smallest eigenvalue of rank one matrix minus diagonal Let $x$ be a $d$-dimensional real vector with $\| x\| = 1$. Define $X := xx^T - \mathrm{diag}(xx^T)$. Is it possible to show that $\lambda_{\mathrm{min}}( X ) \geq - 1/2$? Running a bunch of random trials in python seems to suggest this is true, but I'm not sure how to show it. The best I could come up with is $\lambda_{\mathrm{min}}(X) \geq - \max_{i} |x_i| \geq -1$.
 A: It suffices to prove that
$$2xx^T-2 \textrm{diag}(xx^T)+\mathbb{I}_n\geq 0$$
or equivalently that
$$\left[\matrix{1 & 2x_1x_2 & \cdots & 2x_1x_n\\ 2x_1x_2 & 1 & \cdots & 2x_2x_n\\
\vdots & \vdots & \ddots & \vdots\\ 2x_1x_n & 2 x_2x_n & \cdots & 1}\right]\geq 0$$
Let now $e_i$  the $i$-th column of $\mathbb{I}_n$. Then if we replace the ones  with $\|x\|^2$ and the $2x_ix_j$ with the associated quadratic forms $x^T(e_ie_j^T+e_je_i^T)x$   we have to prove that
$$\left[\matrix{x^Tx & x^T(e_1e_2^T+e_2e_1^T)x & \cdots &x^T(e_1e_n^T+e_ne_1^T)x\\ x^T(e_1e_2^T+e_2e_1^T)x & x^Tx & \cdots & x^T(e_2e_n^T+e_ne_2^T)x\\ \vdots & \vdots & \ddots & \vdots\\ x^T(e_1e_n^T+e_ne_1^T)x & x^T(e_2e_n^T+e_ne_2^T)x & \cdots &x^Tx}\right]\geq 0$$
or equivalently
$$U^TQU\geq 0$$
with 
$$U:=\left[\matrix{x & 0 & \cdots & 0\\0 & x & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & x}\right]$$
and
$$Q:=\left[\matrix{\mathbb{I}_n & e_1e_2^T+e_2e_1^T & \cdots &e_1e_n^T+e_ne_1^T\\ e_1e_2^T+e_2e_1^T & \mathbb{I}_n & \cdots & e_2e_n^T+e_ne_2^T\\ \vdots & \vdots & \ddots & \vdots\\ e_1e_n^T+e_ne_1^T & e_2e_n^T+e_ne_2^T & \cdots &\mathbb{I}_n}\right] $$
The positive semi-definiteness of $Q$ has been shown  here
from which the desired property follows directly. 
