$d \cdot A - u u^t$ is posivite

Question

I am studying for my linear algebra test and I don't know how to solve the following exercise of the problem set:

Given $d>0$, $A_{n \times n}$ a non-negative semidefinite matrix ($A \geq 0$). Given $u \in \mathbb R^n$, prove that $d \cdot A - u u^t$ is non-negative semidefinite if $u^t A^{-1} u \leq d$.

I would thank any hint or approach to the exercise.

Answer 1

Pick any $x\in\mathbb R^n$, then \begin{align} x^\top(d\cdot A-uu^\top)x&=d\cdot x^\top Ax-(u^\top x)^2\\ &\ge (u^\top A^{-1}u)(x^\top Ax)-(u^\top x)^2\\ &\ge 0 \end{align} where the last inequality follows from the well known result that \begin{equation} \sup_{x\in\mathbb R^n}\dfrac{(u^\top x)^2}{x^\top Ax} = u^\top A^{-1}u \end{equation}

Answer 2

Hint. Let $v=(dA)^{-1/2}$. Then the problem reduces to proving that $I-vv^T\succeq0$ if and only if $v^Tv\le1$.