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?

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.