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I've proved that, if $V$ is a real vector space with a symmetric bilinear form $\langle \ , \rangle$ such that $\langle\overset{\rightharpoonup} v, \overset{\rightharpoonup} v\rangle >0$ for some $\overset{\rightharpoonup} v\in V$, and such that $\langle \overset{\rightharpoonup} v, \overset{\rightharpoonup} v\rangle=0$ iff $\overset{\rightharpoonup} v=\overset{\rightharpoonup} 0$, then $\langle \ , \rangle$ is an inner product (i.e., the positivity of just one vector implies the positivity of every nonzero vector).

I can see why my proof doesn't extend to complex vector spaces, since there the inner product is conjugate symmetric, not symmetric. But I haven't been able to think of a specific counterexample on a complex vector space: a conjugate symmetric bilinear form such that one vector is positive, and only the zero vector is orthogonal to itself, but with some nonzero vector which is not positive. In particular, I'm having a hard time thinking of any product that satisfies the definiteness condition besides the standard one...

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The theorem is still true for complex vector spaces, and the proof is essentially the same.

Let $M$ be the matrix representing your inner product. You are assuming that $M$ is Hermitian, and satisfies $v^*Hv \neq 0$ for all nonzero $v$.

Since $M$ is Hermitian, it has a full set of orthogonal eigenvectors $w_i$ with real eigenvalues $\lambda_i$. Clearly $\lambda_i \neq 0$. I claim that, furthermore, all of the $\lambda_i$ have the same sign. Indeed, if WLOG $\lambda_1 > 0$ and $\lambda_2 < 0$, then setting $v = \sqrt{-\lambda_2}v_1 + \sqrt{\lambda_1}v_2$ gives

$$v^*Mv = (\sqrt{-\lambda_2}v_1 + \sqrt{\lambda_1}v_2)^*(\sqrt{-\lambda_2}\lambda_1 v_1 + \sqrt{\lambda_1}\lambda_2 v_2) = -\lambda_1\lambda_2 v_1^*v_1 + \lambda_1\lambda_2 v_2^*v_2 = 0,$$ a contradiction.

But if all of the eigenvalues are positive, then for any vector $v$ we can expand $v$ in the eigenbasis $$v = \sum \alpha_i w_i$$ and get $$v^*Mv = \sum \alpha_i^*\alpha_i \lambda_i \geq 0$$ since $\alpha_i^*\alpha_i \geq 0$, with equality only when $v=0$. Similarly, if all of the eigenvalues are negative, $v^*Mv$ is always negative for nonzero $v$. Therefore if $v^*Mv$ is positive for one vector, it is positive for all nonzero vectors, just as in the real case.

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  • $\begingroup$ Thanks! I guess I wasn't asked to prove it for the complex case because we haven't proved that complex self-adjoint operators are ortogonally diagonalizable yet... $\endgroup$ – Nishant Apr 6 '14 at 22:53

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