How to find the kernel of a linear operator given the representative matrix the standard bases? Our professor gave us this exercise:
Consider a linear operator $T:\mathbb{R}^2→\mathbb{R}^2$ and let $A=\begin{bmatrix} 2 & -2 \\
 2 & -2 \end{bmatrix}$
be its representative matrix on the standard bases. Find ker T.
The solution he gave us is the following:
We have $dim Im T = 1 ⇒ dim ker T = 2 − 1 = 1$. We get
$ker T = \Bigl\{\textbf{x} ∈ \mathbb{R}^2:\textbf{x} = α\begin{bmatrix}1\\1\end{bmatrix} \Bigl\}  $
My problem is that I don't really understand what's the "role" of the representative matrix and how we can use it to solve this type of problems. For example the definition of kernel is the set of vectors $\textbf{x}\in\mathbb{R}^n$ such that $T(\textbf{x})=\textbf{0}$. How does the representative matrix "represent" the linear operator in a way such that to find the kernel (like in this case) we can simply do $A\textbf{x}=\textbf{0}$?
I hope I'm making myself somewhat clear
 A: The representative matrix represents a linear function $f:\mathbb R^2 \to \mathbb R^2$.
We use column vectors like $\begin{pmatrix} x \\ y\end{pmatrix}$ to represent points $(x,y)\in\mathbb R^2$ because we apply functions on the left which corresponds to multiplying the matrix by the column vector.
For your matrix, we can compute the function by $$\begin{pmatrix}
2 & -2 \\
2 & -2 \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 2x-2y \\ 2x-2y\end{pmatrix}$$
In standard function notation this gives $$f(x,y)=(2x-2y, 2x-2y)$$
When you find the kernel, you are looking for points $(x,y)$ such that $f(x,y)=0$.
We get, as you have shown, $$(2x-2y,2x-2y)=(0,0)\implies x=y$$
Therefore, your result shows that any point $(x,x)$ is in the kernel.
A: You can compute $\ker T$ just from the definitions, more directly than your teacher did (and write the final result more rigorously).
Let ${\bf x}=\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb R^2$. By definition of representative matrices,
$$T({\bf x})=A{\bf x}=\begin{bmatrix}2&-2\\2&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2x-2y\\2x-2y\end{bmatrix}.$$
Hence by definition of the kernel (which you correctly recalled),
$$\ker T=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb R^2:2x-2y=0\right\}=\left\{\begin{bmatrix}x\\x\end{bmatrix}:x\in\mathbb R\right\}=\left\{\alpha\begin{bmatrix}1\\1\end{bmatrix}:\alpha\in\mathbb R\right\}.$$
