Find all values $a\in\mathbb{R}$ such that vector space $V=P_2(x)$ is the sum of eigenvectors of linear transformation $L: V\rightarrow V$. Find all values $a\in\mathbb{R}$ such that vector space $V=P_2(x)$ is the sum of eigenvectors of linear transformation $L: V\rightarrow V$ defined as $L(u)(x)=(4+x)u(0)+(x-2)u'(x)+(1+3x+ax^2)u''(x)$. $P_2(x)$ is the space of polynomials of order $2$.
Attempt:
First, we find the matrix of $L$ (choose a standard basis $\{1,x,x^2\}$).
$$L(1)\Rightarrow L(u)(x)=0x^2+1x+4\Rightarrow [L(1)]=        \begin{bmatrix}
        4  \\
        1  \\
        0  \\
        \end{bmatrix}$$
$$L(x)=0x^2+1x-2\Rightarrow [L(x)]=        \begin{bmatrix}
        -2  \\
        1  \\
        0  \\
        \end{bmatrix}$$
$$L(x^2)=(2+2a)x^2+2x+2\Rightarrow [L(1)]=        \begin{bmatrix}
        2  \\
        2  \\
        2+2a  \\
        \end{bmatrix}$$
$$\Rightarrow [L]_{\mathcal{B}}=        \begin{bmatrix}
        4 & -2 & 2 \\
        1 & 1 & 2 \\
        0 & 0 & 2+2a \\
        \end{bmatrix}$$
Next, we find eigenvalues and eigenvectors of $[L]_{\mathcal{B}}$:
$$\det([L]_{\mathcal{B}}-\lambda I)=(2+2a-\lambda)(\lambda-3)(\lambda-2)\Rightarrow$$ eigenvalues are $\lambda_1=2+2a,\lambda_2=3,\lambda_3=2,a\neq 0,a\neq \frac{1}{2}$.
Corresponding eigenvectors are $v_1=\begin{bmatrix}
        1  \\
        1  \\
        a  \\
        \end{bmatrix},v_2=\begin{bmatrix}
        2  \\
        1  \\
        0  \\
        \end{bmatrix},v_3=\begin{bmatrix}
        1  \\
        1  \\
        0  \\
        \end{bmatrix}$.
Question: What does it mean that the space $V$ is the sum of eigenvectors of $L$?
 A: More precisely, $V$ has a basis consisting of eigenvectors. It means the linear map is diagonalizable.
However, your analysis is not complete: if $a\ne0$ and $a\ne1/2$, the matrix has three distinct eigenvalues, so it is diagonalizable.
For $a=0$, you have to see whether the eigenspace relative to $2$ has dimension $2$ (equal to the multiplicity of the eigenvalue). The matrix to compute the rank of is
$$
[L]_{\mathcal{B}}-2I=
\begin{bmatrix}
2 & -2 & 2 \\
1 & -1 & 2 \\
0 & 0 & 0
\end{bmatrix}
$$
Since this matrix has rank $2$, the eigenspace has dimension $3-2=1$, so the matrix is not diagonalizable.
For $a=1/2$, we need to consider the dimension of the eigenspace relative to $3$, so the matrix to look at is
$$
[L]_{\mathcal{B}}-3I=
\begin{bmatrix}
1 & -2 & 2 \\
1 & -2 & 2 \\
0 & 0 & 0
\end{bmatrix}
$$
This matrix has rank $1$, so the eigenspace has dimension $3-1=2$ and the matrix is diagonalizable. I'll leave to you computing a basis in this case.
A: You are asked when $V$ is a direct sum of the eigenspaces of the operator $L$. This means that every vector can be written (uniquely) as a sum of eigenvectors of $L$ associated to different eigenvalues. Alternatively, by the identification of matrices and linear transformations, you need to find when every vector in $\mathbb{R}^3$ can be written (uniquely) as a sum of eigenvectors of the matrix $[L]_{\mathcal{B}}$.
If $a \neq 0$ and $a \neq \frac{1}{2}$, then the operator and the matrix have three distinct eigenvalues and thus the operator and the matrix are diagonalizable. If $a = 0$, then there are two distinct eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 3$. You need to check whether the eigenspace of the matrix associated to the eigenvalue $\lambda_1 = 2$ is two dimensional. In your case, the eigenspace is
$$ \ker \, \begin{pmatrix} 2 & -2 & 2 \\ 1 & -1 & 2 \\ 0 & 0 & 0 \end{pmatrix} $$
and since the matrix has rank two, it has one-dimensional kernel and so the eigenspace is one-dimensional and the matrix (and operator) is not diagonalizable.
Similarly, you need to check what happens if $a = \frac{1}{2}$. 
