# How do I find the transformation matrix T with respect to the following bases?

Let $P_2$ be the vector space of polynomials of degree ≤ 2. A function $T : P_2 \to {\mathbb R}^2$ is defined by: $$T(p(x))=(p(1),p(2))$$ How do I find the transformation matrix with respect to the standard bases $B=\{1,x,x^2\}$ and $S=\{(1,0),(0,1)\}$?

I don't fully understand what that transformation represents and hence I am unable to proceed with the problem.

Hint: for any polynomial $$p(x)=ax^2+bx+c=\begin{pmatrix}x^2 & x & 1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$ we should calculate the values at the points $x=1$ and $x=2$. For example, $$p(1)=\begin{pmatrix}1 & 1 & 1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}.$$ Then $(p(1),p(2))=A(1,0)+B(0,1)$. Can you represent the transformation $T$ from $(a,b,c)$ to $(A,B)$ as a matrix multiplication with some matrix?
1. What you must understand is that every transformation "acts" as the product of a matrix by a vector. Let a linear transformation between two vecorial spaces $T\colon V\to W$, we fix a base for $V$ given by $v_1, \dots, v_m$ and also a base for $W$ given by $w_1, \dots, w_n$. Let $v$ in $V$, note that of the concepts of base and linearity $$T(v) = T\left(\sum_{i=1}^{m}x_iv_i\right) = \sum_{i=1}^m x_iT(v_i)$$ then, just know \begin{align}\label{2} T(v_i)= \sum_{j=1}^n a_{ji}w_j; \quad i=1, \dots, m, \end{align} as a linear combination of the base in $W$, to determine $T(v)$, and therefore $T$. These last equations define a matrix $A=[a_{ij}]$, which depends on the bases of $V$ and $W$, and allow writing \begin{align}\label{3} \sum_{i=1}^m x_i T(v_i) = \sum_{i=1}^m x_i \left(\sum_{j=1}^n a_{ji} w_j \right) = \sum_{j=1}^n \left(\sum_{i=1}^m a_{ji}x_i \right) w_j. \end{align} This suggests that a linear transformation can be thought of as a matrix that transforms the vector $x\in \mathbb{R}^m$ into a vector $y=Ax\in \mathbb{R}^n$ so that $$T(v)=\sum_{j=1}^{n}y_jw_j, \quad \text{if} \quad v=\sum_{i=1}^{m}x_iv_i.$$
2.For your problem: \begin{align*} T(1)=(1,1) &= 1\cdot(1,0)+1\cdot(0,1),\\ T(x)=(1,2) &= 1\cdot(1,0)+2\cdot(0,1),\\ T(x^2)=(1,4) &= 1\cdot(1,0)+4\cdot(0,1), \end{align*} and therefore $$A = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \end{array} \right).$$ Let $p(x)$ a polynomial of degree $\le 2$ then $$p(x)=a_0+a_1x+a_2x^2=a_0\cdot 1+a_1\cdot x + a_2\cdot x^2,$$ and of $$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \end{array} \right) \left( \begin{array}{c} a_0 \\ a_1 \\ a_2 \end{array} \right) = \left( \begin{array}{ccc} a_0+a_1+a_2 \\ a_0+2a_1+4a_2 \end{array} \right)$$ it follows $$T(p(x)) = (a_0+a_1+a_2)\cdot (1,0) + (a_0+2a_1+4a_2)\cdot (0,1) = (a_0+a_1+a_2,a_0+2a_1+4a_2).$$