Hi all i could use a little help. Thanks in advance.
Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.
what ive tried:
(Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)
(x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
(x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
note*[trans = transpose]
(x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
(x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
(x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)
I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.
i dont understand this operator:
I had understood a hermitian matrix to be one such that
Your first step made some leaps i cant quite seem to follow especially this one:
I can mostly follow after this step, but I dont understand this equation. Could you elaborate this step for me? Thank you.
I cant follow this:
When i work it out:
Am I missing the transpose because we are summing across the i direction instead of the j direction? I have a problem when working from (x, Ay)
I keep showing that
For the sake of clarity I will mention that the operator is defined (virtually always, at least in Physics) as the Hermitian conjugate, or transpose conjugate, operator: Acting on a square matrix A, . (Where, to be clear, * is the complex conjugate operator.) So the Hermitian scalar product acting on two vectors x (a row vector) and y (a column vector) is , where is the usual (Euclidean) scalar product.
I feeling pretty slow today....forgot that
Thank you CaptainBlack i can follow your solution now! Both of you were a big help. Now i can move on to my next problem...proving all eigenvalues of a hermitian matrix A are real. Thanks for all your help.