Let $\vec v$ be a non-zero vector in $\Bbb R^n$ such that $\|\vec v\| = 1$ and let $A = I −β\vec v\vec v^T$, with $β > 0$.
(a) Show that if $β ≤ 1$, then $A$ is spd.
Hint: use Cauchy-Schwarz inequality
Proving its symmetric is of course very straightforward. However, I am struggling with figuring out why if $β ≤ 1$ then $A$ is spd.
$x^{T}Ax=x^{T}x-\beta x^{T}vv^{T}x=\|x\|^{2}-\beta \|v^{T}x\|^{2} \geq 0$ because $\|v^{T}x\|\leq \|v\|\|x\|=\|x\|$ by Cauchy - Schwarz inequality.