I have that $V = \mathcal{P}_2(R)$ and $T \in \mathcal{L}(V)$ where $T$ is defined by $T(p) = xp', p \in V.$
In a previous problem, I found that the matrix for $T$ $w.r.t.$ the basis $1, x, x^2$ is $$M(T) = \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{bmatrix}.$$
I am given the matrix for $T$ with respect to an unknown basis (which is the same for the domain and codomain). The matrix is $$M'(T) = \begin{bmatrix} 2 & -3 & 6\\ 0 & 1 & 4\\ 0 & 0 & 0 \end{bmatrix}.$$
I must find a basis of $V$ such that it satisfies the matrix $M'(T)$.
I tried approaching this problem by solving systems of equations for the three elements of the unknown basis, but I got stuck on how to apply this to $M'(T)$. Also, I was unable to solve the system of unknowns as I assumed each element of the basis was of the form $ax^2+bx+c$, where $a,b,c\in F$.
How should I approach this problem? Any help would be greatly appreciated.
Use eigen decomposition.
The characteristic polynomial for $M'$ is
$p(t) = (2-t)(1-t)t$.
The characteristic values are $\lambda = 0, 1, 2$.
Now find the basis of null space of $M', M'-I, M'-2I$.
Those are
$ \begin{bmatrix} 9 \\ 4 \\ -1 \end{bmatrix} $, $ \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} $, $ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $
corresponding to $\lambda=0,1,2$.
Now M' =
$ \begin{bmatrix} 9 & 3 & 1 \\ 4 & 1 & \\ -1 & & \end{bmatrix} \begin{bmatrix} 0 & & \\ & 1 & \\ & & 2 \end{bmatrix} \begin{bmatrix} 9 & 3 & 1 \\ 4 & 1 & \\ -1 & & \end{bmatrix}^{-1} $.
Now denote the old basis as $v_1=1,v_2=x, v_3=x^2$ and new basis for $M'$ as $w_1, w_2, w_3$. Then $(v_1, v_2, v_3)=(w_1, w_2, w_3) \begin{bmatrix} 9 & 3 & 1 \\ 4 & 1 & \\ -1 & & \end{bmatrix}$.
$w_1 = v_3, 3w_1+1w_2 = v_2, 9w_3+4w_1-w_3=v1$.
Can you proceed from here?
(answer: $w_1 = x^2, w_2=x-3x^2, w_3=-1+4x-3x^2$)