My $f$ is: $f(x_{1},x_{2},x_{3})=(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3})$

1. Basis of $ker f$: $$f(x_{1},x_{2},x_{3})=(0,0,0)$$ So I have a matrix: $$\begin{bmatrix} 1 & 2 & -1 \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{bmatrix}$$ Then I have $$x_{1}=-5x_{3}, x_{2}=3x_{3}, x_{3} \in \mathbb R$$ Finally basic of $ker f$ is $(-5,3,1)$

It is my solution but I'm not sure if this is the good job, so please rate.

2. Basis of $im f$: $$im f=\{(x_{1}+2x_{2}-x_{3},x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}+x_{3}): x_{1}, x_{2}, x_{3} \in \mathbb R \}$$ So I have a matrix: $$\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ -1 & 2 & 1\end{bmatrix}$$ However I do not understand what can I do after elementary matrix because in books are only some steps and I do not know where the solution comes from.

  • $\begingroup$ Just to clarify, by "basic" I think you mean "basis". $\endgroup$ – Dave Jan 11 '19 at 1:21
  • $\begingroup$ "math.stackexchange.com/questions/236541/…" refer to this link to find the solution for $Im(f)$ $\endgroup$ – Mohamad Misto Jan 11 '19 at 1:24
  • $\begingroup$ Per Doug M’s answer below, you don’t have to start over to find a basis for the image. It’s spanned by the columns of the matrix and after finding the kernel you know what its dimension must be. $\endgroup$ – amd Jan 11 '19 at 2:33

You have the correct kernel.

You can should verify this by multiplying

$\begin{bmatrix} 1&2&-1\\1&1&2\\2&3&1\end{bmatrix}\begin{bmatrix}-5\\3\\1\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$

As for the image, since the degree of the Kernel is 1.

Choose 2 independent columns, any two will do. They will span the image.


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