Q:

How can we prove that the rank of a matrix is a non-convex function of that matrix?

How can we prove that  $\operatorname{rank}(\mathbf{X})= 1$ is a non-convex function of $\mathbf{X}$.

A:

It seems pretty clear that if we take $X = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$ and $Y = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$, then $\operatorname{rank}(tX+(1-t)Y) = 2$ for $t\ne 0,1$.

