Find a linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ with kernel spanned by $(4,5,6,7)$ and $(8,9,10,11)$ Here's the problem:

Find a linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ with kernel spanned by $(4,5,6,7)$ and $(8,9,10,11)$.

Here's my work:
Thinking of $T$ as a matrix of dimension $3$ by $4$ with columns $c_1, c_2,c_3$ and $c_4$, we have (note. here $0$ is a vector)
(1): $4c_1 + 5c_2 + 6c_3 + 7c_4 =0 $
(2): $8c_1+9c_2+10c_3+11c_4=0$.
I think any matrix whose columns satisfy this system should work ( not quite as then $c_1=c_2=c_3=c_4=0$ works but it just gives the zero matrix. So some sort of extra conditions apply?). I'm not sure how to continue from here... Can anyone help me out? Also is this the "correct/best" way to answer this type of question? Thanks!
 A: Let $e_1,e_2,e_3,e_4$ be the standard basis for $\mathbb{R}^4$. $e_1=(1,0,0,0)$, $e_2=(0,1,0,0)$, $\ldots$ etc.
Let $v_1=(4,5,6,7)$, and let $v_2=(8,9,10,11)$.
Let's start by finding a basis for $\mathbb{R}^4$, that contains $v_1,v_2$. Let $B=\{v_1,v_2,e_3,e_4\}$. $B$ is a basis for $\mathbb{R}^4$ since
$$
\det\,\begin{bmatrix}4&8&0&0\\5&9&0&0\\6&10&1&0\\7&11&0&1\end{bmatrix}=\det\,\begin{bmatrix}4&8\\5&9\end{bmatrix}=-4
$$
Let $T=T_2\circ T_1^{-1}$, where $T_1:\mathbb{R}^4\to\mathbb{R}^4$ is the linear transformation which sends
$$e_1\mapsto v_1$$
$$e_2\mapsto v_2$$
$$e_3\mapsto e_3$$
$$e_4\mapsto e_4$$
and $T_2:\mathbb{R}^4\to\mathbb{R}^3$ is the linear transformation which sends
$$e_1\mapsto(0,0,0)$$
$$e_2\mapsto(0,0,0)$$
$$e_3\mapsto(0,1,0)$$
$$e_4\mapsto(0,0,1)$$
Now note that
$$T\left(v_1\right)=T_2\circ T_1^{-1}\left(v_1\right)=T_2(e_1)=(0,0,0)$$
$$T\left(v_2\right)=T_2\circ T_1^{-1}\left(v_2\right)=T_2(e_2)=(0,0,0)$$
$$T\left(e_3\right)=T_2\circ T_1^{-1}\left(e_3\right)=T_2(e_3)=(0,1,0)$$
$$T\left(e_4\right)=T_2\circ T_1^{-1}\left(e_4\right)=T_2(e_4)=(0,0,1)$$
Note that $T(v_1)=T(v_2)=(0,0,0)$. So the dimension of the kernel of $T$ is at least $2$. Also note that the image of $T$ is the span of $\left\{(0,1,0),(0,0,1)\right\}$, which also has dimension $2$. Since the dimensions of the image of $T$ and kernel of $T$ add up to $4$, the kernel of $T$ must have dimension $2$. Hence the kernel of $T$ is spanned by $v_1,v_2$. $\square$.
Finally, let's note that the matrix for $T$ will be
$$
\begin{bmatrix}0&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\cdot\begin{bmatrix}4&8&0&0\\5&9&0&0\\6&10&1&0\\7&11&0&1\end{bmatrix}
^{-1}=\begin{bmatrix}0&0&0&0\\1&-2&1&0\\2&-3&0&1\end{bmatrix}
$$
A: You’re on the right track. The row space of a matrix is the orthogonal complement of its null space (kernel), and finding the orthogonal complement of the given kernel amounts to solving the system of equations that you’ve come up with. This system of homogeneous linear equations is underdetermined, so it will have an infinite number of solutions. In fact, the solution set is exactly the null space of the coefficient matrix $$\begin{bmatrix}4&5&6&7\\8&9&10&11\end{bmatrix}.$$ Row reducing this matrix yields $$\begin{bmatrix}1&0&-1&-2\\0&1&2&3\end{bmatrix},$$ from which we can read that $(1,-2,1,0)^T$ and $(2,-3,0,1)^T$ form a basis for its null space. For the matrix that represents $T$ we can thus take $$\begin{bmatrix}1&-2&1&0\\2&-3&0&1\\*&*&*&*\end{bmatrix},$$ where the last row is any linear combination of the first two (such as all $0$s). Of course, this isn’t the only possible solution: the rows can be any three elements of the space computed above, as long as two of them are linearly independent.
