Find the linear transformation from kernel and range Find the linear transformation $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4 $ with  
$\ker T = [(1,0,1,0),(-1,0,0,1)] $
$\operatorname{Range}T = [(1,-1,0,2),(0,1,-1,0)] $
So if $v \in \ker T$, then 
$v = a*(1,0,1,0) + b*(-1,0,0,1)$ with $ a,b \in \mathbb{R}$
$T(v) = a*T(1,0,1,0) + b*T(-1,0,0,1) = (0,0,0,0)$
other hand. $T(v) = c* (1,-1,0,2) + d*(0,1,-1,0)$ with $c,d \in \mathbb{R} $
but i can´t  relate $T(1,0,1,0)$ and $T(-1,0,0,1)$ to the problem.. i was think to use the standard base of $\mathbb{R}^4$ and the fact that $T(u)$ with $u \in$ standard base will be a base to the $\operatorname{Range}T$.. but.. i dont know if that is the correct way.
thanks for any help
 A: Hint: The vectors $(1,0,1,0), (−1,0,0,1), (1,−1,0,2), (0,1,−1,0)$ are linearly independent and hence form a basis.


*

*Express the transformation as a matrix in terms of this basis.

*Compose this with a change of basis matrix to get the transformation as a matrix in terms of the standard basis.


(Note that the transformation will not be unique. If $T$ is such a transformation, then so is $kT$ for $k \in \mathbb R\setminus \{0\}$)
A: We seek a matrix $A$ such that $T(\vec x) = A\vec x$ satisfies the given properties. Let's arbitrarily let $x_1,x_2$ be the basic variables and let $x_3,x_4$ be the free variables. Now since we know the range of $T$, we know that $A$ must have the form:
$$ A = \begin{bmatrix}
1 & 0 & a & b \\
-1 & 1 & c & d \\
0 & -1 & e & f \\
2 & 0 & g & h \\
\end{bmatrix}$$
Row reducing to row-reduced echelon form, observe that:
$$ A \sim \begin{bmatrix}
1 & 0 & a & b \\
0 & 1 & a+c & b+d \\
0 & -1 & e & f \\
0 & 0 & g-2a & h-2b \\
\end{bmatrix}
\sim \underbrace{\begin{bmatrix}
1 & 0 & a & b \\
0 & 1 & a+c & b+d \\
0 & 0 & a+c+e & b+d+f \\
0 & 0 & g-2a & h-2b \\
\end{bmatrix}}_B$$
Now consider the kernel. Notice that if $\vec x \in \ker A$, then it must have the form:
$$
\begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix} = \begin{bmatrix}
1 \\ 0 \\ 1 \\ 0
\end{bmatrix}x_3
+ \begin{bmatrix}
-1 \\ 0 \\ 0 \\ 1
\end{bmatrix}x_4
= \begin{bmatrix}
x_3 - x_4 \\ 0 \\ x_3 \\ x_4
\end{bmatrix}
$$
which implies that the row-reduced echelon form of $A$ should be something like:
$$\begin{bmatrix}
1 & 0 & -1 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$$
Comparing this with $B$, we can immediately see that $a = -1$ and $b = 1$. Substituting these into the second row, we see that $c = 1$ and $d = -1$. Continuing, we see that $e = f = 0$ and $g = -2$ and $h = 2$. Thus, we conclude that:
$$ A = \begin{bmatrix}
1 & 0 & -1 & 1 \\
-1 & 1 & 1 & -1 \\
0 & -1 & 0 & 0 \\
2 & 0 & -2 & 2 \\
\end{bmatrix}$$
will work.
