Positive semidefinite versus all the eigenvalues having non-negative real parts

Question

1. Suppose matrix $A$ with all its eigenvalues having non-negative real parts, can we get that $x^TAx\geq0$ holds for any vector $x$?

2. Suppose matrix $A$ is positive semidefinite, $B$ is a positive definite diagonal matrix with the same dimension as $A$. Do all the eigenvalues of $AB$ have nonnegative real parts?

Answer 1

In terms of the second question, this is true.

Since $B$ is a diagonal positive definite matrix, $B^{\frac{1}{2}}$ is invertible. Then $B^{\frac{1}{2}}ABB^{-\frac{1}{2}}=B^{\frac{1}{2}}AB^{\frac{1}{2}}=(B^T)^{\frac{1}{2}}AB^{\frac{1}{2}}$, which is positive semi-definite.

Therefore, all the eigenvalues of $AB$ have non-negative real parts.

Answer 2

For your first question, the answer is negative. A counter-examples is as follows.

Consider $A = \begin{bmatrix}1 & -100\\0 & 1\end{bmatrix}$ which has two non-negative real parts, but we have $x^{\mathrm T}Ax = -98$ when $x = \begin{bmatrix}1\\1\end{bmatrix}$.