A: We have $T\left( \text{span}\{v_1=(4,5,6,7),v_2=(8,9,10,11) \} \right) = (0,0,0)$. First we find a basis for $\mathbb{R}^4$ that includes $v_1,v_2$. We can use elementary row operations for that, we have:
$$    \begin{bmatrix}
    4 & 5 & 6 & 7 \\
    8 & 9 & 10 & 11 \\
    \end{bmatrix} \Rightarrow  
      \begin{bmatrix}
    4 & 5 & 6 & 7 \\
    0 & -1 & -2 & -3 \\
    \end{bmatrix} \Rightarrow
      \begin{bmatrix}
    4 & 0 & -4 & -8 \\
    0 & 1 & 2 & 3 \\
    \end{bmatrix} \Rightarrow
      \begin{bmatrix}
    1 & 0 & -1 & -2 \\
    0 & 1 & 2 & 3 \\
    \end{bmatrix}
$$
Therefore we can add two other vectors to get a whole basis since dimension of domain is $4$.
$$      
    \begin{bmatrix}
    1 & 0 & -1 & -2 \\
    0 & 1 & 2 & 3 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    \end{bmatrix}
$$
Now we consider $T$ to be like 
$$
    \begin{bmatrix}
    a_1 & a_2 & a_3 & a_4 \\
    b_1 & b_2 & b_3 & b_4 \\
    c_1 & c_2 & c_3 & c_4 \\
    \end{bmatrix}
$$
In some basis. We know from rank-nullity theorem that $\text{rank}(T)=\text{dim}(\mathbb{R}^4)-\text{nullity}(T)=4-2=2$. Also the fact $T(1 , 0 , -1 , -2)=T(0 , 1 , 2 , 3 )=0$ and $T(0,0,1,0)\neq0$ and also $T(0,0,0,1)\neq0$. We solve the equations for the first row, the second and third rows have identical equations as first row. The inequalities $T(0,0,1,0)\neq0$ and also $T(0,0,0,1)\neq0$ dictates $a_3 \neq 0$ and $a_4 \neq 0$
Therefore we have the set of first two equations 
$$
    \begin{cases}
    T(1 , 0 , -1 , -2) = a_1 - a_3 - 2a_4=0 \\
    T(0 , 1 , 2 , 3 ) =  a_2 + 2a_3 + 3a_4=0 \\
    \end{cases} \Rightarrow
    \begin{cases}
     a_3 + 2a_4= a_1\\
     2a_3 + 3a_4= -a_2 \\
    \end{cases} \Rightarrow
    \begin{cases}
     a_4= 2a_1 + a_2\\
     a_3 = -\frac{1}{2}(3a_1+2a_2) \\
    \end{cases}
$$
Now we can choose two set of different values for $a_1$ and $a_2$ to get the first two rows of $T$ (with the condition that $a_3 , a_4 \neq 0$) and the third row can be any combination of first and second rows. for example $a_1=1,a_2=0$ and $a_1=0,a_2=1$ yields 
$$
    T \equiv \begin{bmatrix}
    1 & 0 & -3 & 2 \\
    0 & 1 & -2 & 1 \\
    1 & 1 & -5 & 3 \\
    \end{bmatrix}
$$
Where the third row is the sum of the first and second rows. Thats because the rank$(T)=2$ must holds true. Of course this is one possible answer.
A: Since $(4,5,6,7)$ and $(8,9,10,11)$ are independent in $\Bbb R^4$, we can extend to a basis $B=\{(4,5,6,7), (8,9,10,11), b_1, b_2\}$.  
Then define,  say, $T(4,5,6,7)=T(8,9,10,11)=0$ and $T(b_1)=(1,0,0)$ and $T(b_2)=(0,1,0)$.
Then the matrix of $T$ rel $B$ and the standard basis for $\Bbb R^3$ is $\begin {pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix}$, to fit the bill.
