The Wikipedia article for Hermitian matrices gives an alternative charactarization for Hermitian matrices link. I want to prove this, and I am having difficulties completing the proof. The forward direction is fairly straightforward:
$$\langle Ax, y\rangle = (Ax)^*y = x^* A^* y = x^* A y = \langle x, Ay\rangle.$$
For the reverse direction this is how far I got:
It is tempting to say that this implies $A = A^*$ which would prove that $A$ is Hermitian, but I don't think that would be correct, since that would division by vectors which is not defined. Am I on the right track, or is there a different approach that has to be used?
Prove that a matrix is Hermitian iff it is equal to its adjoint.
linear algebra
Best Answer
If the equality $x^*A^*y=x^*Ay$ holds for any two vectors $x,y\in\mathbb C^n$, then apply this equality to two vectors $e_j$ and $e_k$ of the canonical basis. So, you know that $e_j^*A^*e_k=e_j^*Ae_k$, which is equivelent to $e_jA^*e_k=e_jAe_k$, and this means that two certain entries of the matrices of $A$ and $A^*$ with respect to the canonical basis era equal. Can you take it from here?