LMI for maximum singular value

Question

The (maximum singular value) matrix norm constraint $\|A(x)\|<1$, where $A(x) \in \mathbb R^{p \times q}$ depends affinely on $x$, is represented as the following linear matrix inequality (LMI)

$$\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} >0$$

since $\|A(x)\|<1$ is equivalent to $I-AA^T > 0$. Why?

I would like to understand this statement of LMI. Thank you for your help and time.

Answer 1

My previous answer is wrong. Here is the correct one.

$$\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix}= \begin{bmatrix} I & 0 \\ A(x)^T & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & I-A(x)^TA(x) \\ \end{bmatrix} \begin{bmatrix} I & A(X) \\ 0 & I \\ \end{bmatrix} $$

Thus $\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} \gt 0$ is equivalent to $\begin{bmatrix}I-A(x)^TA(x)\end{bmatrix} \gt 0$ and @Ian in his comment proved the second part.

Answer 2

because $\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} $ is positive definite. For any non-empty vector,

$$\begin{bmatrix} z_1 &z_2 \end{bmatrix} \begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$$

$$= ||z_1||^2+||z_2||^2+2||z_1^TA^Tz_2||$$ $$= ||z_1||^2+||z_2||^2+2||z_1||\cdot||A^TA||\cdot||z_2|| \gt 0$$