Q:

Kernel and matrix of linear transformation $(x_1,x_2,x_3,x_4)^T\mapsto(x_2-x_1,x_3-x_2,x_4-x_3)$

Let $f:\mathbb{R}^4\rightarrow\mathbb{R}^3,x = \left(\begin{array},x_1 \\x_2\\x_3\\x_4\end{array}\right)\mapsto\left(\begin{array},x_2-x_1\\x_3-x_2\\x_4-x_3\end{array}\right)$be a linear transformation.
Give its matrix $A\in\mathbb{R}^{3\times4}$ with $f(x)=A\cdot x$, $\ker(f):=\{x\in\mathbb{R}^4|f(x)=0\}$ and $\dim(\ker(f))$.

I'm pretty lost on how to do this right now, the topic is going a bit too fast for me. Could anybody be so kind to show me how to proceed? I apologize for the bad post but I'm really rather clueless.

A:

Here is a post of mine here on Math Stack Exchange where you can read up a bit on the matrix of a linear transformation. 
I'm sure you're well aware that, in order to find the matrix of a linear transformation $T \colon U \to V$, where $U$ and $V$ are vector spaces over the same field $F$, we need to first select an ordered basis for $U$ and an ordered basis for $V$. Thus the matrix of a linear transformation is relative to a specific choice of ordered bases for both the domain and the co-domain space. 
Now for your problem: 
Let us choose the following ordered basis for $\mathbb{R}^4$: 
$$ \left( \ \left[ \begin{matrix} 1 \\ 0 \\ 0 \\ 0  \end{matrix} \right], \left[ \begin{matrix} 0 \\ 1 \\ 0 \\ 0  \end{matrix} \right], \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0  \end{matrix} \right], \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1  \end{matrix} \right] \ \right), \tag{1} $$
and let us choose the following ordered basis for $\mathbb{R}^3$: 
$$ \left(  \ \left[ \begin{matrix} 1 \\ 0 \\ 0   \end{matrix} \right], \left[ \begin{matrix} 0 \\ 1 \\ 0  \end{matrix} \right], \left[ \begin{matrix} 0 \\ 0 \\ 1  \end{matrix} \right] \ \right). \tag{2} $$
These are of course the standard (ordered) bases for $\mathbb{R}^4$ and $\mathbb{R}^3$, respectively. 
We note that 
$$
\begin{align} 
 T \left( \left[ \begin{matrix} 1 \\ 0 \\ 0 \\ 0  \end{matrix} \right] \right) &= \left[ \begin{matrix} -1 \\ 0 \\ 0   \end{matrix} \right] = (-1) \left[ \begin{matrix} 1 \\ 0 \\ 0   \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 1 \\ 0   \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 0 \\ 1   \end{matrix} \right], \\ 
T \left( \left[ \begin{matrix} 0 \\ 1 \\ 0 \\ 0  \end{matrix} \right] \right) &= \left[ \begin{matrix} 1 \\ -1  \\ 0   \end{matrix} \right] = 1 \left[ \begin{matrix} 1 \\ 0 \\ 0   \end{matrix} \right] + (-1)  \left[ \begin{matrix} 0 \\ 1 \\ 0   \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 0 \\ 1   \end{matrix} \right], \\
T \left( \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0  \end{matrix} \right] \right) &= \left[ \begin{matrix} 0 \\ 1 \\ -1   \end{matrix} \right] = 0 \left[ \begin{matrix} 1 \\ 0 \\ 0   \end{matrix} \right] + 1 \left[ \begin{matrix} 0 \\ 1 \\ 0   \end{matrix} \right] + (-1) \left[ \begin{matrix} 0 \\ 0 \\ 1   \end{matrix} \right], \\
T \left( \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1  \end{matrix} \right] \right) &= \left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] = 0 \left[ \begin{matrix} 1 \\ 0 \\ 0   \end{matrix} \right] + 0 \left[ \begin{matrix} 0 \\ 1 \\ 0   \end{matrix} \right] + 1 \left[ \begin{matrix} 0 \\ 0 \\ 1   \end{matrix} \right]. 
\end{align}
$$
Now the matrix of $T$ with respect to the standard ordered bases for $\mathbb{R}^4$ and $\mathbb{R}^3$ will be the one formed by putting the co-efficients in the above representations of each one of the images under $T$ of the basis vectors in (1) above as a linear combination of the basis vectors in (2) above into the columns. 
Therefore the required matrix of the linear transformation $T$ is the matrix $[T]$ given by 
$$[T] \colon= \left[ \begin{matrix} -1 \ & 1 \ & 0 \ & 0  \\ 0 \  & -1 \ &  1 \ & 0 \\ 0 \ &  0 \ & -1 \ &  1 \end{matrix} \right]. $$
Now for any vector 
$$ \mathbf{x} \colon= \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] $$ 
in $\mathbb{R}^4$, we find that 
$$ 
\begin{align} 
[T] \mathbf{x} &= \left[ \begin{matrix} -1 \ & 1 \ & 0 \ & 0  \\ 0 \  & -1 \ &  1 \ & 0 \\ 0 \ &  0 \ & -1 \ &  1 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] \\
&= \left[ \begin{matrix} -x_1 + x_2 \\ -x_2 + x_3  \\ -x_3 + x_4  \end{matrix} \right] \\
&= \left[ \begin{matrix} x_2 -x_1  \\  x_3 -x_2   \\ x_4 - x_3   \end{matrix} \right] \\
&= T( \mathbf{x} ).
\end{align}
$$
A vector $$ \mathbf{x} \colon= \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] $$ 
in $\mathbb{R}^4$ is in the kernel of $T$ if and only if 
$$ T( \mathbf{0} ) = \mathbf{0}_{\mathbb{R}^3}, $$
that is, 
$$ \left[ \begin{matrix} x_2 -x_1  \\  x_3 -x_2   \\ x_4 - x_3   \end{matrix} \right]  = \left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right],   $$
which is the case if and only if 
$$ x_4 = x_3 = x_2 = x_1. $$
Hence the kernel of $T$ is given by 
$$ \ker (T) = \left\{ \ \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right] \in \mathbb{R}^4 \ \colon \ x_1 = x_2 = x_3 = x_4 \  \right\},  $$
which can also be expressed as 
$$ \ker (T) = \left\{ \ \alpha \left[ \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right] \ \colon \ \alpha \in \mathbb{R}  \  \right\}.  $$
Hope this helps. 

