Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.