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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?

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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.

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  • $\begingroup$ If $\dim (im \phi)=\dim \mathbb{Z}^2_3\implies $surjective? $\endgroup$ – Doesbaddel Jun 5 '19 at 13:12
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    $\begingroup$ @Doesbaddel Certainly so. And the dimension of the image is the same as the rank of any associated matrix. $\endgroup$ – egreg Jun 5 '19 at 13:21
  • $\begingroup$ Alright, thanks! $\endgroup$ – Doesbaddel Jun 5 '19 at 13:31
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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 $\phi(0,1,0)=(0,1)$.

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