Question

# Consider the following linear transformation T : P_2 rightarrow

Alternate coordinate systems

Consider the following linear transformation $$T : P_2 \rightarrow P_3$$, given by $$T(f) = 3x^2 f$$'. That is, take the first derivative and then multiply by $$3x^2$$ (a) Find the matrix for T with respect to the standard bases of $$P_n$$: that is, find $$[T]_{\epsilon}^{\epsilon}$$, where- $$\epsilon = {1, x, x^2 , x^n}$$ (b) Find N(T) and R(T). You can either work with polynomials or with their coordinate vectors with respect to the standard basis. Write the answers as spans of polynomials. (c) Find the the matrix for T with respect to the alternate bases: $$[T]_A^B$$ where $$A = {x - 1, x, x^2 + 1}, B = {x^3, x, x^2, 1}.$$

Sulotion: Given that nebce that to the $$\displaystyle{T}:{P}_{{2}}\rightarrow{P}_{{3}}{T}{\left({f}\right)}={3}{x}^{{2}}{f}'$$ a) $$B = \left\{ 1, x, x^2 \right\}, \gamma = \left\{ 1, x, x^2, x^3 \right\} \ T(1) = 0.1 + 0.x + 0.x^2 + 0.x^3 \ T(x) = 3 x^2 \cdot 1 = 0.1 + 0 \cdot x + 3x^2 + 0 \cdot x^3 \ T(x^2) = = 3x^2 \cdot 2x = 6x^3 = 0.1 + 0.x + 0.x^2 + 0 x^3 \ [T]_B^{\gamma} = A = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 6\\ \end{bmatrix}$$ b. That to that the $$N(1) \cdot A x = 0 \ \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 6\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} \ 3x^2 = 0;\ 6x^3 = 0; \ x_2 = 0$$ Let then the $$n(1) = 5 pon | \left\{ 1, 0, x \right\}$$ spon let $$\displaystyle{P}{\left({x}\right)}={7}+{8}{x}+\gamma{x}^{{2}}\in{P}_{{2}}$$ hence that to Let then to $$T(P) = A - \begin{bmatrix} x\\ B \\ \gamma \end{bmatrix} = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 6\\ \end{bmatrix} \begin{bmatrix} x \\ 0B\\ \gamma \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 3B \\ 6 \gamma\\ \end{bmatrix}$$ Then the $$R(T) = span \left\{ (0, 0, 1, 0)^T (0, 0, 0)^T \right\}$$ is spon $$\left\{ x^2, x \right\}$$ c. $$A = \left\{ x^{-1}, x, x^2 + 1 \right\} B = \left\{ x^3, x, x^2 \right\}$$ Let the then $$T(x - 1) = 3 x^2 - 1 = 0.x^3 + 0.x + 3 x^2 + 0.1$$
$$T(x) = 3x^2 - (1) = 0.x^3 + 0.1$$
$$T(x^2 +1) = 3x^2(2x) = 6.x^3 + 0.x + 0.x^2 + 0.1$$
$$[T]_{A}^{B} = \begin{bmatrix}0 & 0 & 6 \\ 0 & 0 & 0\\ 3 & 3 & 0\\ 0 & 0 & 0\\ \end{bmatrix}$$