For example, let $\phi:\mathbb{Z}^3_3\to\mathbb{Z}^2_3$ be a linear map with
$$\phi\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}:=\begin{pmatrix}x_1+x_3\\x_2-x_1+x_3\end{pmatrix}$$
How do I determine, if $\phi$ is injective/surjective? Can we use the fact that if (edited) $\dim(\operatorname{ker}\phi)=0$, then the linear map is injective?
Best Answer
A linear map is injective if and only if $\dim\ker\phi=0$ (not when $\le1$).
The matrix of the linear map with respect to the canonical bases is $$ \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \end{bmatrix} $$ A standard Gaussian elimination yields the reduced row echelon form (RREF) $$ \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} $$ so the matrix has rank $2$. Therefore the map $\phi$ is surjective. It cannot be injective, because the nullity is $3-2=1$. Indeed a basis of the kernel of $\phi$ is given by the single vector $$ \begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} $$ obtained as a nonzero solution of the linear system $\phi(x)=0$, by looking at the RREF.