Decomposes the following linear operator in the direct sum of linear operators Let be $T : \mathbb{P}_2(\mathbb{R}) \longrightarrow \mathbb{P}_2(\mathbb{R})$ the linear operator $$ T(at^2 + bt + c) = (2a-b+c)t^2 + (a+c)t + 2c. $$ Write $T$ like a direct sum of operators.
My attempt:
I know that $T = T_1 \oplus T_2$, where $T_1$ is the linear operator restricted to the subspace $Ker$ $T^m$ and $T_2$ is the linear operator restricted to the subspace $Im$ $T^m$ with $m$ being the lowest such that $Ker$ $T$ = $Ker$ $T^m$. Knowing this, I computed $Ker$ $T$ and I found $Ker$ $T$ = $\{ 0 \}$. After, I found the matrix of $T$ in the canonical basis $$ [T]_{can} = \left( \begin{array}{crc} 2 & -1 & 1\\
1 & 0 & 1\\
0 & 0 & 2 \end{array} \right). $$
Next, I computed $[T]^2_{can}$ and found $$ [T]^2_{can} = \left( \begin{array}{crc} 3 & 2 & 3\\
2 & -1 & 3\\
0 & 0 & 2 \end{array} \right) $$ and $Ker$ $T^2 =\{ 0 \} $, but now I don't know how to find the matrices of $Ker$ $T^2$ and $Im$ $T^2$. How can I find this matrices? In this case, the matrix of $Ker$ $T$ = $ (0) $?
Thanks in advance!
 A: If $Tp = \lambda p$ where $p\ne 0$ then we have the system of equations
\begin{align}
\lambda a &= 2a-b+c\\
\lambda b &= a+c\\
\lambda c &= 2c.
\end{align}
Assuming $c\ne0$ yields $\lambda=2$, from which $b=c$ and hence $a=b$. So the eigenspace corresponding to $\lambda=2$ is $$\Lambda_2 = \left\{\alpha (t^2 + t + 1) : \alpha\in\mathbb R \right\}. $$
Assuming $a,b\ne0$ and $c=0$ yields $\lambda^2 b-2\lambda b + b=0$, from which $\lambda=1$ and hence $a=b$. So the eigenspace corresponding to $\lambda=1$ is $$\Lambda_1 = \left\{\alpha (t^2 + t):\alpha\in\mathbb R\right\}. $$
Now, the geometric multiplicity of the eigenvalue $1$ is less than its algebraic multiplicity, so we need to find a generalized eigenvector. Let $q(t) = t^2+t$. If $(T-I)^2 p = 0$ and $(T-I)p\ne 0$, then as $(T-I)q = 0$, we see that it suffices to solve $(T-I)p = q$. If $p(t)=at^2+bt+c$ then the system of equations is
\begin{align}
a-b+c&=1\\
a-b+c&=1\\
c&=0,
\end{align}
from which $c=0$ and $b=a-1$. Setting $a=1$, we have $q(t)=t^2$ as a generalized eigenvector corresponding to $\lambda=1$. So we have the generalized eigenspace
$$\tilde\Lambda_1 = \{\alpha(t^2+t) + \beta t^2:\alpha,\beta\in\mathbb R\}, $$
and hence $T$ can be written as the direct sum
$$T = \left.T\right|_{\tilde\Lambda_1} \oplus \left.T\right|_{\Lambda_2}. $$
