$V$ 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 $\tau$.
Inner product inequality involving linear operator
inner-productslinear algebra
Best Answer
Hint: Use $\langle v, \tau v \rangle = \overline{\langle \tau v, v \rangle}$ and the Cauchy-Schwarz inequality.