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!
 A: Simply define $\;T:\Bbb R^4\to\Bbb R^3\;$ by  
$$T(1,1,1,1)=(1,2,1)\,,\,\,T(1,0,1,0):=(2,1,0)$$
and then complete  $\;\{v_1=(1,1,1,1)\,,\,\,v_2=(1,0,1,0)\}\;$ to a basis of $\;\Bbb R^4\;$ and on the two other vectors define $\;T\;$ to be zero. For example, take $\;\{v_3=(0,1,0,0)\,,\,\,v_4=(0,0,1,0)\}\;$ . Check now that $\;A:=\{v_1,v_2,v_3,v_4\}\;$ is a basis of $\;\Bbb R^4\;$ , and (again), define:
$$Tv_1=(1,2,1)\,,\,Tv_2=(2,1,0)\,,\,\,Tv_3=Tv_4=(0,0,0)$$
and extend the above definition by linearity, thus obtaining a linear map. And now represent this map wrt the above basis in $\;\Bbb R^4\;$ and say the standard one $\;B\;$ in $\;\Bbb R^3\;$, getting :
$$[T]_A^B=\begin{pmatrix}
1&2&0&0\\
2&1&0&0\\
1&0&0&0\end{pmatrix}$$
A: Find two vectors $b_3,b_4$ such that with $b_1 = (1,1,1,1)^T, b_2 = (1,0,1,0)^T$, the
vectors $b_1,...,b_4$ span $\mathbb{R}^4$.
Note that ${\cal R} \phi = \operatorname{sp} \{ b_1,b_2\}$, so you must have $\phi(b_3), \phi(b_4) \in {\cal R} \phi$,
otherwise this would lead to a contradiction.
So you can choose $\phi(b_3), \phi(b_4)$ arbitrarily as long as they lie in ${\cal R} \phi$. (Choosing the zero vector is an easy one.)
Once you have chosen these, this defines $\phi $ completely and the matrix
representation $A$ is straightforward to obtain.
We have $y_k = \phi(b_k) = A b_k = AB e_k$, where $B$ is the matrix with columns
$b_1,...,b_4$. From this we get $Y=AB$, where $Y$ is the matrix with columns $y_1,...,y_4$, and so
$A = Y B^{-1}$.
A: Hint:
Let the matrix of $\varphi$ be the matrix
$$A=\begin{pmatrix}
a_1&b_1&c_1&d_1\\
a_2&b_2&c_2&d_2 \\
a_3&b_3&c_3&d_3
\end{pmatrix}.$$
The conditions on $\varphi$ can be written as
$$\begin{cases}
\varphi(e_1+e_2+e_3+e_4)=\scriptsize\begin{pmatrix}
1\\ 2 \\ 1
\end{pmatrix}\\[1ex] 
\varphi(e_1+e_3)=\scriptsize\begin{pmatrix}
-1\\ 1 \\ 1
\end{pmatrix}
\end{cases}\iff\begin{cases}
\varphi(e_2+e_4)=\varphi(e_2)+\varphi(e_4)=\scriptsize\begin{pmatrix}
1\\ 2 \\ 1
\end{pmatrix}\\[1ex]
\varphi(e_1+e_3)=\varphi(e_1)+\varphi(e_3)=\scriptsize\begin{pmatrix}
2\\ 1 \\ 0
\end{pmatrix}
\end{cases}$$
So one obtains the linear systems
$$\begin{cases}
a_1+c_1=2,\\ a_2+c_2=1\\a_3+c_3=0
\end{cases},\qquad \begin{cases}
b_1+d_1=-1,\\ b_2+d_2=1,\\b_3+d_3=1
\end{cases}$$
Can you proceed?
A: I can offer you a hint for an ad hoc solution for this exercise specifically (rather than all such questions in general).
One way to construct the matrix of such a linear transformation is to know the images under $\varphi$ of the standard basis vectors $e_1=(1,0,0,0)$, $e_2=(0,1,0,0)$, $e_3=(0,0,1,0)$, and $e_4=(0,0,0,1)$. Then you will put them down as the columns of the desired matrix:
$$M = \begin{bmatrix} \varphi(e_1) & \varphi(e_2) & \varphi(e_3) & \varphi(e_4) \\ \end{bmatrix}.$$
Note that you already know $\varphi(e_1)+\varphi(e_3)=\varphi(1,0,1,0)=(2,1,0)$, as it's given to you. And you can also find $\varphi(e_2)+\varphi(e_4)=\varphi(0,1,0,1)=\varphi(1,1,1,1)-\varphi(1,0,1,0)$.
From here, you can make up as many examples satisfying the given conditions as you want. Pick any two vectors in $\mathbb{R}^3$ that add up to $(2,1,0)$ to be the values of $\varphi(e_1)$ and $\varphi(e_3)$. And pick any two vectors in $\mathbb{R}^3$ that add up to what you need to be the values of $\varphi(e_2)$ and $\varphi(e_4)$.
A: 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.
