Is the symmetric matrix $vv^{T} - hI$ is strictly negative definite for all small enough $h>0$? Is $\det(vv^{T} - hI)<0$ for small positive $h?$

Question

Let $v\in \mathbb{R}^n, v\ne 0$ be a column vector and $h>0.$

QUESTION: Is the symmetric matrix $B(h):=vv^{T} - hI$ is always strictly negative definite for all $n, v$ and for all small enough $h>0$?

What I tried so far to check the sigh of the determinant:

Case $n=2:$ for sign of the determinant:

I just checked that when $n=2,$ the matrix:

$$B(h):=vv^{T} - hI_2$$

is strictly negative definite. It's easy to check, as a simple calculation of its determinant shows that

$$\det(B(h))= -h(v_1^2 + v_2^2 - h) < 0, v=(v_1, v_2)^{T},$$

Case $n=3:$ for the sign of the determinant

I also did a similar calculation for $n=3, v=(v_1,v_2,v_3)^{T},$ and here we get:

$$\det(B(h))= (v_1^2-h)(-h(v_2^2 + v_3^2 - h)) + v_1^2v_2^2h + v_1v_3h^2,$$

which, upon simplification becomes:

$\det(B(h))= -hv_1^2(v_2^2 + v_3^2) + v_1^2v_2^2h + v_1^2v_2^2h + O(h^2) = -h^2v_1^2v_3^3 + O(h^2)$

which is strictly negative when $h>0$ is small enough.

My question: So it seems to me that $\det(B(h))$ is strictly negative definite for any dimension $n$. Is it correct? If yes, is there an upper bound on $h$ that guarantees that $\det(B(h))$ is strictly negative?

Furthermore, how can we show that $B(h)$ is indeed strictly negative definite?

Answer 1

This is false. Take $v = (1,0)^T.$ Then $v v^T - hI$ is $$ v v^T - hI = \left( \begin{array}{cc} 1-h & 0 \\ 0& -h \end{array} \right) $$ which is indefinite for small $h > 0$

Answer 2

Rephrasing, for which values of $h > 0$ does $h \, {\bf I}_n - {\bf v} {\bf v}^\top \succ {\bf O}_n$ hold?

Dividing both sides by $h > 0$ and using the Schur complement, we conclude that $h \, {\bf I}_n - {\bf v} {\bf v}^\top \succ {\bf O}_n$ is equivalent to

$$ \begin{bmatrix} {\bf I}_n & {\bf v} \\ {\bf v}^\top & h \end{bmatrix} \succ {\bf O}_{n+1} $$

Looking for the value of $h > 0$ at which the determinant vanishes,

$$ \det \begin{bmatrix} {\bf I}_n & {\bf v} \\ {\bf v}^\top & h \end{bmatrix} = h - {\bf v}^\top {\bf v} = h - \| {\bf v} \|_2^2 = 0$$

Hence, $h \, {\bf I}_n - {\bf v} {\bf v}^\top \succ {\bf O}_n$ holds for $\color{blue}{h > \| {\bf v} \|_2^2}$.