# 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.

• It is, since clearly $e_1$ and $e_2$ are in its image. And yes, in this case the rank is 2.
– user622002
Mar 27, 2020 at 8:04

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.
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.
• What is the domain of your second map? In particular, I am wondering what $uA$ denotes, is this vector-matrix multiplication? Mar 27, 2020 at 17:06
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 ....