Let T be linear map on $P_2(R)$. Find the eigenvalues of T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix. Let T be linear map on V. Find the eigenvalues of T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix.
$V = P_2(R), T(f(x)) = xf'(x)+f(2)x+f(3)$.
I was able to create a basis for the set of eigenvectors, $\alpha = \{x−3,4x^2−13x−3,x+1\}$, but how would I construct $[T]_\beta$ so that it is a diagonal matrix?
 A: An eigenvalue of a linear operator $T$ consists of a scalar $\lambda$ such that $Tv = \lambda v$, for some $v\in V\backslash\{0\}$.
This is the same as to require that $\ker(T - \lambda I) \neq \{0\}$. This happens iff $\det([T]_{\mathcal{B}} - \lambda I) = 0$ for some (arbitrary) matricial representation of $T$.
At your case, $V = P_{2}(\textbf{R})$. If we use the basis $\mathcal{B} = \{1,x,x^{2}\}$ and $f(x) = a + bx + cx^{2}$, then
\begin{align*}
T(f(x)) & = xf'(x) + f(2)x + f(3)\\\\
& = x(b + 2cx) + (a + 2b + 4c)x + (a + 3b + 9c)\\\\
& = a + 3b + 9c + (a + 3b + 4c)x + 2cx^{2}
\end{align*}
Consequently, $T(1) = 1 + x$, $T(x) = 3 + 3x$ and $T(x^{2}) = 9 + 4x + 2x^{2}$, whence
\begin{align*}
[T]_{\mathcal{B}} = 
\begin{bmatrix}
1 & 3 & 9\\
1 & 3 & 4\\
0 & 0 & 2
\end{bmatrix} \Longrightarrow \det([T]_{\mathcal{B}} - \lambda I) & = \begin{vmatrix}
1 - \lambda & 3 & 9\\
1 & 3 - \lambda & 4\\
0 & 0 & 2 - \lambda
\end{vmatrix} = \lambda(\lambda-4)(2-\lambda)
\end{align*}
Once the characteristic polynomial splits and the algebraic multiplicity of each eigenvalue equals one, we conclude that $T$ is diagonalizable.
Now it remains to determine the corresponding eigenspaces.

First case: $\lambda = 0$

\begin{align*}
\begin{bmatrix}
1 & 3 & 9\\
1 & 3 & 4\\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \Longleftrightarrow (x,y,z) = (-3t,t,0)
\end{align*}

Second case: $\lambda = 4$

\begin{align*}
\begin{bmatrix}
-3 & 3 & 9\\
1 & -1 & 4\\
0 & 0 & -2
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \Longleftrightarrow (x,y,z) = (t,t,0)
\end{align*}

Third case: $\lambda = 2$

\begin{align*}
\begin{bmatrix}
-1 & 3 & 9\\
1 & 1 & 4\\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \Longleftrightarrow (x,y,z) = (-3t,-13t,4t)
\end{align*}
Therefore $\mathcal{B}' = \{(-3,1,0),(1,1,0),(-3,-13,4)\}$ is a basis of eigenvectors.
Thence we conclude that $[T]_{\mathcal{B}'}$ is a diagonal matrix, just as desired.
