The problem states, "Prove that a unitary matrix preserves the sum of the modulus squares of a column vector multiplied on its right."
I don't understand the solution below:
I think there is a typo after the first equality. Shouldn't that be $\sum_j \big\lvert \sum_k (U[j,k] \times X[k] \big\rvert^2$? I think you want $X[k]$ to be incrementing up instead of $X[j]$ for the matrix-vector multiplication to work.
What allows the second equality? How do you pull the summation $\sum_k$ out of the modulus? I'm not convinced in general that $\lvert\sum_k x_k \rvert^2 = \sum_k \lvert x_k \rvert^2$.
I agree with your observations. It is also not clear how this supposed proof uses the fact that $U$ is unitary. Here's a correct version of the proof. \begin{align} \sum_j |(U\star X)[j]|^2 &= \sum_j \left|\sum_k U[j,k]\times X[k]\right|^2 \\ & = \sum_j \left(\sum_k U[j,k]\times X[k]\right)\left(\overline{\sum_\ell U[j,\ell]\times X[\ell]}\right) \\ & = \sum_{j,k,\ell} U[j,k]\overline{U[j,\ell]} X[k]\overline{X[\ell]} \\ & = \sum_{k,l}X[k]\overline{X[\ell]}\sum_{j} U[j,k]\overline{U[j,\ell]} \\ & = \sum_{k,l}X[k]\overline{X[\ell]} \delta[k,\ell] = \sum_k X[k]\overline{X[k]} = x. \end{align} Here, $\delta[k,\ell]$ is a Kronecker delta, which arises using the fact that the columns of a unitary matrix are orthonormal.