Consider the real matrices $V, W, X, Y$. Define the function $L(W, V) := \frac 1 2 \| Y - V W X \|_2^2$.
How can it be shown that $L$ non-convex, as claimed here?
Tnx.
I have tried to look first at the one-dimensional case. I calculated $L$'s Hessian but wasn't sure how to prove that it's not PSD. Also, I plotted $L$ to gain some intuition (see below). In the contour plot, it's easy to see that the function is non-convex.
To see that the one dimensional case is non-convex, let $L(w, v) = \frac{1}{2}(y - vwx)^2$. Even though it is possible to make the OP's approach with computing the Hessian work, there is a more convenient characterization of convexity for this problem.
Specifically, $L$ is convex if and only if it is convex ``along every line'' in its domain. More formally, $L$ is convex if and only if for any choice of vectors $(w_{0}, v_{0}), (w_{1}, v_{1}) \in \mathbb{R}^{2}$, the one dimensional function $g : \mathbb{R} \to \mathbb{R}$ given by $g(t) = L((w_{0},v_{0}) + t(w_{1}, v_{1}))$ is convex.
Now to establish that $L$ is non-convex, let $(w_{0}, v_{0}) = (0,0)$ and $(w_{1}, v_{1}) = (u,1)$, where $u \in \mathbb{R}$ is a free parameter. In light of the above discussion, let $g_{u}(t) = L(ut, t) = \frac{1}{2} (y - t^2 u x)^2$. The second derivative of $g_{u}$ equals $$ g_{u}''(t) = \frac{(6t^2ux - y)^2 - y^2}{6t^2} $$ If $x = 0$ then $L$ is a constant function, hence convex. If $y = 0$, then $L(w,v) = \frac{1}{2}x^2w^2v^2$, is readily seen to be non-convex by computing the Hessian at $w=v$.
Now suppose that $x \neq 0$ and $y \neq 0$. It remains a simple exercise to find appropriate values of $u, t \in \mathbb{R}$ so that $g_{u}''(t) < 0$, proving that $L$ is indeed non-convex.