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?
$$ A = \begin{pmatrix} 1 & 3 & 10 \\ 2 & 3 & 14 \end{pmatrix} $$
My lecture slide says this does not have a full rank because any multiple of $x'=\begin{bmatrix} 2 & 1 & -1/2 \end{bmatrix}$ will give $Ax=0$
$$ \begin{pmatrix} 1 & 3 & 10 \\ 2 & 3 & 14 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -1/2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$
I don't think this is correct but may I check?
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.