TITLE: Can a non-square matrix have a full rank?
QUESTION [6 upvotes]: 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?

REPLY [11 votes]: 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.