The question I'm given is:
For each matrix $A$ find a basis for the kernel and image of $T_A$, and find the rank and nullity of $T_A$.
$$A=\begin{matrix} 2 & 1 & 0 \\ 1 & -1 & 3 \\ 1 & 2 & -3 \\ 0 & 3 & -6\\ \end{matrix}$$
I know the answer is given by the following:
$$[-1,2,1]^T , \{[1,0,1,1]^T,[0,1,-1,-2]^T\}; 2,1$$
Answers being given in the order they are asked.
I don't know how to get to these answers, I tried looking through my text book and all the examples are very sparse with their detail so I couldn't get a hold of the process for getting a hold of these answers. Could somebody possibly walk me through how this is done?
What is the rank of $A$? How many independent column vectors are there? If all three columns are independent then the kernel is trivial.
If the rank is less than the number of columns then the dimension of the kernel equals this difference.
Whatever row operations that eliminate a column will point us to a vector in the kernel.
If we label the columns $v_1,v_2, v_3$ we see that $v_3 = v_2 - 2v_2$ or $v_1 - 2v_2 - v_3 = 0$
And this is tells us our kernel $\begin{bmatrix} 1\\-2\\-1\end {bmatrix}$
The way I found it gave the negative of what you show above, but they span the same space.
Now that we know that there are 2 independent column vectors, any 2 will span the image.