When is $v^T A w > 0$?

Question

Consider two vectors $v,w \in \mathbb{R}^n$ and a matrix $A \in \mathbb{R}^{n \times n}$. What conditions need to be imposed on the matrix $A$ such that $v^T A w > 0$?

I understand that if $A$ is positive-definite then $v^T A v > 0, \ \forall$ non-zero $v$. However not having the same vector on both the sides may not result in a positive answer. (I checked with some examples).

In search of this answer, I have been going in circles between positive definiteness of a matrix and that of a bilinear form. Any help is deeply appreciated!

Answer 1

There is no matrix $A$ for which $v^TAw > 0$ holds for all choices of $v,w$. In particular, for any choice of $w$, the vector $v = -Aw$ is such that $v^TAw \leq 0$.

Answer 2

Assuming you're talking about a fixed pair of $(v,w)$, if $A$ is positive definite, and if $v,w$ can be written as a positive linear combination of eigenvectors of $A$ (e.g. $v=a_1u_1+\cdots$, $a_i>0$), then the assertion will hold. More generally, you'd need $\sum_ia_i\lambda_i$ to have the same sign, for each of $u,v$.

Answer 3

$$\mathrm v^\top \mathrm A \, \mathrm w = \frac12 \mathrm v^\top \mathrm A \, \mathrm w + \frac12 \mathrm w^\top \mathrm A^\top \, \mathrm v = \frac12 \begin{bmatrix} \mathrm v\\ \mathrm w\end{bmatrix}^\top \underbrace{\begin{bmatrix} \mathrm O & \mathrm A\\ \mathrm A^\top & \mathrm O\end{bmatrix}}_{=: \mathrm B} \begin{bmatrix} \mathrm v\\ \mathrm w\end{bmatrix}$$

Hence, the characteristic polynomial of (symmetric) block matrix $\mathrm B$ is

$$\det \begin{bmatrix} s \,\mathrm I & -\mathrm A\\ -\mathrm A^\top & s \,\mathrm I\end{bmatrix} = \det \left( s^2 \mathrm I - \mathrm A \mathrm A^\top \right)$$

and, thus, the eigenvalues of (symmetric) block matrix $\rm B$ are of the form $\color{blue}{\pm \sigma (\rm A)}$, where $\sigma (\rm A) \geq 0$ is a singular value of matrix $\rm A$. Therefore, $\mathrm v^\top \mathrm A \, \mathrm w$ is always an indefinite quadratic form.