The function T is given by
T(M) = M$\begin{bmatrix}1, 2\\0, 1\end{bmatrix}$ – $\begin{bmatrix}1, 2\\0, 1\end{bmatrix}$M
this is one part of a question (part c), and the part before this (part b) asks for the basis matrix B of T with respect to the basis
basis = ($\begin{bmatrix}1, 0\\0, 1\end{bmatrix}$, $\begin{bmatrix}0, 1\\0, 0\end{bmatrix}$, $\begin{bmatrix}1, 0\\0, -1\end{bmatrix}$)
for which I found B = $\begin{bmatrix}0, 0, 0\\0, 0, 4\\0,0,0\end{bmatrix}$ by plugging the basis into T and taking constant multiples out of each product so that it was some constant c times each one of the bases, and those constants c were then the columns of B, if that makes sense..
My main problem is I don't know what to find the kernel of (i.e. do I find the kernel of B since that's the basis matrix of T? or do I need some other matrix?) and I'm also kind of thrown off by the basis in part b, if that has anything to do with this part.
I appreciate any help at all! Thanks in advance!
Best Answer
It is possible to find the kernel of $T$ directly. However, it should be noted that basis was specifically chosen in order to make it easier to find the kernel using $B$. So, with that said:
We begin by finding the nullspace of the matrix $B$. That is, we want the solutions to the equation $Bx = 0$. Without too much effort, you should find that the elements of the nullspace are the vectors of the form $x = x_1(1,0,0) + x_2(0,1,0) = (x_1,x_2,0)$ (the vectors here are column vectors written sideways for my convenience).
Now, let's translate this solution into a statement about $T$. Let $M_1,M_2,M_3$ denote the matrices of the given basis. The entire point of the matrix $B$ is that we have $$ T(x_1 M_1 + x_2 M_2 + x_3 M_3) = y_1 M_1 + y_2 M_2 + y_3 M_3 \iff B \pmatrix{x_1\\x_2\\x_3} = \pmatrix{y_1\\y_2\\y_3}. $$ So, by finding the solutions to $Bx = 0$, we have found all $x_1,x_2,x_3$ for which $M = x_1 M_1 + x_2 M_2 + x_3 M_3$ satisfies $T(M) = 0$. In other words, we have described all matrices $M$ of the kernel of $T$. In particular, we can say that $M$ is an element of the kernel iff its coordinate vector has the form $(x_1,x_2,0) = x_1\cdot (1,0,0) + x_2\cdot(0,1,0)$, which holds iff $$ M = x_1 M_1 + x_2 M_2 = \pmatrix{x_1&x_2\\0&x_1}. $$ Likewise, we can use the column space of $B$ to describe the image of $T$. The column space of $B$ consists of elements of the form $(0,t,0)$ for some $t \in \Bbb R$. Correspondingly, the image of $T$ consists of elements of the form $t M_2$ for some $t \in \Bbb R$.