How to verify if a linear system of symmetrical matrix blocks has solution? I have the matrix:

  • $\left[M\right]_{p \times p}$, symmetrical
  • $\left[G\right]_{p \times q}$

and then, I would like to solve the following linear system:

$$ \underbrace{\begin{bmatrix} \left[M\right] & \left[G\right] \\ \left[G^T\right] & \left[0\right] \end{bmatrix}}_{\left[A\right]} \cdot \underbrace{ \begin{bmatrix} \left[\mu\right] \\ \left[\lambda \right] \end{bmatrix}}_{\left[X\right]} = \underbrace{ \begin{bmatrix} \left[F_{\mu}\right] \\ \left[F_{\lambda}\right] \end{bmatrix}}_{\left[B\right]} $$

So far, I found this article Solve linear system with bordered positive definite matrix that explains how to solve this problem, and in some steps we have to calculate $M^{-1}$ and $H^{-1}$, where

$$H = G^T \cdot M^{-1} \cdot G$$

So, the restrictions to exist a solution are:

  • $M^{-1}$ exists $\Leftrightarrow \det{M} \ne 0$
  • $H^{-1}$ exists $\Leftrightarrow \det H \ne 0$

So, I would like to know if there's a easier way to known if the system $AX = B$ has a solution than:

  • Calculate $\det M$, if it's $= 0$, stop because the system doesn't have solution
  • Calculate $M^{-1}$
  • Calculate $H = G^T M^{-1} G$
  • Calculate $\det H$, and if it's $=0$, the system doesn't have solution.

For my specific problem, a computational problem, I know that besides $M$ being symmetric, it's also positive-definite. So, I formulated the two hypothesis to get a easier conclusion, but I don't know how to prove it and even if it's true: Suppose that $M$ is inversible, symmetric and positive-definite, if $\det M \ne 0$, so:

  • $\det H = 0 \Leftrightarrow \det G^T \cdot G = 0 $
  • $\det H = 0 \Leftrightarrow \det G \cdot G^T = 0 $

And the last question: does always $\det \left(G^T \cdot G\right) = \det \left(G \cdot G^T\right)$?

  • 1
    $\begingroup$ If $p \neq q$, then one of $G^T G$ and $G G^T$ has a kernel. OTOH, by looking at the projection operators, you see that it is possible to arrange the other to be the identity. And hence the two determinants are not equal. $\endgroup$ – Willie Wong Feb 5 at 21:28

You find various conditions in Section 3 of the classical review paper by Benzi-Golub-Liesen on this kind of problems, which are known as saddle-point problems.


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