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I'm trying to understand the solution to the question below. I warrant I'm probably confused over the notation.

In the question (attached below) it says that transformation T(p)[x] = xp(x-3), with standard basis for P2 and P3.

I assume you need to break it down into:

Po = 1 + 0x + 0x^2

P1 = 0 + 1x + 0x^2

P2 = 0 + 0x + x^2

And apply the transformation to each Pn. However, I have no idea how you would do that stepwise when the transformation is defined by xp(x-3).

Any help would be greatly appreciated!

Thanks

Polynomial Question

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2 Answers

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When you want to change basis, the columns of the transformation matrix are made up by applying the transformation on each element of the basis of your domain space, to get the corresponding column of your transformation matrix when expressed in the basis of $P3$.

Now the standard domain of $P2$ is $\{(1,0,0), (0,1,0) ,(0,0,1)\}$ where each element represents the coefficient of $1, x, x^2$ respectively

Now, let these "vectors" be $P_1, P_2, P_3$ (as each represents a polynomial)

Hence, $T(P_1) = xP_1(x-3) = x = (0,1,0,0)$

$T(P_2) = xP_2(x-3) = x(x-3) = x^2-3x = (0,-3,1,0)$

$T(P_3) = xP_3(x-3) = x((x-3)^2) = x^3 - 6x^2 + 9x = (0,9,-6,1)$

Hence, we can write the standard matrix representation of the transformation T as

$$\begin{bmatrix}0 & 0 & 0 \\ 1 & -3 & 9 \\ 0 & 1 & -6 \\ 0 & 0 & 1\end{bmatrix}$$

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The transformation is defined by $$T(p)(x)=xp(x-3),$$and we need to find the image of the standard basis vectors (of $\mathbf P^2$) under $T$, in terms of the standard basis of the image space $\mathbf P^3$. So we need to separately compute $T(p_0),T(p_1),T(p_2)$ and see how we can express this as a linear combination of $\{p_0,p_1,p_2,p_3\}$. For example, $$T(p_1)(x)=xp_1(x-3)=x(x-3)=x^2-3x=p_2(x)-3p_1(x).$$(Because $p_2(x):=x^2$ and $p_1(x):=x$.) Can you figure out how to do the rest?

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