Conditions to solve linear system with matrix blocks

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 $$\det \left(G^T \cdot G\right) = \det \left(G \cdot G^T\right)$$?

• 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. – Willie Wong Feb 5 at 21:28