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?
Best Answer
$P_{B\to C} = P_{B\to S}P_{S\to C}$
$P_{S\to C} = P_{C\to S}^{-1}$
$P_{B\to C} = \begin{bmatrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 2 \\ 1 & -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 3 & -1 \\ 2 & -4 & 3 \\ 1 & 3 & 2 \end{bmatrix}$
$[p(t)]_{B}= \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}$
$[p(t)]_{C} = P_{B\to C} [p(t)]_B = \begin{bmatrix} 0 & 3 & -1 \\ 2 & -4 & 3 \\ 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}=\begin{bmatrix} 9 \\ 19 \\ -13 \end{bmatrix}$
or,
$p(t) = (1+t+t^2) - 2(2t-t^2) + 3(t+t^2) = -3 +4t + 6t^2 = -9(1-t) + 19(1+t^2) - 13(1-t+t^2)$