Can a non-square matrix have a full rank?

Can a non-square matrix have a full rank?

I always see cases with square matrix with full rank but seldom with non-square matrix. Can anyone help on this?

For example, is the following matrix full rank?

 A =( 1  3   10)
( 2  3   14)


My lecture slide says this does not have a full rank because any multiple of x'=[2 1 -1/2] will give Ax=0

 (1   3   10)(  2 )  ( 0 )
(2   3   14)(  1 )==( 0 )
(-1/2)


I don't think this is correct but may I check?

• Let $A$ be an $n \times m$ matrix. By definition, $A$ has full rank if and only if $A$ has rank equal to $\min \{ n,m \}$. Does this answer your question? – Crostul Sep 27 '15 at 17:41
• How are you defining "full rank" ? – pjs36 Sep 27 '15 at 17:41
• Thanks then may I ask whether the 2x3 matrix above in my example a full rank? – Eric Sep 27 '15 at 17:47
• This is a $2 \times 3$ matrix of rank 2: by definition it is a full-rank matrix. – Crostul Sep 27 '15 at 17:48
• I see. I got it. That's what I expected. Then my lecture note is wrong. Bad professor... – Eric Sep 27 '15 at 17:49

If a matrix is $m \times n$, then we say it has full row rank if the rank is at least $m$ and it has full column rank if the rank is at least $n$. Unless the matrix is square, it is impossible for both to occur.
We could say that the matrix is "full rank" if the rank is $\min \{ m,n \}$. I would understand this usage, even though I don't think I've actually seen it in practice.
• Non-square matrices with full rank have an important role in differential geometry (implicit function theorem in dimension $>1$, and in looking for singularities in varieties defined by equations). – Crostul Sep 27 '15 at 17:46
• @Eric It has rank 2, which is $\min \{ 2,3 \}$. So yes. – Ian Sep 27 '15 at 17:47