How can we construct a non-square isometry matrix $U\in \mathcal{M_{n,m}}$; that is, all columns of $U$ are orthonormal and $U U^T=I_{n,n}$?


1 Answer 1


Suppose such a matrix does exist. Then, since the columns of $U$ are linearly independent and $n\neq m$, it must be the case that $m<n$ (otherwise we'd have $n<m$ and $\operatorname{rank}(U)<m$, contradicting the fact that the columns of $U$ are linearly independent).

However it is well known that $\operatorname{rank}(AB)\leq \min (\operatorname{rank}(A),\operatorname{rank}(B))$, therefore $n=\operatorname{rank}(I_n)=\operatorname{rank}(UU^T)\leq \min (m,m)=m$: contradiction.

If you were to ask about $U$ such that its lines were orthonormal, then you could take $U=(u_{ij})$ defined by $u_{ij}=1$ if $i=j\leq n$ and $(u_{ij})=0$ otherwise. In this case $n<m$ and everything is good.


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