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

131 Views Asked by At

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?