find a linear transformation problem

Question

The problem goes like that: "find a linear transformation $T: \mathbb{R} _{4} \left[ x \right] \rightarrow\mathbb{R} _{4} \left[ x \right] $ so that $imT=kerT=sp \left\{1-x,x-x^3 \right\} $. wrote a formula for: $T \left( ax^3+bx^2+cx+d \right) $ " I did find that for base $B= \left\{ 1-x,x-x^3,x^2,1 \right\} $ my transformation can be defined like that: $T \left( 1-x \right)=T \left( x-x^3 \right)=0 $ and $T \left(x^2 \right)=1-x $, $T \left(1 \right)=x-x^3 $ . but I'm struggling to write formula for $T \left( ax^3+bx^2+cx+d \right) $

Answer 1

A general polynomial $ax^3 + bx^2 + cx + d$ can be written in terms of your basis as

$$ (-(a+c)) \cdot (1 - x) + (-a) \cdot (x - x^3) + b \cdot x^2 + (a + c + d)\cdot 1 $$

(compare the coefficients to see that this indeed holds). Hence,

$$ T(ax^2 + bx^2 + cx + d) = (-(a + c))T(1 - x) + (-a) T(x - x^3) + b \cdot T(x^2) + (a + c + d)T(1) \\ = b(1 - x) + (a + c + d)(x - x^3) = b + (a + c + d - b)x - (a + c + d)x^3. $$

Answer 2

If $T(1)=x-x^3$ and $T(1-x)=0$, then $T(x)=T(1)-T(1-x)=x-x^3$. And if $T(x)=x-x^3$ and $T(x-x^3)=0$, then $T(x^3)=T(x)-T(x-x^3)=x-x^3$. So\begin{align}T(ax^3+bx^2+cx+d)&=aT(x^3)+bT(x^2)+cT(x)+dT(1)\\&=a(x-x^3)+b(1-x)+c(x-x^3)+d(x-x^3)\\&=(-a-c-d)x^3+(a-b+c+d)x+b.\end{align}