Is $x^TAy$ convex or concave, for $x,y$ not identically equivalent?

Question

It is well known if we had something like:

$f(x) = x^TQx$

A quadratic form, is positive semidefinite of $Q$ is positive semidefinite

How is the structure of

$f(x,y) = x^TQy$

analyzed? i.e. what conditions is required for this function to be convex or concave?

Answer 1

Note that $$x^\top Q \, y = \begin{pmatrix}x\\y \end{pmatrix}^\top \begin{pmatrix}0 & \frac12 \, Q \\ \frac12 \, Q^\top & 0 \end{pmatrix}\begin{pmatrix}x\\y \end{pmatrix}.$$ Hence, your function $f$ of $(x,y)$ is convex iff $$\begin{pmatrix}0 & \frac12 \, Q \\ \frac12 \, Q^\top & 0 \end{pmatrix}$$ is positive semi-definite. This is the case iff $Q = 0$. Otherwise, take $(x,y)$ with $x^\top Q y < 0$, which yields $$x^\top Q \, y = \begin{pmatrix}x\\y \end{pmatrix}^\top \begin{pmatrix}0 & \frac12 \, Q \\ \frac12 \, Q^\top & 0 \end{pmatrix}\begin{pmatrix}x\\y \end{pmatrix} < 0.$$

Hence, your function is convex iff $Q = 0$.