Let $T:\ P_2 \to P_1$ be a linear transformation defined by
$$T(a_0 + a_1 x + a_2 x^2) = (a_0 + a_1) – (2a_1 + 3a_2)x.$$
Find the matrix $T$ with respect to the standard bases $B = \{1, x, x^2\}$ and $B' = \{1, x\}$ for $P_2$ and $P_1$.
My Attempt
I'm not sure what to start with on this problem. I think that I should be plugging in $B$. But, then I'm not sure where $B$ belongs within the equation. Do I need to plug in the entire vector $B$ at once? For example, let $1 = a_0$, $x = a_1$, and $x^2 = a_2$? That would leave me with the answer $1 + x – 2x^2 – 3x^3$.
Best Answer
ZettaSuro has given you the explanation but let me add a bit more to help you understand the problem deeply. Consider the following example of a transformation.
$ T: R^{2}\rightarrow R^{2} \mbox{ given by } T(x_{1},x_{2})=2x_{1}+3x_{2} $
All this transformation is doing is taking a vector in the plane and scaling it in each respective direction i.e $x$ and $y$. Now there is one idea that needs to be emphasized behind these types of transformations.
Any linear transformation can be represented by a matrix or a matrix is really a linear transformation from one space to another
If we change our example to the one below
$ T: R^{2}\rightarrow R^{2} \mbox{ given by } T(x_{1},x_{2})=2x_{1}^{2}+3x_{2} $
then we no longer have a linear transformation(check for yourself) and there is no representation of the transformation using a matrix.
So how do you find the matrix that corresponds to a linear transformation? All you need to do is consider what the transformation does to the basis vectors. Let's see the first example above. In this case the basis vectors are $\begin{bmatrix}0\\1 \end{bmatrix}$ and $\begin{bmatrix}1\\0 \end{bmatrix}$.
$ T \begin{bmatrix}0\\1 \end{bmatrix}=\begin{bmatrix}0\\3 \end{bmatrix} $ and $ T \begin{bmatrix}1\\0 \end{bmatrix}=\begin{bmatrix}2\\0 \end{bmatrix} $
Now, here is the key. Consider any other vector $\begin{bmatrix}a\\b \end{bmatrix}$ on the plane and apply the transformation $$ T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=T\left(a\begin{bmatrix}1\\0 \end{bmatrix} +b\begin{bmatrix}0\\1 \end{bmatrix} \right)\\ T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=aT\begin{bmatrix}1\\0 \end{bmatrix} +bT\begin{bmatrix}0\\1 \end{bmatrix} \\ T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=\begin{bmatrix} T\begin{bmatrix}1\\0 \end{bmatrix}& T\begin{bmatrix}0\\1 \end{bmatrix} \end{bmatrix} \begin{bmatrix} a\\ \\b \end{bmatrix} $$
Now we can just read the matrix from the calculation and the matrix is \begin{bmatrix} 2 & 0\\ 0 & 3\\ \end{bmatrix}
You can apply this for your case and get the matrix below \begin{bmatrix} 1 & 1 & 0\\ 0 & -2 & 3\\ \end{bmatrix}