Does $M^2$ have an LU decomposition?

Question

It's well known that matrices of the form $M^T M$ have an (unpivoted, of course) LU decomposition. In fact, because they are positive semidefinite, they have a Cholesky decomposition. The field here is the real numbers.

My question is: Do matrices of the form $M^2$ have an LU decomposition? I suspect the answer is no, but the usual counterexamples I know don't work.

[edit]

Even though the answer is easy, I'll keep this question undeleted in case someone else Googles this very same question.

Answer 1

They do not. In particular, if $M^2$ is a 90-degree rotation about the origin of $\mathbb R^2$, it is indeed the square of a 45-degree rotation, but some of its leading principal minors are zero. Therefore, this $M^2$ does not have an LU decomposition.