# Outer products & positive (semi-) definiteness

1. Oct 30, 2015

### economicsnerd

Let $u,v$ be vectors in the same Euclidean space, and define the symmetric matrix $M = uv'+vu'$, the sum of their two outer products.

I'm interested in whether or not $M$ is positive (semi)definite.

Does anybody know of any equivalent conditions that I might phrase "directly" in terms of the vectors $u,v$?

2. Oct 30, 2015

### andrewkirk

An equivalent condition is $(x\cdot u)(x\cdot v)\geq 0$ for all vectors $x$.

3. Oct 30, 2015

### fzero

Let $z$ be an arbitrary vector, then we are interested in the properties of
$$z^\dagger M z = \langle z, u \rangle \langle v, z \rangle + \langle z, v \rangle \langle u, z \rangle .$$
Assume that $v\neq u$, then we can write
$$z = a u + b v + z_\perp,~~~\langle z_\perp, u \rangle = \langle z_\perp, v \rangle =0.$$
Then, with $A = (a~~b)^T$, we have
$$z^\dagger M z = A^\dagger Q A,$$
where $Q$ is a Hermitian 2x2 form determined in terms of the inner products of $u$ and $v$. Positivity properties of $M$ are reduced to those of $Q$ for which it is simple to compute the eigenvalues.