In a prehilbert space $E$ of finite dimension $n$ I have an operator $A$ and an orthonormal basis formed by $n$ vectors: $(v_1, v_2, …, v_n)$. $A$ is defined such that:
$A(v_i) = v_{i-1}$ for $i = 2, …, n$ and $A(v_1) = v_1$ for $i = 1$.
The question is, how can I prove that this operator is not hermitian? My first course of action was to figure out the matrix associated to the operator and directly find out if the property holds; however, I've been having trouble identifying it since as you can see the operator isn't as simple as, say: $G(x, y) = (2x, x+y)$. This last example has a clear-cut matrix, but the $A$ operator in my problem is much more confusing to me. Any help in finding this matrix or generally finding a proof for it not being hermitian is appreciated.
Best Answer
A Hermitian operator has to satisfy $$\langle x,A y\rangle = \langle A x, y\rangle.$$ However, if you choose $x = v_1$ and $y = v_2$. You would have $$\langle x,A y\rangle = \langle v_1, v_1\rangle = 1.$$ On the other hand, $$\langle A x, y\rangle = \langle v_1, v_2\rangle = 0.$$ Therefore, $A$ is non-Hermitian.