# Determine whether a given linear map is injective/surjective

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?

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.
• If $\dim (im \phi)=\dim \mathbb{Z}^2_3\implies$surjective? – Doesbaddel Jun 5 '19 at 13:12
The map is injective if and only if $$\ker\phi=\{0\}$$. But it follows from the rank-nullity theorem that $$\dim\ker\phi\geqslant1$$ and therefore $$\phi$$ is not injective.
But it is surjective; this follows from the fact the both $$(1,0)$$ and $$(0,1)$$ belong to the image of $$\phi$$: $$\phi(1,1,0)=(1,0)$$ and $$\phi(0,1,0)=(0,1)$$.