# 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

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.
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.
(Note that the transformation will not be unique. If $T$ is such a transformation, then so is $kT$ for $k \in \mathbb R\setminus \{0\}$)