Given that $A$ is a complex $n\times n$ Matrix. How do I prove that $\langle A(u + v) , u + v\rangle − \langle A(u − v), u − v \rangle = 2\langle Au, v \rangle + 2\langle Av, u \rangle$ ?
I am stuck at the step $2\langle A(u+v), u+v \rangle + \langle A(u+v),A(u+v) \rangle + \langle u+v,u+v \rangle + 2\langle A(u-v), u-v \rangle + \langle A(u-v),A(u-v) \rangle - \langle u-v,u-v \rangle$ and do not know where does $2\langle Au, v \rangle + 2\langle Av, u \rangle$ comes from.
Also, why does if $\langle Au,v \rangle = 0$, for all $u,v$ are input of complex matrix $n$, then $A = 0$ is true/exists ? Or the other way round, that if $\langle Av,u \rangle =0$, then $A = 0$? I have so much problem when it comes to proving polarisation identity. Would be glad if someone helps me out!
Let $f_{A}(u,v) = \langle A(u +v), u + v \rangle = \langle Au, u+v \rangle + \langle Av, u + v \rangle = \langle Au, u\rangle + \langle Au, v \rangle + \langle Av, u \rangle + \langle Av, v \rangle$.
Consequently, $f_{A} (u, -v) = \langle Au, u\rangle - \langle Au, v \rangle - \langle Av, u \rangle + \langle Av, v \rangle$.
Hence, $f_{A}(u,v) - f_{A}(u, -v) = 2 \langle Au, v\rangle + 2 \langle Av, u \rangle$.
If $Au_{0} \neq 0$, then take $v = Au_{0}$ and by the assumption you have $0 = \langle Au_{0} , Au_{0} \rangle = \|Au_{0}\|^{2}$, which leads to a contradiction.