It is well-known that
$$\mathrm{Diag}(x)- xx^\top \succeq 0 \iff x_i \geq 0,\ i=1,\ldots,n,\ \sum_{i=1}^nx_i \leq 1, $$
where $x \in \mathbb{R}^n$, and $\mathrm{Diag}(x)$ is the diagonal matrix whose $i$-th diagonal element is $x_i$.
Let us have another variable $y \in \mathbb{R}^n$ such that $x_i \geq y_i \geq 0$ for all $i=1,\ldots,n$. Obviously, if $y=x$, then we have $\mathrm{Diag}(y)- xx^\top \succeq 0$ by the equivalence above. My question is, can we say that for sure $\mathrm{Diag}(y) – xx^\top \not \succeq 0 $ if $y \neq x$?
As an attempt, I tried to see what happens when $n=2$. $\mathrm{Diag}(y) – xx^\top$ becomes:
$$\begin{pmatrix} y_1 – x_1^2 & – x_1x_2 \\ -x_1x_2 & y_2 – x_2^2 \end{pmatrix}$$
so I should show that if $y_1 < x_1$ or $y_2 < x_2$, then assuming diagonal is non-negative, I should show:
$$(y_1 – x_1^2)(y_2 – x_2^2) < (x_1x_2)^2.$$
I tried many ways, and I am unable to prove this.
Best Answer
This is not true. Suppose $x>0$ and $\sum_ix_i<1$. Then $\operatorname{diag}(x)-xx^T>0$ and hence $\operatorname{diag}(y)-xx^T>0$ when $y$ is slightly smaller than $x$.