Matrix-vector inner products

Question

everyone! I am doing my homework of matrix algebra. The task is to proof that if $$ U \in \mathbb C^{n\times n} $$ unitary matrix, and: $$ \{x_1,x_2,...,x_n\} $$ are orthonormal vectors, then: $$\{Ux_1,Ux_2,...,Ux_n\} $$ is orthonormal as well. So my decision is: $$ \langle Ux_i, Ux_j \rangle = \sum U^*\bar x_i \times Ux_j = U^*U \sum \bar x_ix_j = I \sum \bar x_ix_j $$ Can anyoune pleas tell me is that way possible and I wondering is it possible to get $$ U^*U \sum ... $$ out of summation? Thanx

Answer 1

Since $U$ is unitary, we have $U^*U=I$, hence

$$<Ux_i, Ux_j>=<x_i, U^{*}Ux_j>=<x_i, x_j>= \delta_{ij},$$

and you are done.

Answer 2

Well, we have $\langle Ux_i,Ux_j\rangle = UU^*\langle x_i,x_j\rangle = I \langle x_i,x_j\rangle = \langle x_i,x_j\rangle =\delta_{ij}$, where $\delta$ is the Kronecker-delta.

Answer 3

We can also prove the result by expressing the inner product in terms of matrices. Since$$(Ux_i)^\dagger Ux_j=x_i^\dagger U^\dagger Ux_j=x_i^\dagger I_nx_j=x_i^\dagger x_j,$$where $I_n$ is the $n\times n$ identity matrix, $x_i^\dagger x_j=\delta_{ij}\implies (Ux_i)^\dagger Ux_j=\delta_{ij}$.