Self-adjoint operator and eigenvalues Let the real vector $P_2(\mathbb{R})$ consisting of real polynomials of degree $\le1$. In the following we define $P_2(\mathbb{R})$ as an inner product space by the inner product 
$$ \langle p,q \rangle = p(0)q(0)+p(1)q(1)$$
for $p,q\in P_2(\mathbb{R})$. Let the mapping be: 
$$L: P_2(\mathbb{R}) \to P_2(\mathbb{R})$$
defined by 
$$ L(\alpha+\beta X) = (8\alpha  + 2\beta)+(\beta-3\alpha)X$$
for $\alpha, \beta \in \mathbb{R}.$
1) Show that L is self-adjoint.
2) Explain why $L$ is orthonormal diagonalizable and find all eigenvalues for $L$ and all the basis for the corresponding eigenspace.

1) In order to show that the linear operator is self-adjoint it must be shown that $\langle v,L(w)\rangle = \langle L(v),w \rangle$. Now let $v = (\alpha+\beta X)$ and let $w = (c+dX)$. 
\begin{align*}\langle v,L(w)\rangle  &= \langle (\alpha + \beta X),(8c + 2d) + (d-3c)X \rangle \\&= 8c\alpha + 2d\alpha + 8c\alpha + 2d\alpha + d\alpha - 3c\alpha + 8c\beta  + 2d\beta  + d\beta -3c\beta \\&= 13c\alpha +5d\alpha + 5c\beta + 3d\beta\end{align*}
\begin{align*}\langle L(v),w \rangle &= \langle (8\alpha  + 2\beta)+(\beta-3\alpha)X,(c+dX)\rangle  \\&= 8c\alpha + 2c\beta+  8c\alpha + 8d\alpha + 2c\beta + 2d\beta+c\beta+d\beta-3c\alpha-3d\alpha \\&= 13c\alpha+5d\alpha+5c\beta+3d\beta\end{align*}
This proves that $L$ is a self-adjoint operator. 
2) I guess that when I have to prove that $L$ is orthonormal diagonalizable, I can use the spectral theorem, since $L$ is self-adjoint by 1) there exists an orthonormal basis for $V$ consisting of eigenvectors for $L$ with real values, which implies that $L$ is orthonormal diagonalizable. 
However, how do I find all eigenvalues of $L$ and all the basis for the corresponding eigenspace? 
 A: We can solve
$$L(\alpha + \beta X) = \lambda (\alpha + \beta X).$$
For most values of $\lambda$, you'll only get $\alpha = \beta = 0$. For at most two values, you'll find other solutions. Using the definition of $L$,
$$(8 \alpha + 2 \beta) + (\beta - 3\alpha)X = \lambda(\alpha + \beta X).$$
Equating coefficients,
\begin{align*}
8\alpha + 2 \beta &= \lambda \alpha \\
\beta - 3 \alpha &= \lambda \beta.
\end{align*}
From the first equation, we get
$$\beta = \frac{\lambda - 8}{2} \alpha.$$
Note therefore that we require $\alpha \neq 0$, otherwise $\beta = 0$, and we get only the zero solution (which is not allowed when searching for eigenvectors). Substituting this into the other equation gives us
$$\frac{\lambda - 8}{2} \alpha - 3 \alpha = \lambda \frac{\lambda - 8}{2} \alpha$$
Since $\alpha \neq 0$, divide by $\alpha$:
$$\frac{\lambda - 8}{2} - 3 = \lambda \frac{\lambda - 8}{2}.$$
Cleaning up,
$$\lambda^2 - 9\lambda + 14 = 0,$$
which factorises to
$$(\lambda - 7)(\lambda - 2) = 0.$$
So the only possible eigenvalues are $\lambda = 2$ or $\lambda = 7$. To verify these eigenvalues, we can find the eigenvalues. For example, when $\lambda = 2$, the system of equations turns into
\begin{align*}
8\alpha + 2 \beta &= 2 \alpha \\
\beta - 3 \alpha &= 2 \beta.
\end{align*}
When you rearrange the equations, you'll find that they are both multiples of the equation
$$\beta + 3\alpha = 0.$$
Let $t = \alpha$. Then $\beta = -3t$. So, the solution is parameterised by
$$\alpha + \beta X = t - 3t X = t(1 - 3X),$$
hence the eigenvectors for $\lambda = 2$ are all multiples of $1 - 3X$.
The eigenvectors for $\lambda = 7$ follow similarly.
