Is $ (1+ |x|^2)\operatorname{Id} - x \otimes x $ positive definite?

Question

How can I prove the matrix $$ (1+ |x|^2)\operatorname{Id} - x \otimes x $$ where $x$ is a vector in $\mathbb{R}^n$, is positive-definite?

Answer 1

For any $v \ne 0$ we have

$$\left\langle (1+\|x\|^2)v - \langle v,x\rangle x, v\right\rangle = (1+\|x\|^2)\|v\|^2 - |\langle v,x\rangle|^2 = \underbrace{\|v\|^2}_{>0}+\underbrace{\|x\|^2\|v\|^2 - |\langle v,x\rangle|^2}_{\ge 0 \text{ by Cauchy-Schwarz}} > 0$$

Answer 2

Hint: Verify that $$ y^T\left[(1+ |x|^2)\operatorname{id} - x \otimes x\right]y = (1+ |x|^2)|y|^2 - (x^Ty)^2 $$ If $y$ is non-zero, then this quantity must be positive.