Find ranges and kernels of a linear transformation $ T \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix}  x-z \\ x-y \\ 0 \end{pmatrix}$
Basis for Range = { $ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},  \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}    $ }
Basis for Kernel = { $ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $ }
I am not sure if the bases for the subspaces: range and kernel, are correct?
 A: Your kernel seems fine. For your range you should have two vectors since it is two dimensional. For the range your vectors are not linearly independent. Using the standard $e_1$ and $e_2$ vectors would suffice to be a basis of range, since they both lie outside of the kernel and are linearly independent. (Here $e_j$ is the vector with all zeros and 1 in the $j$th position)
A: Again it is important that you remember the definition of $\color{blue}{image}$ (what you call range) and the definition of $\color{blue}{kernel}$ of a linear transformation.
Below, I will give a detailed answer to the problem and leave the calculation details to you.
First, let's start by remembering the Image definition of a linear transformation and then the definition of kernel by a linear transformation.

Image of a linear transformation:
Let $V$ and $W$ vector spaces and let $T:V \to W$ a linear transformation, so  the image of a linear transformation is defined by $$\mathbf{im}(T)=\left\{w\in W, \exists v \in V: T(v)=w\right\}$$
Kernel of a linear transformation:
Let $V$ and $W$ vector spaces and let $T:V \to W$ a linear transformation, so  the kernel of a linear transformation is defined by $$\mathbf{ker}(T)=\left\{v\in V: T(v)=0_{W}\right\}$$

Now, in your problem you have a linear transformation defined by $T: \mathbb{R}^{3}\to \mathbb{R}^{3}$ such that $$T\begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} x-z \\ x-y \\ 0 \end{pmatrix}$$
So, using the definition of $\color{blue}{image}$ we have, $$\mathbf{im}(T)=\left\{\begin{pmatrix} a \\ b \\ c \end{pmatrix}\in \mathbb{R}^{3}, \exists \begin{pmatrix} x \\ y \\ z \end{pmatrix}\in \mathbb{R}^{3}:T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} a \\ b \\ c \end{pmatrix} \right\}$$
Now, you need to solve
$$\mathbf{im}(T)=\left\{\begin{pmatrix} a \\ b \\ c \end{pmatrix}\in \mathbb{R}^{3}, \exists \begin{pmatrix} x \\ y \\ z \end{pmatrix}\in \mathbb{R}^{3}:\underbrace{T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} a \\ b \\ c \end{pmatrix}}_{\color{blue}{solve}} \right\}$$
Now, using the definition of $\color{blue}{kernel}$ so, you have
$$\mathbf{ker}(T)=\left\{\begin{pmatrix} x\\ y \\ z \end{pmatrix} \in \mathbb{R}^{3}: T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \right\}$$
Now, you need to solve,
$$\mathbf{ker}(T)=\left\{\begin{pmatrix} x\\ y \\ z \end{pmatrix} \in \mathbb{R}^{3}: \underbrace{T\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}}_{\color{blue}{solve}}\right\}$$

Note that:

*

*If you solve (that is, find the conditions that fall on the associated systems of linear equations) and always look for the consistency of the system of linear equations (in the case of the image) and the solutions (in the case of the kernel), then you have solved the requested problem.


*Also, notice that in the problem you are not asked to find bases for the vector subspaces, they ask you to find the vector subspaces themselves.


*However if you want to find a basis for the kernel and a basis for the image, you start with $\mathbf{im}(T)$ and $\mathbf{ker}(T)$ as vector subspaces with their respective conditions, then you can easily find a basis for the vector subspaces.
