What would be a characterization of a definite operator?

Question

Let $V$ be an $n$-dimensional inner product space and let's call $T\in \mathcal L (V)$ definite if $$\forall x \neq0: \langle Tx,x\rangle \neq 0. $$

An obvious sufficient condition for $T$ to be definite is when it's self-adjoint with $\lambda_j>0$ (i.e. positive-definite). I wonder if there is a definite but not positive-definite $T$? If yes, what would be a characterization of a definite operator?

Answer 1

What's wrong with the idea of a negative-definite operator? If $A$ is positive definite with spectral decomposition $PDP^T$, why don't we replace $D$ with $-D$?

Answer 2

I will assume that we restrict our interest to real symmetric matrices $T$.

Both Positive Definite (PD) and Negative Definite (ND) matrices satisfy your definition of definiteness. In the sequel, we show that these are the only two families of symmetric matrices that satisfy the definition. Equivalently, for any symmetric matrix $T$ that is neither PD, nor ND, there exists $x$ such that $\langle Tx,x \rangle=0$.

Assuming $T$ is neither PD nor ND, one of the following cases must be true:

• $T$ has an eigenvalue $\lambda_{i}=0$. In that case, $T$ is clearly not definite: for $x = u_{i}$, the eigenvector corresponding to $\lambda_{i}$, $\langle Tx,x \rangle=0$.
• $\lambda_{i} \neq 0$ $\forall i \in [n]$. In that case, $T$ is indefinite; it has both positive and negative eigenvalues. Without loss of generality, let $\lambda_{1}>0$, $\lambda_{n} < 0$, and let $u_{1}$, $u_{n}$ be the respective eigenvectors. Let $x = \frac{1}{\sqrt{\lambda_{1}}}u_{1} +\frac{1}{\sqrt{|\lambda_{n}|}}u_{n}$. Then, $$ Tx = U\Lambda U^{T} x = \sqrt{\lambda_{1}}\cdot u_{1} - \sqrt{|\lambda_{n}|}\cdot u_{n}, $$ and $$ \langle Tx, x\rangle = \langle \sqrt{\lambda_{1}}\cdot u_{1} - \sqrt{|\lambda_{n}|}\cdot u_{n}, \frac{1}{\sqrt{\lambda_{1}}}u_{1} +\frac{1}{\sqrt{|\lambda_{n}|}}u_{n}\rangle =1 -1 = 0. $$