0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$
  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 \begin{equation} T(v) = T\left(\sum_{i=1}^{m}x_iv_i\right) = \sum_{i=1}^m x_iT(v_i) \end{equation} 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 $$ T(p(x)) = (a_0+a_1+a_2)\cdot (1,0) + (a_0+2a_1+4a_2)\cdot (0,1) = (a_0+a_1+a_2,a_0+2a_1+4a_2). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.