# Inner product inequality involving linear operator

$$V$$ is an inner product space and $$\tau \in \mathcal {L}(V)$$. I want to show that for any unit vector $$v\in V$$, $$\langle \tau v, v\rangle\langle v,\tau v\rangle\leq\langle \tau v,\tau v\rangle$$. I tried to extend the eigenvectors of $$\tau$$ to a basis for $$V$$ and write $$v$$ as a linear combination of this basis. But I don't know how to deal with the vectors that are in the basis but are not the eigenvectors of $$\tau$$.

Hint: Use $$\langle v, \tau v \rangle = \overline{\langle \tau v, v \rangle}$$ and the Cauchy-Schwarz inequality.
• OMG, thank you. So the equality holds when $\tau v$ and $v$ are linearly dependent, right? This is to say that $v$ is an eigenvector of $\tau$. – Jiexiong687691 Feb 26 at 5:01