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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....

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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.

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  • $\begingroup$ Thank you for your clear intuition and wonderful answer......... You save my life!! Thank you!! $\endgroup$ – Hwang May 30 '20 at 9:54
  • $\begingroup$ @Hwang if his answer saved your life, then kindly accept his answer. $\endgroup$ – user550103 May 31 '20 at 6:12
  • $\begingroup$ @user550103 How can I accept his answer? I don't know how to do that..... Is there any button in the page? $\endgroup$ – Hwang Jun 2 '20 at 4:24
  • $\begingroup$ @user1551 I have one further question. How can I guarantee I-X^T X is positive definite? $\endgroup$ – Hwang Jun 2 '20 at 8:17
  • $\begingroup$ @Hwang This is a given condition in your question. $\endgroup$ – user1551 Jun 2 '20 at 8:25

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