bounds of Lyapunov operator Let $A$ be a nonsingular matrix of $\mathbb{R}^{n\times n}$,
and let $ L_AP=A^TPA$.
Let $\beta\in\mathbb{R}$ be such that $L_AP\geqslant \beta P$ holds for all $P>0$. Let $\lambda_i,\sigma_i$ are the $i$-th eienvalue and sigular value of $A$, respectively.
I have two problems in the following:
Prob 1:
Suppose the algebraic multiplicity of $\lambda_i$ is $1,\forall~i$.
I wonder if the upper bound of $\beta$ is $\min_i|\lambda_i|^2$.
Prob 2:
Suppose $A$ has have eienvalue whose algebraic multiplicity is more than $1$.
I wonder if there is a upper bound of $\beta$. If there is, is it $\min_i\sigma_i^2$?
Any help is appreciated!
 A: In all cases, $\min_i |\lambda_i|^2$ is an upper bound to $\beta$ (regardless of the multiplicity of the eigenvalues of $A$).
First, Suppose that $\lambda \in \Bbb R$ is an eigenvalue of $A^T$ and that $v \in \Bbb R^n$ is an associated eigenvector. For $t \in \Bbb R$, let $P_t = vv^T + tI$. Note that $P_t > 0$ for all $t > 0$. We note that
$$
L_A(P_t) - \beta P_t = A^Tvv^TA - \beta vv^T + t(A^TA - \beta I)
\\ = (\lambda^2 - \beta) vv^T + t(A^TA - \beta I).
$$
This matrix must be positive semidefinite for all $t > 0$. Because the limit of positive semidefinite matrices is positive semidefinite, taking the limit of the above as $t \to 0^+$ allows us to conclude that $(\lambda^2 - \beta)vv^T$ is positive semidefinite, which can only occur if $\beta \leq \lambda^2$.
Now, suppose that $\lambda = a + bi$ with $b \neq 0$ is an eigenvalue of $A^T$, and that $v = x + iy$ (with $x,y \in \Bbb R^n$) is an associated eigenvector. Note that
$$
A^T(x + iy) = (a + bi)(x + iy) = (ax - by) + i(bx + ay) \implies\\
A^Tx = ax - by, \quad A^Ty = bx + ay.
$$
Using this fact, verify that
$$
L_A(xx^T + yy^T) = (a^2 + b^2)(xx^T + yy^T) = |\lambda|^2 (xx^T + yy^T).
$$
With that established: for $t > 0$, let $P_t = xx^T + yy^T + tI$. We find that
$$
L_A(P_t) - \beta P_t = (|\lambda|^2 - \beta)(xx^T + yy^T) + t(A^TA - \beta I).
$$
This matrix must be positive semidefinite for all $t > 0$. Because the limit of positive semidefinite matrices is positive semidefinite, taking the limit of the above as $t \to 0^+$ allows us to conclude that $(\lambda^2 - \beta)vv^T$ is positive semidefinite, which can only occur if $\beta \leq |\lambda|^2$.
Thus, we conclude that $\beta \leq |\lambda_i|^2$ for all $i$, which means that $\min_i |\lambda_i|^2$ is indeed an upper bound for $\beta$.
