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

  • 1
    $\begingroup$ think first as an equation on $Y = EV$ $\endgroup$ – Daniel Aug 13 at 15:07

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.

  • $\begingroup$ 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? $\endgroup$ – mrkwjc Aug 13 at 16:05
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Aug 13 at 16:08
  • $\begingroup$ See my latest edit. $\endgroup$ – Omnomnomnom Aug 13 at 16:32
  • $\begingroup$ 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? $\endgroup$ – mrkwjc Aug 13 at 16:48
  • $\begingroup$ 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. $\endgroup$ – 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?


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