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$V$ 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 $\tau$.

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Hint: Use $\langle v, \tau v \rangle = \overline{\langle \tau v, v \rangle}$ and the Cauchy-Schwarz inequality.

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  • $\begingroup$ 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 $\tau$. $\endgroup$ – Jiexiong687691 Feb 26 at 5:01
  • $\begingroup$ Welcome! And yes, you're right. $\endgroup$ – Minus One-Twelfth Feb 26 at 5:03

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