Given a transformation matrix $L$, how can I find a basis $B$ for which the same transformation is the identity matrix $I$?

Question

$L: V \to W$ where $V$ is the vector space of all polynomials with degree $n \leq2$, $W = \mathbb R^3$ and $L$ is the linear transformation that maps each polynomial $p \in V$ to the vector $w \in W$ where $w_1, w_2, w_3$ correspond to $p$ evaluated at $-1, 0, 1$ respectively.

For the basis $(x^2,x,1)$ the matrix $L$ should be equal to $\begin{pmatrix}1 & -1 & 1 \\0 & 0 & 1 \\1 & 1 & 1 \\\end{pmatrix}$. Now I need to find a basis $B$ such that $L'=I$ where $I$ is the identity matrix.

My reasoning is that we could have a change of basis matrix $A$ and then $A^{-1}LA=I$, if we find $A^{-1}$ we can find $B$ since $B=A^{-1}$. If we multiply $A^{-1}LA=I$ by $A$ on both sides we get $LA=A$ and hence $L=I$ which is a contradiction. Where is my error?

Answer 1

Hint:

Think of Lagrange's interpolation polynomial: you have to find quadratic polynomials $p_1,p_2,p_3$ such that $$\begin{cases}p_1(-1)=1,& p_1(0)=p_1(1)=0, \\[1ex] p_2(0)=1,& p_2(-1)=p_2(1)=0, \\[1ex] p_3(1)=1,& p_1(-1)=p_1(0)=0. \end{cases}$$

Answer 2

You can't find such a change of basis, since then the only eigenvalue would be $1$. In fact the characteristic polynomial is $(x-1)^2x-1-1-x=x^3-2x^2-2$, so $1$ is not an eigenvalue at all (much less the only one).

You and @uniquesolution correctly noted that if there were such a change of basis, then your original matrix would be the identity, since we would have $ALA^{-1}=I\implies L=A^{-1}IA=I$.

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However, as @Bernard notes, it is possible if you allow a different basis for the domain and the range. Change the basis for $V$ to $\{1/2x(x-1), -(x+1)(x-1),1/2x(x+1)\}$, so that the change of basis matrix is $L^{-1}$.