Is this proof that $BA \prec 0$ and $B \succ 0$ implies $A \prec 0$ correct?

Question

Consider the $n \times n$ positive-definite matrix $B$ and the $n \times n$ matrix $A$. I want to check if $BA$ being negative-definite and $B$ being positive definite implies that $A$ is negative definite. To do this, I'll try to show that all the eigenvalues of $A$ are strictly negative.

Each eigenvalue $\lambda$ of $A$ satisfies the equation $$ Av = \lambda v \tag{1} $$ for some eigenvector $v$ of $A$. Then, pre-multiplying both sides of $(1)$ by $v^T B$ yields $$ v^T B A v = \lambda v^T B v \tag{2} $$ Because $B$ is positive definite, then $v^T B v > 0$ for every $v$. Therefore, we can divide both sides of $(2)$ by $v^T B v$ to get $$ \lambda = \frac{v^T B A v}{v^T B v} \tag{3} $$ Because the numerator and denominator of $(3)$ are respectively negative and positive, then $\lambda < 0$, and so $A$ is negative-definite. Is this proof correct?

Answer 1

The matrix $$X=\begin{pmatrix}1& x\\ 0&1\end{pmatrix}$$ is positive definite iff $|x|<2.$ Indeed if $|x|<2$ then for $v\neq 0$ we get $$v^TXv=v_1^2+v_2^2+xv_1v_2\\ \ge v_1^2+v_2^2-{|x|\over 2}(v_1^2+v_2^2)>0$$ On the other hand if $|x|\ge 2$ then for $v=(1,\pm 1)^T$ we get $v^TXv=2\pm x$, hence the outcome is negative for $(1,1)^T$ or for $(1,-1)^T.$

Let $$C=\begin{pmatrix}1& c\\ 0&1\end{pmatrix}\quad B=\begin{pmatrix}1& b\\ 0&1\end{pmatrix} $$ Then $$BC=\begin{pmatrix}1& b+c\\ 0&1\end{pmatrix}$$ Thus for $|b|<2,$ $|c|>2,$ $|b+c|<2$ the matrices $B$ and $BC$ are positive definite while $C$ is not positive definite. For example it occurs when $b=-{5\over 3},$ $c={5\over 2},$ as $b+c={5\over 6}.$ Let $A=-C.$ Then $A$ is not negative definite, while $BA$ is negative definite and $B$ is positive definite.

Answer 2

To add to @RyszardSzwarc's answer, it turns out that there are a some subtle problems in my proof. First, $A$ is not guaranteed to be symmetric, and so an eigenvalue $\lambda$ of $A$ could be complex. This would mean that the numerator in $(3)$, which is $v^T BAv$, could become $\lambda \cdot v^T Bv$ (since $Av = \lambda v$ for some $v$), where $\lambda$ is complex. This would mean that the expression in $(3)$ is complex in general, and so the argument that $\lambda$ is negative would no longer hold.

Similarly, it could be the case that $v$ is an eigenvector for both $B$ and $A$, and since $B$ is also not necessarily symmetric, then $v^T B v$ becomes $\gamma \cdot v^T v$, where $\gamma$ is a complex eigenvalue associated with $v$. Again, this would mean that the argument associated with $(3)$ no longer holds.

Finally, as shown in @RyszardSzwarc's answer, if a square matrix is not symmetric, then we can longer say that it is positive-definite if and only if all of its eigenvalues are positive. @RyszardSzwarc gave some examples of matrices with positive eigenvalues but that are not positive-definite. Therefore, even if we do show that all of $A$'s eigenvalues are negative, it does not mean that $A$ is negative-definite.

For more information about this perspective, see this answer.