Suppose $U$ is an $m\times n$ orthogonal matrix. Show that $m \geq n$.

I'm having trouble with this proof --

I understand that the columns of $~U~$ can only be linearly independent in the cases where

$(i) ~~~m > n~$ and

$(ii)~~~ m = n~$,

but how do I go on to discuss whether or not this indicates that the column vectors themselves are orthogonal or not?

And why this is not the case when $~m < n~$?

  • 1
    $\begingroup$ Usually orthogonal matrices are defined to be square. What is your definition of orthogonal, then? Just that $A^\top A = I_n$? (In that case, use that $\operatorname{rank}(A B) \leq \operatorname{min}(\operatorname{rank}(A),\operatorname{rank}(B))$.) $\endgroup$ – Travis Willse Aug 14 '19 at 1:37

Orthogonal matrices are by definition square matrices?

Edit: Recall that the $rank(U) \leq min(m, n)$. Then note that since it must have linearly independent columns since each column is by definition orthogonal to one another, it must be at least $n$. Therfore, $n \leq rank(U) \leq min(n, m) \leq m$.

  • $\begingroup$ My understanding of the question (which I did not write might i add haha) is to prove why a nonsquare matrix cannot be an orthogonal matrix (using reasoning beyond the exact definition and moreso by means of the definition of orthogonal column vectors and linear independence) $\endgroup$ – mbus2sus Aug 14 '19 at 1:01
  • $\begingroup$ @mbus2sus see edit, maybe it helps answer your question. $\endgroup$ – aerupons Aug 14 '19 at 1:49

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