Non-positivity of a matrix Let $A\in M_{n\times n}(\Bbb R)$ be stable (all
real parts of the eigenvalues of $A$ are negative). $a,b\in\Bbb R^n, a>0.\ C$ is a symmetric matrix given by
$AC+CA^T=-ab^T$
Is it possible to show that if $b\ngeq0$ (has at least one negative
element) then $C$ is not positive semi-definite?
 A: Update. The OP has changed their question and the following answer no longer applies.

We shall prove a stronger result:

Lemma. If $AC+CA^T=P$ for some real stable matrix $A$, real symmetric matrix $C$ and some real nonzero and positive semidefinite matrix $P$, then $C$ is not positive semidefinite.

Proof. For any $x\in\ker(C)$, we have $x^\ast Px=x^\ast(AC+CA^T)x=0$ and in turn $Px=0$. Hence $\ker(C)\subseteq\ker(P)$ and $\operatorname{range}(P)\subseteq\operatorname{range}(C)$. Also, as $CA^Tx=(AC+CA^T)x=Px=0$, we see that $A^Tx\in\ker(C)$, i.e. $\ker(C)$ is an invariant subspace of $A^T$.
So, by a change of orthonormal basis (over $\mathbb R$), we may assume that
$$
C=\pmatrix{D\\ &0},\ A^T=\pmatrix{B^T&0\\ \ast&\ast},\ P=\pmatrix{S\\ &0}
$$
where $D$ is a nonsingular symmetric matrix, $B$ is stable and $S$ is positive semidefinite. Note that $D$ is non-empty, otherwise $C$ will be zero, but that is impossible because $P$ is nonzero. The equality $AC+CA^T=P$ now reduces to $BD+DB^T=S$. Let $(\lambda,x)$ be any eigenpair of $B^T$ over $\mathbb C$. Then
$$
2\Re(\lambda)x^\ast Dx=x^\ast(BD+DB^T)x=x^\ast Sx\ge0
$$
and hence $x^\ast Dx\le0$. Therefore $D$ is not positive definite and $C$ is not positive semidefinite.$\ \square$

Now, return to your question. Since $AC+CA^T$ is symmetric, so is $-ab^T$. As $b$ has a negative element, it must be a negative multiple of $a$. Hence $-ab^T$ is a rank-$1$ positive semidefinite matrix. So, by the lemma above, $C$ is not positive semidefinite.
