= Full row rank
{title2=$\operatorname{rank}A=m$}
An $m\times n$ <matrix> has <full row rank> when its <matrix rank> is $m$, equivalently when its rows are <linearly independent vectors>. This requires $m\leq n$. Its <column space> is all of $\mathbb F^m$, so the represented <linear map> is <surjective> and every <linear system> $Ax=b$ is solvable. If $n>m$, its <nullity> is $n-m>0$, so solutions are not unique.<Interactive Linear Algebra by Dan Margalit and others/One-to-one and Onto Transformations>[Interactive Linear Algebra, §3.2.2, Onto matrix transformations theorem and Tall matrices do not have onto transformations]{ref}
For example, the <matrix transpose> of the <matrix> in <full column rank> is
$$
B^T=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}.
$$
It has <full row rank> two. Every output $(p,q)^T$ has solutions $(p-t,q-t,t)^T$, with $t\in\mathbb R$. A <square matrix> of size $n$ has <full row rank> and <full column rank> together exactly when its <matrix rank> is $n$; then it is an <invertible matrix>. A nonsquare <matrix> cannot have both properties, and has no two-sided inverse: such an inverse would make its <linear map> a <linear isomorphism> between <vector spaces> of different <dimensions>.
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