Why do the columns of a unitary matrix form an orthonormal basis?

Question

So I'm trying to understand why the columns of a unitary matrix form an orthonormal basis. I know it has something to do with the inner product, but I don't fully understand that either (we learned all of this together this past week).

I've searched here and done a google search, and everything I found seems to assume I would understand the connection between the inner product and why it would be important, or they rely on eigenvalues/vectors, which we haven't explicitly learned about yet.

If anyone is able to help with this, I would appreciate it!

Answer 1

First of all, you have $UU^*=U^*U=I$, so $U^{-1}=U^*$, which means that the columns of $U$ are linearly independent. Now, let $U_i$ be the $i$th column of $U$ and think about what the elements of $U^*U$ are: $$[U^*U]_{ij}=\sum_k u^*_{ik}u_{kj} = U_i^*U_j=\delta_{ij},$$ but this is just the inner product $\langle U_j,U_i\rangle$. So, the columns of $U$ are pairwise orthogonal, and $\langle U_i,U_i\rangle = \|U_i\|^2 = 1$, i.e., they’re all unit vectors. Put that together and you’ve got an orthonormal basis.

Answer 2

Let $U$ be a unitary matrix of order $n$. Let $e_1, e_2 ..., e_n$ be the columns of $U$. That is $U = (e_1, e_2, ..., e_n)$.

Condition $UU^T = I$ equals $(e_i, e_j) = 0$ for $i \ne j$; and $(e_i, e_i) = 1$ for $i=1..n$.

It means that $e_1, ..., e_n$ - orthonormal basis.