Finding the matrix representation of a linear operator $T$ for $T(f(x))=2(f(x))-f'(x)$ in $\mathbb{P}_2(\mathbb{R})$

Question

Using the standard basis $\{1,x,x^2\}$ gives me a matrix however it doesn't work when working towards Jordan Canonical Form with the matrix. What I get is $\begin{pmatrix} 2 \\ 0 \\ 0\end{pmatrix}\begin{pmatrix} -1 \\ 2 \\ 0\end{pmatrix}\begin{pmatrix} 0 \\ -2 \\ 2\end{pmatrix}$ as my columns. Where did I make a mistake?

Answer 1

Your matrix representation for the operator $T$ is correct. This operator has 2 as eigenvalue with algebraic multiplicity 3, but geometric multiplicity 1. In other words: the eigenspace has dimension one and the matrix is not diagonalizable. Thus the Jordan normal form is: $$ \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}. $$