Change of basis for a polynomial

Question

So I am given the following.

$B=\{1+t+t^2,2-t^2,t+t^2\}$ and $C=\{1+t,1+t^2,1-t+t^2\}$ are two bases for $\mathbb P_2$. I have to find

a) The standard basis for $\mathbb P_2$ is $S=\{1,t,t^2\}$. I am told to find the matrices $P_{S\leftarrow B}$ and $P_{S\leftarrow C}$

b) The change of basis from $P_{C\leftarrow B}$

c) If $[p(t)]_{B}=\begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix} $, what is $[p(t)]_{C}$ ?

d) If $[q(t)]_{B}=\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix} $, what is $q(t)$ ?

Work for part $a)$ :

$P_{S\leftarrow B}=\begin{bmatrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & 1 \end{bmatrix} $

and

$P_{S\leftarrow C}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{bmatrix} $

I think that's okay. Since I am changing the basis back to the standard one, nothing special needs to be done.

Work for part $b)$ :

If both of these bases span $\mathbb P_2$ then I should be able to write the bases as a linear combination of each other I think.

$i)$ $1+t+t^2 = a_{1} (1+t) + a_{2} (1+t^2) + a_{3} (1-t+t^2)$

$ii)$ $2-t^2$ = $b_{1} (1+t) + b_{2} (1+t^2) + b_{3} (1-t+t^2)$

$iii)$ $t+t^2$ = $c_{1} (1+t) + c_{2} (1+t^2) + c_{3} (1-t+t^2)$

If I solve the coefficients, I get:

$a_{1}=0 , a_{2}=2 , a_{3}=-1$

$b_{1}=-3 , b_{2}=2 , b_{3}=-3$

$c_{1}=-1 , c_{2}=3 , c_{3}=-2$

So the matrix would just be:

$P_{C\leftarrow B}=\begin{bmatrix} 0 & 3 & -1 \\ 2 & 2 & 3 \\ -1 & -3 & -2 \end{bmatrix} $

Is that right? I was told something with inverses but I am not sure how that is related.

I have no idea what part $c)$ and $d)$ is asking though. Can someone guide me through what it's even asking? Thanks!

Also in general, what is a change of basis? How is it useful?

Answer 1

$P_{B\to C} = P_{B\to S}P_{S\to C}$

$P_{S\to C} = P_{C\to S}^{-1}$

$P_{B\to C} = \begin{bmatrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 2 \\ 1 & -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 3 & -1 \\ 2 & -4 & 3 \\ 1 & 3 & 2 \end{bmatrix}$

$[p(t)]_{B}= \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}$

$[p(t)]_{C} = P_{B\to C} [p(t)]_B = \begin{bmatrix} 0 & 3 & -1 \\ 2 & -4 & 3 \\ 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}=\begin{bmatrix} 9 \\ 19 \\ -13 \end{bmatrix}$

or,

$p(t) = (1+t+t^2) - 2(2t-t^2) + 3(t+t^2) = -3 +4t + 6t^2 = -9(1-t) + 19(1+t^2) - 13(1-t+t^2)$

Answer 2

If $M_1$ is a matrix achieving the change of basis $B \to S$ and $M_2$ achieves $C \to S $ then the change $S \to C$ is given by the inverse $M_2^{-1}$

Thus, $B \to C$ can be achieved in 2 steps as $B \to S \to C$ with the matrix $M_2^{-1}M_1$

(c) and (d) appear to give the polynomial in one basis and are asking you to work it out in the new basis.

Basis sets are minimal sets that allow you to describe fully the space you are working in e.g. the polynomials you have can be described by various basis sets . They of course appear in other places, commonly say in describing all position vectors in 3d where the standard cartesian basis is $\{\vec{i}, \vec{j}, \vec{k}\}$ However if you were describing e.g. crystals with particular oblique unit cells then that orthogonal basis might not be the most suitable and a change of basis to a non-orthogonal basis would simplify the description. The simplified description is why they can be useful.