Here is the problem

$V$ is an inner product space, and a is a linear transformation from $V$ to $V$. Prove that for any unit vector $v$ belongs to $V$, we have $\langle av, v\rangle \langle v, av\rangle \le \langle av, av\rangle$.

The textbook I am reading on talks nothing about the product of two inner products. But I assume that $\langle v, v\rangle \langle v, v\rangle = |\langle v, v\rangle|^2$, is that right?

I think this problem has to discuss two circumstances, one is where $\langle av, v\rangle = 0$, so both sides equal to $0$, that is an equivalent, but what about the other case when $\langle av, v\rangle$ is not equal to $0$, where the left hand side less than the right hand side?

  • $\begingroup$ By the way, what info can I have from "unit vector"? ‹v, v› = 1? $\endgroup$ – PixieBlade Feb 26 at 22:49
  • $\begingroup$ $V$ is a space over what field? $\mathbb R$ or $\mathbb C$? $\endgroup$ – J. W. Tanner Feb 26 at 22:55
  • $\begingroup$ You wrote |<v,v>|^2; did you realize <v,v> must be a non-negative real number? $\endgroup$ – J. W. Tanner Feb 26 at 23:00
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    $\begingroup$ You may find this helpful: math.stackexchange.com/questions/3127021/…. $\endgroup$ – Minus One-Twelfth Feb 26 at 23:07
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    $\begingroup$ @DonAntonio because I am new here, not knowing how to type greek letter, sorry $\endgroup$ – PixieBlade Feb 27 at 0:01

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