<s>[INST] Find basis of im, ker and dim im, dim ker verification In my homework, I've to find basis of im and ker of linear transformation
$\varphi:\mathbb{R}^{3}\rightarrow\mathbb{R}^{2}, \varphi((x_{1},x_{2},x_{3}))=(2x_{1}+x_{2}-3x_{3},x_{1}+4x_{2}+2x_{3})$
My solution:
Kernel of $\varphi$ is described by a matrix representing set of equations  
$\left[ \begin{matrix}2 & 1 & -3 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & -7 & -7 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & 1 & 1 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right]$
General solution : lin (2, -1, 1) - that's the ker $\varphi$ , dim ker $\varphi$ = 1
$M(\varphi)_{st}^{st}=\left[ \begin{matrix}2 & 1 & -3\\ 1 & 4 & 2 \end{matrix}\right]$
$(M(\varphi)_{st}^{st})^{T}=\left[ \begin{matrix}2 & 1\\ 1 & 4\\ -3 & 2 \end{matrix}\right] \sim\left[ \begin{matrix}1 & 4\\ 0 & 1\\ 0 & 0 \end{matrix}\right]$
that's im $\varphi$
dim im $\varphi$ = 2
Is idea of my solution correct?
 [/INST] Kernel of $\varphi$ is described by a matrix representing set of equations  
$\left[ \begin{matrix}2 & 1 & -3 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & -7 & -7 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & 1 & 1 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right]$
General solution : lin (2, -1, 1) - that's the ker $\varphi$ , dim ker $\varphi$ = 1
$M(\varphi)_{st}^{st}=\left[ \begin{matrix}2 & 1 & -3\\ 1 & 4 & 2 \end{matrix}\right]$
$(M(\varphi)_{st}^{st})^{T}=\left[ \begin{matrix}2 & 1\\ 1 & 4\\ -3 & 2 \end{matrix}\right] \sim\left[ \begin{matrix}1 & 4\\ 0 & 1\\ 0 & 0 \end{matrix}\right]$
that's im $\varphi$
dim im $\varphi$ = 2
Posted as answer. Thanks for verification!
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