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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?

Rodrigo de Azevedo
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mhdadk
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  • Equivalently: $BC\ge 0$ and $B\ge 0$ implies $C\ge 0$ by substituting $C=-A.$ The conclusion is not true. – Ryszard Szwarc Dec 25 '23 at 11:29
  • @RyszardSzwarc wouldn't it be "$BC > 0$ and $B > 0$ implies $C > 0$ by substituting $C = -A$" instead? – mhdadk Dec 25 '23 at 11:31
  • The conclusion is true in the complex case. But I doubt it holds in the real case. – Ryszard Szwarc Dec 25 '23 at 11:34
  • @mhdadk Positive definiteness is for quadratic forms, i.e., quadratic polynomials in which every monomial is of degree 2. A quadratic form can be written as $x^\top A x$. However, the skew-symmetric part of $A$ contributes nothing to the quadratic form and, thus, one usually considers only symmetric matrices $A$ – Rodrigo de Azevedo Dec 25 '23 at 12:20
  • @mhdadk I think you may write something like "positive definite real matrix $B$ (which is not necessarily symmetric or diagonalizable)". – River Li Dec 25 '23 at 14:46
  • @RiverLi thanks for the feedback. Does restricting $B$ to be real ensure that "$BA \succ 0$" is a valid statement? Or is this for something else? – mhdadk Dec 25 '23 at 15:34
  • @mhdadk You can also assume $B$ is complex, as long as the problem is clearly addressed. – River Li Dec 25 '23 at 15:39

2 Answers2

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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.

Ryszard Szwarc
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  • Nice counterexample! Does this still work if all matrices are assumed to be symmetric? – mhdadk Dec 25 '23 at 16:23
  • Wait, why is $$X=\begin{pmatrix}1& x\0&1\end{pmatrix}$$ positive definite iff $|x| < 2$? For any vector $z \in \mathbb R^2 \setminus {0}$, the quadratic form associated with $X$ is $z_1^2 + z_2^2 + xz_1z_2$. For any value of $x$, I can choose $z_1$ positive enough and $z_2$ negative enough (or vice versa) for this quadratic form to be negative... – mhdadk Dec 25 '23 at 16:30
  • If all three matrices are symmetric the conclusion holds. – Ryszard Szwarc Dec 25 '23 at 16:32
  • Thanks. Could you clarify my comment regarding your claim for the positive definiteness of $X$ iff $|x| < 2$? – mhdadk Dec 25 '23 at 16:34
  • Nevermind. I made a mistake on my end. – mhdadk Dec 25 '23 at 16:36
  • Thank you the additional clarifications in your answer! – mhdadk Dec 25 '23 at 16:58
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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.

mhdadk
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