0
$\begingroup$

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 $~m < n~$?

$\endgroup$
  • 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 Aug 14 at 1:37
1
$\begingroup$

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, $n \leq rank(U) \leq min(n, m) \leq m$.

$\endgroup$
  • $\begingroup$ 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 at 1:01
  • $\begingroup$ @mbus2sus see edit, maybe it helps answer your question. $\endgroup$ – aerupons Aug 14 at 1:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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