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I know these 2 statements to be true:

1) An $n$ x $n$ matrix U has orthonormal columns iff. $U^TU=I=UU^T$.

2) An $m$ x $n$ matrix U has orthonormal columns iff. $U^TU=I$.

But can (2) be generalised to become "An $m$ x $n$ matrix U has orthonormal columns iff. $U^TU=I=UU^T$" ? Why or why not?

Thanks!

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    $\begingroup$ If $m\ne n$ then $A^tA=I=AA^t$ is impossible, as can be seen by considering the rank. $\endgroup$ – Gerry Myerson Aug 14 '12 at 6:12
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    $\begingroup$ Square matrices with orthonormal columns also have orthonormal rows. Think about it: $U^TU$ and $UU^T$ aren't even of the same size. $\endgroup$ – Frenzy Li Aug 14 '12 at 6:12
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    $\begingroup$ @FrenzYDT. Not sure why being the same size would be a matter of concern. Over some (necessarily noncommutative) rings, there exists an $m\times n$ matrix $A$ and $n\times m$ matrix $B$ with $m\neq n$ and $AB=I_n$ and $BA=I_m$. It's weird but not unheard of! $\endgroup$ – rschwieb Aug 14 '12 at 13:42
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The $(i,j)$ entry of $U^T U$ is the dot product of the $i$'th and $j$'th columns of $U$, so the matrix has orthonormal columns if and only if $U^T U = I$ (the $n \times n$ identity matrix, that is). If $U$ is $m \times n$, this requires $m \ge n$, because the rank of $U^T U$ is at most $\min(m,n)$. On the other hand, $U U^T$ is $m \times m$, and this again has rank at most $\min(m,n)$, so if $m > n$ it can't be the $m \times m$ identity matrix.

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