Find a matrix of (any) linear map $ \varphi : \mathbb{R}^{4} \rightarrow \mathbb{R}^{3} $, so that conditions apply I'm trying to find a matrix of (any) linear map $ \varphi : \mathbb{R}^{4} \rightarrow \mathbb{R}^{3} $, for which the following conditions apply:


*

*the dimension of the image $  \varphi = 2$

*$ \varphi(1,1,1,1) = (1,2,1)$

*$ \varphi(1,0,1,0) = (2,1,0)$
I already determined, that the dimension of the kernel $\varphi$ will be $2$. What should be my next steps? I think I know how I would find such a matrix, that the dimension of the image is 2, but I don't know what to do about the second and the third point.
Thanks!

Let's pretend for a moment that points two and three require that:


*

*$\phi(e_1) = w_1$

*$\phi(e_2) = w_2$
Where $e_i$ are the elements of the canonical basis for $\mathbb R^4$, $w_1 = (1,2,1)$ and $w_2 = (2,1,0)$. An immediate solution for this problem would be the linear function defined by the associations above and sending $e_3$ and $e_4$ to the zero vector in $\mathbb R^3$, represented by the matrix:
$$ A= \left(\begin{matrix}1&2&0&0\\2&1&0&0\\1&0&0&0\end{matrix}\right) $$
The kernel of this function has obviously dimension 2. Now to solve the original problem, let $v_1=(1,1,1,1)$ and $v_2=(1,0,1,0)$, and consider the change of basis matrix below:
$$ B= \left(\begin{matrix}1&1&0&0\\1&0&0&0\\1&1&1&0\\1&0&0&1\end{matrix}\right) $$
You can see that $Be_1 = v_1$ and $Be_2=v_2$; so inverting the matrix we have $B^{-1}v_1=e_1 \implies AB^{-1}v_1=w_1$, and similarly $AB^{-1}v_2=w_2$. $B^{-1}$ is invertible, so $\dim \ker AB^{-1} = \dim\ker A$, hence the linear function represented by $AB^{-1}$ satisfies your requirements.
