I am studying orthogonal columns and matrices right now and I have encountered the following theorem:

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> **Theorem** An $m \times n$ matrix $U$ has orthonormal columns if and only if $U^T U = 1$.
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Is it even possible to have a matrix $U$ of the size $m\neq n$ here? 

I ask because $U^TU=I\iff U^T=U^{-1}$, which means $U$ must be $n\times n$.

Also, I tried using it on some examples.

For instance, let $\displaystyle y=(4,8,1), u\_1=\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right), u\_2=\left(-\frac{2}{3},\frac{2}{3},\frac{1}{3}\right), W=\text{span}(u\_1,u\_2)$ and $U$ be the matrix formed by $u\_1$ and $u\_2$ as columns.

So $\{u\_1,u\_2\}$ is an orthonormal set (I checked it is an orthogonal set and $u\_1\cdot u\_1=u\_2\cdot u\_2 = 1$).

But I am not able to use this theorem computing $U^TU$ and $UU^T$ (i.e. $UU^T\neq I$!). I am definitely misunderstanding something here. 

If $u\_1\cdot u\_1=1$, would $u\_1$ be an orthonormal column?