Find matrix $M\in\mathbb{C}^n\times\mathbb{C}^n$ that minimizes $||v^*(M-M^2)v||$ for some $v\in\mathbb{C}^n$

Question

Let $v\in\mathbb{C}^n$, want to find two symmetric matrices $M,N\in\mathbb{C}^n\times\mathbb{C}^n$ such that

1. $M+N=I$
2. $<Mv, Nv>$ is minimized

This boils down to finding matrix $M\in\mathbb{C}^n\times\mathbb{C}^n$ that minimizes $||v^*(M-M^2)v||$ for some $v\in\mathbb{C}^n$

There is a series of simple solutions, playing on the diagonal of $I$, such as

$M=diag([0,1,0,1...]), N=diag([1,0,1,0,...])$

What if we exclude this solution from the problem? What can we say about it then? Is this a known problem? If so, what are the references? If not, any suggestions as to which direction I should look into?

Answer 1

There are a lot of issues in this original posting.

OP wrote "This boils down to finding matrix $M\in\mathbb{C}^n\times\mathbb{C}^n$ that minimizes $||v^*(M-M^2)v||$ for some $v\in\mathbb{C}^n$" but the RHS is positive definite (in particular bounded below by zero). Suppose I select $M=2I$, then $N=-I$, then, for any $v \neq \mathbf 0$

$\langle Mv, N v\rangle= \langle 2Iv, -I v\rangle = -2 \langle v, v\rangle = -2\big \Vert v\big \Vert_2^2 \lt 0 \leq \big \Vert v^*(M-M^2)v\big \Vert_2^2$

which is a contradiction. It also suggests the problem doesn't have a minimum. The problem only requires N and M be "symmetric" though I infer Hermitian was meant.

If the problem is changed to require that $M$ and $N$ are Hermitian Positive semi-definite, then we only need $v$ to be in the nullspace of $M\cdot (I-M)$ i.e. one of those matrices has it with eigenvalue 1 and the other eigenvalue 0. That's it for any fixed $v \neq \mathbf 0$. There is no requirement that $M$ be a projector, though $M(I-M)=\mathbf 0$ is the minimal polynomial for projections (outside of the special cases of $M$ being the identity matrix or zero matrix).