If $U^TU$ is the identity matrix, then the columns of $U$ form an orthonormal set?

Question

I have the following statement: "If $U^TU$ is the identity matrix, then the columns of $U$ form an orthonormal set".

I want to figure out if it is true or not. By the way I know that it works with the square matrix, but is it right for any matrix $U$??

Answer 1

It is true in general:

By the very definition of matrix multiplication, the $(i,j)$-entry of $U^TU$ is the scalar product of the $i$th row of $U^T$ with the $j$th column $u_j$ of $U$, so that $$(U^TU)_{i,j}=\langle u_i, u_j\rangle$$ and thus it is the identity matrix iff $u_i$ is an orthonormal system (but not necessarily a basis).

Answer 2

Let us suppose that $U\in\text{M}_{m\times n}(\textbf{F})$. Then we can express it as $U = [u^{T}_{1},u^{T}_{2},\ldots,u^{T}_{n}]$, where each $u^{T}_{j}\in\textbf{F}^{m}$. Consequently, one has that $U^{T} = [u^{T}_{1},u^{T}_{2},\ldots,u^{T}_{n}]^{T}$. Finally, due to the assumption $U^{T}U = I_{n}$, we conclude that $u_{j}u^{T}_{j} = 1$. That is to say, $\|u_{j}\| = 1$.

On the other hand, based on the same assumption, one has that $u_{i}u^{T}_{j} = 0$ for $i\neq j$. This means the set $\{u_{1},u_{2},\ldots,u_{n}\}$ is mutually orthogonal, just as it has been claimed.

Hopefully this helps.