Constructing a symmetric, positive semi-definite with a given nullspace

Question

Suppose I'm given a vector $\vec x$. Is there a straightforward way to construct a symmetric postive semidefinite matrix A such that $\vec x$ is its nullspace?

That is, given $\vec x$ I need to create an $A$ such that $A\vec x=0$.

If $A$ doesn't have constraints its not difficult to construct such an $A$: namely, take all the rows of $A$ to be orthogonal to $\vec x$. But satisfying the symmetric, positive semidefinite condition makes this more challenging.

Answer 1

You can always take $A = 0$. For a non-trivial construction, assuming $n > 1$, you can choose some vector $\vec{y}$ which is orthogonal to $\vec{x}$ and take $A = \vec{y} \cdot \vec{y}^T$. This is a symmetric rank-one matrix and $$ A\vec{x} = \vec{y} \cdot \vec{y}^T \cdot \vec{x} = \left< \vec{y}, \vec{x} \right> \vec{y} = 0 \cdot \vec{y} = \vec{0}. $$

Answer 2

Normalise $x$ first. Then you can take $A=(I-xx^T)P(I-xx^T)$ where $P$ is any positive semidefinite matrix. Every positive semidefinite matrix $A$ with $x\in\ker(A)$ can be written in this form. In particular, if you take $P=I$, then $A=I-xx^T$ is the orthogonal projection onto $\{x\}^\perp$.