# Matrix non-singular proof

I have one question of how to derive the nonsingularity of one matrix. Here's the matrix I'm interested in:

\begin{align} A = I + SHFG, \end{align} where $$A \in \mathcal{R}^{m \times m}$$, $$I\in \mathcal{R}^{m \times m}$$ is an identity matrix, $$S \in \mathcal{R}^{m \times n}$$, $$H \in \mathcal{R}^{n \times r}$$, $$F \in \mathcal{R}^{r \times r}$$, and $$G \in \mathcal{R}^{r \times m}$$ are appripriate constant matrices.

I have seen the proof process from one journal that the matrix $$A$$ is a non-singular matrix if the following inequality is satisfied: \begin{align} S H F G G^T F^T H^T S^T < I. \end{align} Why nonsingularity of matrix $$A$$ is guaranteed if the above inequality is satisfied?

I have tried to understand the proof process by using my knowledge... however, I don't know how to get the conclusion....

Write $$A=I+X$$ where $$X=SHFG$$. The given condition means that $$I-X^TX$$ is positive definite. Now suppose $$Av=0$$. Then $$(I+X)v=0, Xv=-v, v^TX^TXv=v^Tv$$ and hence $$v^T(I-X^TX)v=0$$. Since $$I-X^TX$$ is positive definite, $$v$$ must be zero. Hence $$A$$ is nonsingular.