# relationship between rows and columns of an orthogonal matrix

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 $$~m < n~$$?

• 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 $\operatorname{rank}(A B) \leq \operatorname{min}(\operatorname{rank}(A),\operatorname{rank}(B))$.) – Travis Aug 14 at 1:37

## 1 Answer

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

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