# Vector Space, Two bases related by Unitary Matrix

1. Nov 18, 2013

### Zatman

1. The problem statement, all variables and given/known data
The vector space V is equipped with a hermitian scalar product and an orthonormal basis e1, ..., en. A second orthonormal basis, e1', ..., en' is related to the first one by

$\mathbf{e}_j^{'}= \displaystyle\sum_{i=1}^n U_{ij}\mathbf{e}_i$

where Uij are complex numbers. Show that Uij = <ei, ej'>, and that the matrix U with entries Uij is unitary.

2. The attempt at a solution
I have done the first part, by simply taking the scalar product with some ek of each side of the defining equation for ej', and using the fact that each term is zero unless i=k, which leads to the first result.

I'm not quite sure how to go about the second part. I've attempted to show that UU* = I; but all I can see is that the diagonal elements of UU* are:

$\displaystyle\sum_{j=1}^n \left\langle \mathbf{e}_i, \mathbf{e}_j^{'} \right\rangle ^2$

for i = 1:n. I think that this probably could be shown to be 1 for all i using the definition given.

I'm not sure how the remaining elements are going to be zero though, which is necessary for UU* = I.

Any nudges in the right direction would be greatly appreciated!

2. Nov 19, 2013

### vela

Staff Emeritus
I haven't worked this out, so this is just a suggestion. It seems you haven't used the fact that the second basis is orthonormal yet. What do you get if you use the fact that $\langle \mathbf{e}'_i, \mathbf{e}'_j \rangle = \delta_{ij}$?

3. Nov 19, 2013

### Zatman

Sorry, but I'm not sure how that can be used. The elements of U are scalar products of the ith basis vector and the jth primed basis vector, and the only relationship between the two bases is U itself.

What you wrote was used in a previous part of the question which I didn't bother typing out (it's not relevant to this part). Thank you, anyway.

4. Nov 19, 2013

### PeroK

U looks like a linear transformation that maps

$e_i \ to \ e'_i$, so it preserves scalar products.

Then show that <U*Ux.y> = <Ux.Uy> = <x.y>

5. Nov 19, 2013

### vela

Staff Emeritus
No, it wasn't. I wrote the inner product of $e'_i$ and $e'_j$, both primed, not one primed and one unprimed.