Linear application of $T(x,y,z)=(x-y,x+z,x+2y)$ I have a problem with four potential linear applications. I just want to learn how to do it for one and then apply it to the others.

Problem: $T:\textbf{R}^3 \to \textbf{R}^3$, $T(x,y,z)=(x-y,x+z,x+2y)$ 
  
  
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*Is the application linear?
  
*If it is find (a) A matrix representation with respect to a basis of your choice, (b) A basis of Ker(T), and (c) A basis of Im(T).
  

 A: 1) Yes, it is linear. Just prove the conditions for linear maps:
A map $T:\mathbb R \rightarrow \mathbb R$ is linear if for all $v,w \in \mathbb R$,$\lambda \in \mathbb R$ holds: 
$T(v+w)=T(v)+T(w)$ and $T(\lambda v)=\lambda T(v)$
2) a)
Take a closer look at to map and the matrix operations. You can easily see that the matrix is:
$\begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 0  \end{pmatrix}  $
Because 
$\begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 0  \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} =T(x,y,z)$
Question: Which basis did i choose?
b)
You have to find the set of all $(x,y,z) \in \mathbb R^3$ such that $T(x,y,z)=(0,0,0)^T$
We consider 
$\begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 0  \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
With elementary-row operations on the matrix above you will obtain:
$\begin{pmatrix} 1 & -1 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & -3  \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$  
Question: Can you tell me know which $(x,y,z)$ fulfills the condition?
c)
If we just want to have a set of vectors which span the image, then it would be enough to take the column-vectors. But we want to have linear-independent vectors, which span our image. 
What we have to do now: Consider $T^{T}$tranposed matrix of T and apply the Gaussian-elemination-algorithm on it. The columns which you will obtain then form a basis of the image.
Now its up to you to do that and show us your result.
