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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?

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    $\begingroup$ Do you know the dimension theorem, aka rank-nullity theorem? It tells you at once that something's wrong as $\,1+1\neq 3\,$...yet your idea of work is correct in general. $\endgroup$ – DonAntonio Jan 8 '13 at 12:23
  • $\begingroup$ Yes, form Kronecker-Capelli theory dim im + dim ker = dim $\varphi$. That't was a typo. Thanks! $\endgroup$ – Jonny Jan 8 '13 at 12:32
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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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