$xx \in \mathbb{R}^{n \times n}$ is positive semi-definite for all $x \in \mathbb{R}^n$

Question

Prove or disprove:

For all $x \in \mathbb{R}^n$ is $xx \in \mathbb{R}^{n \times n}$ positive semi-definite.

As I did not manage to find a counter example yet, I'd say this is true. Note that the task description does not differ between $x$ and $x^\top$, vectors are flipped appropriately such that they fit into the context. My definition of Matrix $A$ being semi-definite is that $A$ is symmetric and $\forall x \in \mathbb{R}^n \setminus \{0\} : xAx \ge 0$.

Any advice is greatly appreciated.

Answer 1

Note that $v^T(xx^T)v = \|x^Tv\|^2\geq 0$, thus $xx^T$ is positive semi-definite.

Besides, the eigenspace associated to $0$ has dimension $n-1$ if $x\neq 0$.

Answer 2

I will use $x^\intercal$ for the sake of clarity.

Let $x=(a_i)_{i=1}^n$. If we define $A=x x^\intercal$, the element $a_{i,j}$ of this matrix is $a_ia_j$ so $A$ is symmetric.

Now $y^\intercal A y = (y^\intercal x)(x^\intercal y) = (y \cdot x)^2 \geq 0$ so $A$ is positive semi-definite.