Is this matrix surjective? Textbook dispute \begin{bmatrix} 
1 & -2 & 3 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
Is this matrix surjective? My textbook says no, but by all logic I've read it should be yes. For example, Check if $f$ is injective / surjective (matrix) says that if Rank is equal to number of rows then a matrix is surjective. 
 A: Yes, this is the matrix of a surjective linear map. Look at the first and the fourth columns: $\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0\\1\end{smallmatrix}\right]$ respectively. It follows from this that the vectors $\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0\\1\end{smallmatrix}\right]$ belong to the range and that therefore the range is $k^2$, where $k$ is the field that you are working with.
A: Just in case, check the conventions your textbook is using, and specifically whether matrices conventionally act from left (on column vectors) or from right (on row vectors).
In particular, for your matrix $$A = \begin{bmatrix}1 & -2 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},$$ the map $v \mapsto Av$ (where $v$ is a column vector) is surjective but not injective, while the map $u \mapsto uA$ (where $u$ is a row vector) is injective but not surjective.
A: For surjectivity of a linear map $A: U\rightarrow V$, you must have 
$rank(A)=dim(V)$.
In your case $rank(A)=2=dim(V)$ so $A$ is ....
