# linear transformation dimZ

What i should do? I have to find $\dim (Z)$.

I have a map $\psi(( x_{1},x_{2},x_{3},x_{4}))=(x_{1}+2x_{2}-x_{3}+3x_{4} , 2x_{1}+5x_{2}+x_{3}+7x_{4}, x_{1}+x_{2}-4x_{3}+2x_{4})$

And $\displaystyle Z= \{ \omega \in L(\mathbb{R}^3,\mathbb{R}^3) : \omega \circ\psi=0_{L(\mathbb{R}^3,\mathbb{R}^3)} \wedge \operatorname{im} \omega \subseteq \operatorname{lin}\bigl((1,0,0),(1,2,2)\bigr)\}$

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Note that $\operatorname{im}\phi\subseteq \ker \omega$ and $\operatorname{im}\omega\subseteq\operatorname{lin}\bigl((1,0,0),(1,2,2)\bigr)$
Let us begin by looking at the matrix associated with $\phi$
$$\left(\begin{matrix} 1&2&-1&3\\ 2&5&1&7\\1&1&-4&2 \end{matrix}\right).$$
We need to calculate its image, which is the span of the rows. To this we row reduce and get $$\left( \begin{array}{cccc} 1 & 0 & -7 & 1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$
and note that the rank is therefore $2$. We have no choice, but to send this image to $0$. Therefore, we have one more a choice over where $\omega$ sends the last dimension, but it must be in $\operatorname{lin}\bigl((1,0,0),(1,2,2)\bigr)$. Therefore, this means that $Z$ is two dimensional as it allows for $\omega$ which send the one free dimension to the available two-dimensional target space.