer :) Commented Aug 11, 2020 at 7:27

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.

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