Linear Matrix Inequality of a matrix $T$ with $\| T \|_2 < \rho$

Question

Assume $T \in \mathbb{R}^{d\times d}$. Let $\rho(T)$ be the spectral radius of $T$ and $\rho \ge 0$. Prove that

$\rho(T) < \rho$ if and only if there exists a $P \succ 0$ satisfying $T^TPT - \rho^2 P \prec 0$.

I can prove the Necessity: If $P = I$, for any $x \in \mathbb{R}^d$,

\begin{equation} x^T T^T P T x - \rho^2 x^T P x = \| Tx \|^2 - \rho^2 \| x\|^2 < 0 \end{equation} so $T^T PT - \rho^2 P \prec 0$

I get stuck in proving the Sufficiency.

Answer 1

Let $\lambda \in \text{sp}(T)$ and $x$ be an associated eigenvector. We have:

$$x^T T^T P T x - \rho^2 x^T P x < 0 \implies (Tx)^T P (Tx) < \rho^2 x^T P x$$

and

$$(Tx)^T P (Tx) = (\lambda x)^T P (\lambda x ) = \lambda ^2 x^T P x$$

As $x^T P x>0$, we get $\lambda ^2 < \rho^2$, i.e. $|\lambda| < \rho$.

Which shows $\rho(T) < \rho$

Answer 2

Sufficiency:

Suppose that there exists such a $P$. Then we have $$ \rho^2 P - T^TPT \succ 0 $$ Since $P$ is positive definite, we can find a positive definite $X = P^{-1/2}$. Conjugating both sides yields $$ \rho^2 X^TPX - X^TT^TPTX \succ 0 \implies\\ \rho^2 I - (X^{-1}TX)^T(X^{-1}TX) \succ 0 $$ So, we conclude that $$ \rho(T) = \rho(X^{-1}TX) \leq \|X^{-1}TX\| < \rho $$ as desired.