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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?

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  • $\begingroup$ Are you looking at real or complex $V$? $\endgroup$
    – Jose27
    Mar 23, 2015 at 6:39
  • $\begingroup$ @Jose27: I left it unspecified on purpose because I'm interested in both and one never knows in advance if those two cases will be analogous or not quite. $\endgroup$
    – Leo
    Mar 23, 2015 at 6:41

2 Answers 2

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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 $$ \langle Tx, x\rangle = \langle \sqrt{\lambda_{1}}\cdot u_{1} - \sqrt{|\lambda_{n}|}\cdot u_{n}, \frac{1}{\sqrt{\lambda_{1}}}u_{1} +\frac{1}{\sqrt{|\lambda_{n}|}}u_{n}\rangle =1 -1 = 0. $$
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  • $\begingroup$ I guess that over reals we indeed can look only at symmetric matrices since only the symmetric part of $T$ is passed to its quadratic form. Is there anything what can be said about definiteness of a non-hermitian $T$? (+1 and thank you!) $\endgroup$
    – Leo
    Mar 23, 2015 at 7:56
  • $\begingroup$ @Leo: What do you mean "only the symmetric part of $T$ is passed to it's quadratic form"? For instance any rotation by an angle less than $\pi/2$ is definite (in fact "positive") and not symmetric. $\endgroup$
    – Jose27
    Mar 23, 2015 at 18:15
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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 $-D$?

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  • $\begingroup$ Could you elaborate, I don't quite see how it answers the question. $\endgroup$
    – Leo
    Mar 23, 2015 at 5:46
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    $\begingroup$ In your question you ask whether there are definite operators which are not positive-definite. The idea that I'm presenting here is that if you have a positive-definite matrix, there is an associated negative-definite matrix by just taking the negative of each eigenvalue. This gives an example of a definite, but not positive-definite, matrix. I do see how this may not be in the spirit of what you were asking for. In which case I can remove my answer. $\endgroup$
    – Mnifldz
    Mar 23, 2015 at 5:49
  • $\begingroup$ Indeed, thank you for this clear example! So my main question was how to characterize a definite $T$ (I changed the title accordingly.) $\endgroup$
    – Leo
    Mar 23, 2015 at 5:54

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