Is $v v^T \succeq 0$?

Question

Suppose $v \in \mathbb{R}^m$ is an $m$-dimensional vector. Then is it true that $v v^T$ is positive semi-definite?

Answer 1

Yes, this is true. To verify that this is the case, note that for any complex vector $x$: $$ x^*(vv^*)x = (x^* v)(v^* x) = |\langle x,v \rangle|^2 \geq 0 $$ Or, restrict $x$ to real vectors and note that $vv^T$ is symmetric.

Answer 2

Let $z$ be a vector. Then $z^Tvv^T z = (zv^T)^T(zv^T) = \langle z, v \rangle^2$. This is just a square of a real number. What do you know about squares of real numbers?

Answer 3

Yes; it's true since

$$\left\langle vv^Tx,x\right\rangle=\left\langle v^Tx,v^Tx\right\rangle=\left|\left|v^T x\right|\right|^2\ge0 .$$

Answer 4

If $\mathrm v \in \mathbb R^n \setminus \{0_n\}$, then $\mathrm v \mathrm v^T$ is a symmetric, rank-$1$ matrix with a single nonzero eigenvalue

$$\lambda = \mbox{tr} (\mathrm v \mathrm v^T) = \mbox{tr} (\mathrm v^T \mathrm v) = \|\mathrm v\|_2^2 > 0$$

Thus, $\mathrm v \mathrm v^T$ is positive semidefinite.