Does $-4xx^T+(\|x\|_2^2+1)I_n$ have Eigenvalue 0 for any $x\in\mathbb R^{n,1}$?

Question

Is there any $x\in\mathbb R^{n,1}$ such that 0 is an eigenvalue of $A_x:=-4xx^T+(\|x\|_2^2+1)I_n$? If $A_x$ was of shape $\alpha xx^T+\beta I_n$ with $\alpha\geq0$ and $\beta>0$, this would be a piece of cake, but the $-4$ makes this harder than I thought. Any hints, how to approach this?

Answer 1

You want there to be a nonzero vector $v$ such that $$ - 4 x x^T v + (x^T x + 1) v = 0$$ The first term is a scalar multiple of $x$, so $v$ and $x$ must be linearly dependent. If $x = t v$ and $\|v\| = 1$, the condition becomes

$$ - 4 t^2 v + (t^2 +1) v = 0$$

which is satisfied if $t = \pm 1/\sqrt{3}$. Thus $x$ can be any vector of norm $1/\sqrt{3}$.