A matrix $M$ acting on an inner product space $V = \mathbb{C}^n$ is called positive-semidefinite if, for every $v \in V$, $\langle Mv,v \rangle \ge 0$.
Consider the inner product space $V \otimes V$ with $\langle v_1 \otimes v'_1 , v_2 \otimes v'_2 \rangle = \langle v_1 , v_2 \rangle\langle v_1' , v'_2 \rangle$.
Let $N$ be a matrix acting on $V \otimes V$ such that $\langle N w , w \rangle \ge 0$, for all $w=v_1 \otimes v'_1$ with $v_1,v_1' \in V$.
Question: Is it true that $N$ is positive-semidefinite? If not, can you provide a counter-example?
Note that an element of $V \otimes V$ is of the form $\sum_i v_i \otimes v'_i$, not just $v_1 \otimes v'_1$ (as $w$ above) so the question in not empty.
Best Answer
The answer is no. These concerns often arise in the context of the study of positive maps (linear maps over matrices that preserve positive semidefiniteness) and the study of "entangled ensembles" and "entanglement witnesses" in quantum information theory.
Within these contexts, a matrix $N$ satisfying this weaker condition is sometimes called a "block-positive" matrix. It turns out that $N$ satisfies this condition if and only if it is the Choi matrix of a positive map, whereas positive definite matrices are specifically Choi matrices of completely positive maps.
As a counterexample, consider the matrix $$ N = \left[ \begin{array}{cc|cc} 1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\ \hline 0&1&0&0\\0&0&0&1 \end{array} \right]. $$ This matrix is not positive definite (in fact, it has a negative determinant). However, for $x=(x_1,x_2),y=(y_1,y_2) \in \Bbb C^2$, we have $$ \langle N (x \otimes y), x \otimes y\rangle = |x_1y_1|^2 + |x_2y_2|^2 + 2\operatorname{Re}(x_1\overline{x_2 y_1}y_2) = |\langle x,y\rangle|^2 \geq 0. $$ Correspondingly, $N$ is the Choi representation of the positive map $X \mapsto X^T$, which fails to be completely positive.