# How to solve $M - EV - (EV)^T = 0$ against $E$?

I have a matrix equation: $$M - EV - (EV)^T = 0$$ which i want to solve against $$E$$. Is this possible? How to do that?

Remarks: matrices are square, $$E$$, $$M$$ are symmetric, $$V$$ is invertibile.

Regards, Marek

• think first as an equation on $Y = EV$ – Daniel Aug 13 at 15:07

## 2 Answers

We can write this equivalently as $$EV + (EV)^T = M.$$ Because the (linear) operator $$E \mapsto EV + (EV)^T$$ is not invertible, this equation will have infinitely many solutions. In particular, we can always take $$E = \frac 12 MV^{-1}$$. Indeed, plugging this $$E$$ in yields $$EV + (EV)^T = \frac 12 MV^{-1}V + (\frac 12 MV^{-1}V)^T = \frac 12 M + \frac 12 M^T = M$$ As it turns out, every solution to this equation can be written in the form $$E = \frac 12 (M + S)V^{-1}$$ where $$S$$ is skew symmetric, which is to say that $$S$$ satisfies $$S^T = -S$$.

Now, requiring that $$E$$ is symmetric amounts to requiring that $$[(M + S)V^{-1}]^T = (M + S)V^{-1} \implies\\ V^{-T}(M - S) = (M + S)V^{-1} \implies\\ V^{-T}S + SV^{-1} = V^{-T}M - MV^{-1}.$$ In other words: there exists a symmetric solution $$E$$ if and only if the Sylvester equation (more specifically Lyapunov equation) $$V^{-T}S + SV^{-1} = V^{-T}M - MV^{-1}$$ has a solution $$S$$ in which $$S$$ is also skew-symmetric. I don't see a way of simplifying this condition.

It is notable that if the eigenvalues of $$V$$ have all positive real part or negative real part, then there exists a unique solution $$S$$. If the eigenvalues of $$V$$ have only negative real part, then this unique $$S$$ can be written explicitly as the integral $$S = \int_0^\infty e^{V^{-T}\tau}[V^{-T}M - MV^{-1}]e^{V^{-1}\tau} d\tau.$$ This $$S$$ is necessarily skew-symmetric, which is to say that our original equation has a (unique) symmetric solution $$E$$ in this case.

Another approach: if we're looking specifically for a symmetric solution $$E$$, then we can rewrite the original equation as $$EV + V^TE^T = M \implies V^TE + EV = M.$$ Now, if the eigenvalues of $$V$$ all have negative real parts, then this equation has a unique solution that can be written as $$E = \int_0^\infty e^{V^{T}\tau}\,M\,e^{V\tau} d\tau,$$ which is necessarily a symmetric matrix.

• Ok, but it should exist only one $E$ which is symmetric, isn't it? It seems the question transforms to: how to find $S$ for which $E$ is symmetric? – mrkwjc Aug 13 at 16:05
• I missed that $E$ is supposed to be symmetric. I don't think that we can necessarily guarantee that a symmetric solution $E$ exists, but I'll think about it and add something. – Omnomnomnom Aug 13 at 16:08
• See my latest edit. – Omnomnomnom Aug 13 at 16:32
• Thanks man, I'm enlightened now :) But... i think that the original equation is also a Lyapunov one and maybe we can write such integral directly for E? – mrkwjc Aug 13 at 16:48
• Technically it isn't because the second $E$ is transposed. However, since we're looking for a symmetric solution we can make it one. I'll add something. – Omnomnomnom Aug 13 at 17:16

In one dimensional case $$M - EV - (EV)^T = M - 2EV = 0 \implies E = \frac{M}{2V}.$$

So, maybe you would like to check if $$E = \frac{1}{2} MV^{-1}$$ in general case?