I need to find the matrix associated with the linear transformation:
$$T(a_{2}t^2+a_{1}t+a_{0})=4(a_{2}-a_{0})t+2a_{1}$$
with respect to the basis $B’$ (in the domain) and $B$ (in the target)
where $B=(1, 2t, 4t^2-1)$ and $B'=(1,t,t^2)$.
So the first thing I write is the required matrix, which is
$M=[[T(1)B'][T(t)B'][T(t^2)]]$
The answer sheet states that I need to find
\begin{align} T(1) &= -4t \\ T(t) &= 2 \\ T(t^2) &= 4t. \end{align}
and then solve them in order to find the matrix columns and thus the required matrix.
I know how to solve these three $T$’s in order to find a matrix, however my doubt is:
Where are these numbers $\boldsymbol{(-4,2,4)}$ coming from? Why does it state that I need to find $T(1)=-4$ or $T(t)=2$?
I do not understand, for example, why it is equating $T(1)=-4$. Where is the minus $4$ coming from? Where is the $2$ from $T(t)$ coming from? This is my doubt, so an answer on why $T(1)$ has the value of $-4$ or why $T(t)$ is equal to $2$ would be greatly appreciated.
This is what the answer sheet states:
$$T(\mathbf1)=\mathbf{-4t}=0\cdot\mathbf1-2\cdot\mathbf{2t}+0\cdot\mathbf{4t^2-1}\implies[T(\mathbf1)]_{\cal B'}=\begin{bmatrix}0\\-2\\0\end{bmatrix}$$ $$T(\mathbf t)=\mathbf2=2\cdot\mathbf1+0\cdot\mathbf{2t}+0\cdot\mathbf{4t^2-1}\implies[T(\mathbf{2t})]_{\cal B}=\begin{bmatrix}2\\0\\0\end{bmatrix}$$
$$T(\mathbf{t^2})=\mathbf{4t}=0\cdot\mathbf1+2\cdot\mathbf{2t}+0\cdot\mathbf{4t^2-1}\implies[T(\mathbf1)]_{\cal B'}=\begin{bmatrix}0\\2\\0\end{bmatrix}$$
Best Answer
It looks like you don't know yet how to read such a formula for a linear transformation.
The given linear transformation maps a quadratic polynomial of the form $a_2t^2+a_1t+a_0$ to the linear polynomial $4(a_2-a_0)t+2a_1$, which is given to you by way of the formula: $$T\left(\color{blue}{a_2}t^2+\color{purple}{a_1}t+\color{green}{a_0}\right)=4(\color{blue}{a_2}-\color{green}{a_0})t+2\color{purple}{a_1}$$ If you want to find the image of, for example, the polynomial $\color{blue}{2}t^2\color{purple}{-3}t+\color{green}{5}$, note that this is exactly of the form $\color{blue}{a_2}t^2+\color{purple}{a_1}t+\color{green}{a_0}$ with $\color{blue}{a_2=2}$, $\color{purple}{a_1=-3}$ and $\color{green}{a_0=5}$ and hence the image becomes: $$T\left(\color{blue}{2}t^2\color{purple}{-3}t+\color{green}{5}\right)=4(\color{blue}{2}-\color{green}{5})t+2(\color{purple}{-3}) = -12t-6$$ Carefully study this example; I used coloring to help see what numbers go where.
If you got this, can you find $T(1)$? Notice that "$1$" is also a polynomial of the form $a_2t^2+a_1t+a_0$, namely with $a_2=0$, $a_1=0$ and $a_0=1$; and $T$ maps it to...?