If $L=\{t^2 , t , 1\}$ and $L_1=\{t^2 + 1, t-2 , t+3\}$ are basis for $\text{Poly}_2$ and $w=[1,2,3]_{L_1}$ (sub $L_1$). Find the coordinates of $w$ with respect to L directly and by using $P^{-1}$ ($P$-inverse). Where $P$ is the coordinate change matrix from the basis L to the basis $L_1$.
Unfortunately what you're about to read is as far as I got. I just remember from my Linear Algebra class there was something with setting $w$ equal to $L_1$ and being able to solve. I've been looking at my old Linear Algebra book and have ordered $3$ additional books through the school library, and I'm beyond lost. I really appreciate your help/input/interest. This is my first time, so I'm unsure if you even want to know what's going on on my paper when such a disaster as mine.
Do I set $w$ equal to $L_1$, since that's its base
$1=a(t^2+1)+b(t-2)+c(t+3)$
$2=a(t^2+1)+b(t-2)+c(t+3)$
$3=a(t^2+1)+b(t-2)+c(t+3)$
After finding $a,b,c$ need to set numbers equal to $L_1$ ,
$a=t^2+1$
$b=t-2$
$c=t+3$ ;
from here can solve for $t^2 , t , 1$ (yeah not sure about solving for $1$ either if that's what you're thinking)
Not sure about $P$ inverse, technique since having problems finding actual $P$.
By definition:
$w = [1,2,3]_{L_1} = 1(t^2 + 1) + 2(t - 2) + 3(t + 3)$
$= t^2 + 1 + 2t - 4 + 3t + 9 = t^2 + (2 + 3)t + (1 - 4 + 9)$
$= t^2 + 5t + 6 = 1(t^2) + 5(t) + 6(1) = [1,5,6]_L$.
The matrix for $P^{-1}$ is actually easier to write down than the matrix for $P$, its columns are just the coordinates of the $L_1$-basis in $L$-coordinates, so:
$P^{-1} = \begin{bmatrix}1&0&0\\0&1&1\\1&-2&3\end{bmatrix}$
Thus $[w]_L = P^{-1}([w]_{L_1})$
$= \begin{bmatrix}1&0&0\\0&1&1\\1&-2&3\end{bmatrix}\begin{bmatrix}1\\2\\3\end{bmatrix} = \begin{bmatrix}1\\5\\6\end{bmatrix}$
as we found before.