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

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!

• What definition of "unitary matrix" are you using? Commented Mar 8, 2016 at 20:09
• U is Unitary if UU = UU = I Commented Mar 8, 2016 at 20:16
• That’s not the usual definition of a unitary matrix. You’re missing a transpose or adjoint there.
– amd
Commented Mar 8, 2016 at 20:26
• @amd: She forgot to put ^ before *, so she meant $U^*U=UU^*=I,$ which is the usual definition. Commented Sep 29, 2018 at 6:23

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.

• Why does this mean that the columns are linearly independent ? (sorry, we just learned what that is this week as well)? And sorry, I'm still missing the connection with the inner product. Commented Mar 8, 2016 at 20:39
• @Rebecca Linear independence means that there aren’t any “extra” vectors in the set, that is, that none of them can be written as a non-trivial linear combination of the others. Formally, this is usually stated as $\sum_i c_iU_i=0$ can only happen if all of the scalars $c_i$ are equal to zero. This is one of the requirements for a set of vectors to be a basis. The connection to the inner product is as I’ve stated: each element of the matrix $U^*U$ is equal to the inner product of a pair of columns of $U$.
– amd
Commented Mar 8, 2016 at 20:44

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.

• I think that the U transpose, actually its conjugate transpose, should left multiply U and result in I to claim what follows. Commented Jun 17, 2020 at 14:28

Lemma: Let $$A,B$$ be two square matrices, then $$AB=I\implies BA=I.$$ (you can find many proofs on this site so I omit it.)

Let $$Q=(b_1,b_2,\dots, b_n),$$ where $$\beta=(b_i)_{i=1}^n$$ is an orthonormal basis of $$F^n,$$ then $$Q,Q^*$$ are square matrices and $$Q^*Q=I_n,$$ by lemma we have $$QQ^*=I_n.$$

"Why does this mean that the columns are linearly independent ?" By definition the columns are orthogonal vectors, since their dot products are zero. This means they are at right angles in n-space. Dot products of vectors within a matrix are inner products. You can think of them as transporting a"thing" to a new coordinate system $$(x,y,z)$$; they are unit-length vectors along each of the axes in a new coordinate system, represented in the former coordinate system. Intuitively we know (and want) $$x$$, $$y$$, and $$z$$ to be ortho-normal, or orthogonal to each other.