What is special about $x^T P y$?

Question

We know that if a matrix $P$ is positive definite, then for any vector $x \ne 0$, $x^T P x > 0$. But what about $x^T P y$ where $x \ne y$?

Answer 1

$$\rm x^{\top} P \, y = \frac 12 \begin{bmatrix} \mathrm x\\ \mathrm y\end{bmatrix}^{\top} \begin{bmatrix} \mathrm O & \mathrm P\\ \,\mathrm P^{\top} & \,\mathrm O\end{bmatrix} \begin{bmatrix} \mathrm x\\ \mathrm y\end{bmatrix}$$

The characteristic polynomial of the block matrix above is

$$\det \begin{bmatrix} s \mathrm I & -\mathrm P\\ -\mathrm P^{\top} & s\mathrm I\end{bmatrix} = \det( s^2 \mathrm I - \mathrm P \mathrm P^{\top} )$$

where $\mathrm P \mathrm P^{\top}$ is positive semidefinite. The spectrum of the block matrix contains the positive and negative square roots of the (nonnegative) eigenvalues of $\mathrm P \mathrm P^{\top}$. Hence, the block matrix

$$\begin{bmatrix} \mathrm O & \mathrm P\\ \,\mathrm P^{\top} & \,\mathrm O\end{bmatrix}$$

is indefinite, and so is the bilinear form $\rm x^{\top} P \, y$.