I'm working on the convex relaxation of a problem, and I came across the following question.
Suppose I have a vector $x \in \mathbb{R}^n$ where $-1 \leq x_i \leq 1$ and a matrix $X$ whose diagonal elements are $1$. How can I prove the following inequality:
$$X \succeq x x^T$$
I've checked it's true for $n=1$, but even for $n=2$ I'm not sure how to show it. Thank you.
This seems like it might not be true. Consider $X = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Doesn't $X - x x^T$ have a negative eigenvalue? What is the matrix notion of $\geq$ here?