On the standard Matrix of a transformation from $\mathcal{P}_2$ to $\mathcal{P}_2$ in a given basis

Question

$\mathcal{P}_2$ is the space of all the polynomials $p(t)$ of degree $\leqslant 2$.

Problem: Transformation $F: \mathcal{P}_2 \rightarrow \mathcal{P}_2$ is defined as $$ F(\mathbf{p})=F(p(t))=p(t-1) \text { where } p(t)=x_1+x_2 t+x_3 t^2 . $$

1. Determine the standard matrix of $F$ in the basis $$ B=\left\{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3\right\}=\left\{t, t+1, t^2-1\right\} $$

Solution :

$F\left({b}_1\right)=b_1(t-1)=t-1= 2b_1 -b_2 $

$F(b_2)={b_2}(t-1)=(t-1)+1=t=b_1$

$F(b_3)=b_3(t-1)=(t-1)^2-1 = -3 b_1+b_2+b_3$

$A=\left[\begin{array}{ccc} 2 & 1 & -3 \\ -1 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]$

Question:

Why it isn't correct to find the standard matrix by calculating it like this :

$$\begin{aligned} & P_2 \rightarrow P_2 \\ & \begin{aligned} F(p) & =F(p(t))=p(t-1) \\ \end{aligned} \\ &\begin{array}{rl} F(p) & =x_1+x_2(t-1)+x_3(t-1)^2 \\ & =x_1+x_2 t-x_2+x_3 t^2+x_3-2 x_3 t\\ & [F]_B X =x_1\left(\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right)+x_2\left(\begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right)+x_3\left(\begin{array}{c} -4 \\ 2 \\ 1 \end{array}\right) \end{array} \end{aligned}$$

And so we would have :

$[F]_B=\left(\begin{array}{ccc} -1 & 2 & -4 \\ 1 & -1 & 2 \\ 0 & 0 & 1 \end{array}\right)$

But this is incorrect and I don't see why given that in a similar manner for another transformation it would be correct :

If we take another example of a transformation: $F: \mathcal{P}_2 \rightarrow \mathbb{R}^3$ defined by $$ F(p)=\left(\begin{array}{c} p(0) \\ p(0)+p(1) \\ p^{\prime}(-1) \end{array}\right) \quad \text { for } \quad p(t)=x_1+x_2 t+x_3 t^2 \in \mathcal{P}_2 $$

It would be correct to find it this way :

We have : $\begin{aligned} & p(t)=x_1+x_2 t+x_3 t^2 \\ & p^{\prime}(t)=x_2+2 x_3 t \end{aligned}$

So: $p(0)=x_1 \\ p(1)=x_1+x_2+x_3\\ p^{\prime}(-1)=x_2-2 x_3$

Then : $F(p)=\left(x_1, 2 x_1+x_2+x_3, x_2-2 x_3\right)$

The standard matrix of $F$ would be : $[F]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 1 & -2 \end{array}\right]$

Answer 1

Your old basis is $e=(1, t, t^2)$ and a new one is $e'=(t, t+1, t^2-1)$.

The image of the transformation in the old basis: $F(p)=(x_1-x_2+x_3)+(x_2-2x_3)t+x_3t^2$. So the coordinates of the image $F(p)$ in old basis is given by $\begin{pmatrix} x_1-x_2+x_3 \\ x_2-2x_3 \\ x_3 \end{pmatrix}$.

To find the coordinates of the image in a new basis you need to find the change of basis matrix from old to new basis and take its transpose or directly find the change of basis matrix from new to old basis. Let's use the first approach. The matrix from old to new is found in the next two steps:

1. Find coordinates of a new basis vectors in an old basis:

$t=0\cdot 1+ 1\cdot t + 0\cdot t^2$

$t+1=1\cdot 1+ 1\cdot t + 0\cdot t^2$

$t^2-1=-1\cdot 1+ 0\cdot t + 1\cdot t^2$

2. Therefore, the change of basis matrix from old to new is:

$C=\begin{pmatrix} 0&1&-1 \\ 1&1&0 \\ 0&0&1 \end{pmatrix}$

To find coordinates of an image in a new basis use the following formula:

From $[Fp]_{e}=C[Fp]_{e'}$ it follows that

$[Fp]_{e'}=C^{-1}[Fp]_{e}$

Therefore:

$[Fp]_{e'}=\begin{pmatrix} -1&1&-1 \\ 1&0&1 \\ 0&0&1 \end{pmatrix}\begin{pmatrix} x_1-x_2+x_3 \\ x_2-2x_3 \\ x_3 \end{pmatrix} = \begin{pmatrix} -x_1+2x_2-4x_3 \\ x_1-x_2+2x_3 \\ x_3 \end{pmatrix}$

Thus, applying these coordinates to a new basis you get:

$Fp=(-x_1+2x_2-4x_3)t+(x_1-x_2+2x_3)(t+1)+x_3(t^2-1)$

You could have obtained the same result using the second approach by computing the change of basis matrix from new to old:

1.

$1=-1\cdot t + 1\cdot (t+1) + 0\cdot (t^2-1)$

$t=1\cdot t + 0\cdot (t+1) + 0\cdot (t^2-1)$

$t^2 = -1 \cdot t + 1 \cdot (t+1) + 1 \cdot (t^2-1)$

2. The change of basis matrix from new to old

$C'=\begin{pmatrix} -1&1&-1 \\ 1&0&1 \\ 0&0&1 \end{pmatrix}$

Therefore, using the formula

$[Fp]_{e'}=C'[Fp]_{e}$ we get

$[Fp]_{e'}=\begin{pmatrix} -1&1&-1 \\ 1&0&1 \\ 0&0&1 \end{pmatrix}\begin{pmatrix} x_1-x_2+x_3 \\ x_2-2x_3 \\ x_3 \end{pmatrix} = \begin{pmatrix} -x_1+2x_2-4x_3 \\ x_1-x_2+2x_3 \\ x_3 \end{pmatrix}$