Inner product propety with Scalar Matrix (Proof)

[RJLiberator]

Homework Statement

Let A be an nxn matrix, and let |v>, |w> ∈ℂ. Prove that (A|v>)*|w> = |v>*(A†|w>)

† = hermitian conjugate

Homework Equations

The Attempt at a Solution

Struggling to start this one. I'm sure this one is likely relatively quick and painless, but I need to identify the trick behind it.

Are there any tips?

My only current thought is to use the properties of inner products. Should I look at each element of the matrix A one by one?

 

[RJLiberator]

How is this for a solution:

Writing in sum notation, we note that matrix A has corresponding eigenvalue such that A|v_i> = t|v_i>

We replace it in the left hand side so that
t|v_i>|w_i> = |v_i>A†|w_i>

Now, since the hermitian conjugate takes the transpose and the complex conjugate, we see that

A^T^*|w_i> = t*|w_i>

Applying properties of inner product spaces, to take the scalar t* out of the right hand side, we take its complex conjugate.

So we are left with
t|v_i>|w_i> on both sides

 

[RJLiberator]

The problem that I am having right now is this:

A^T^* (|w_i>) = t*|w_i>
where T is transpose and * is complex conjugate

I know that the eigenvalues of a matrix and it's transpose are equal. But that does mean that the eigenvalues of a complex matrix and its conjugate are off by conjugation?

If so, this proof is concluded.

 

[DrClaude]

Writing in sum notation, we note that matrix A has corresponding eigenvalue such that A|v_i> = t|v_i>

I don't understand why you are assuming that |v> is an eigenvector of A.

I would do the proof by considering the elements of the matrix and the vectors. For instance, defining |u> = A|v>, we have
$$
u_i = \sum_j A_{ij} v_j
$$
and so on.