# $V$ is a inner product space, prove $\langle av, v\rangle \langle v, av\rangle \le \langle av, av\rangle$

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?

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