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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.

Rodrigo de Azevedo
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Todd Chavez
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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?

leslie townes
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  • I have just realized that maybe the other direction is true. In other words, if $X \geq x x^T$, with $X_{ii}=1$, then the components of $x$ have absolute value less than or equal to 1. What do you think of this case? – Todd Chavez Feb 17 '21 at 07:04
  • Yes. This comes from the fact that the diagonal entries of a positive semidefinite matrix are nonnegative (e.g. because they are inner/dot products $\langle A e_i, e_i \rangle$ with $A$ positive semidefinite and $e_i$ the $i$th standard basis vector), and because a short calculation shows that the $i$th diagonal entry of the assumed-to-be-positive-semidefinite matrix $X - xx^T$ is $X_{ii} - x_i^2$. Because this diagonal entry is $\geq 0$ by assumed positive semidefiniteness of $X - xx^T$, we deduce that $x_i^2 \leq X_{ii}$ for all $i$. Applying absolute values and $X_{ii} = 1$ you've got it. – leslie townes Feb 17 '21 at 15:30
  • Great point! Thank you – Todd Chavez Feb 17 '21 at 19:04