Let $\mathbb{R}[X]_{\leq 1}$ be the vector space of polynomials of degree less than or equal to 1, and $$ \langle p, q \rangle := \int_{0}^{1} p(X) q(X) \, \mathrm{d}X $$ be a dot product on $\mathbb{R}[X]_{\leq 1}$.
(a) Show that the linear map $f: \mathbb{R}[X]_{\leq 1} \rightarrow \mathbb{R}[X]_{\leq 1}$, given by the Assignment $$ p(X) \mapsto\left(X-\frac{2}{3}\right) \cdot p(0) $$ is self-adjoint with respect to the dot product $\langle\cdot, \cdot\rangle$.
(b) Is the linear map $$ f: \mathbb{R}[X]_{\leq 1} \longrightarrow \mathbb {R}[X]_{\leq 1} $$ from part (b) diagonalizable? Justify your answer.
My attempt for (a):
Left-hand side: $\begin{array}{l}\langle f(p), q\rangle=\int \limits_{0}^{1}\left(X-\frac{2}{3}\right) \cdot p(0) \cdot(c X+e) d X \end{array}$
Now, integrate this expression:
$\begin{array}{l} =p(0) \int \limits_{0}^{1} (cX^2 +eX - \frac{2}{3}cX - \frac{2}{3}e) dX \\\ =p(0) \left[\frac{1}{3}c X^{3}+\frac{e}{2} X^{2}-\frac{c}{3}X^2-\frac{2}{3}eX \right]_{0}^{1} \\\ = -p(0) \frac{e}{6} \end{array}$
Now, for the right-hand side:
$\begin{array}{l}\langle p, f(q)\rangle=\int \limits_{0}^{1}\left(X-\frac{2}{3}\right) \cdot q(0) \cdot(a X+b) d X \end{array}$
$\begin{array}{l} =q(0) \int \limits_{0}^{1} (aX^2 +bX - \frac{2}{3}aX - \frac{2}{3}b) dX \\\ =q(0) \left[\frac{1}{3}a X^{3}+\frac{b}{2} X^{2}-\frac{a}{3}X^2-\frac{2}{3}bX \right]_{0}^{1} \\\ = -q(0) \frac{b}{6} \end{array}$
So $f$ is self adjoint if $e = b$ and I think it's not the point of the task to restrict the polynomials to some parameters making it self adjoint.
(b)
A linear operator in a finite dimensional vector space is diagonalizable iff it has a basis containing eigenvectors. Consider the basis $1,x$
$f(p(x)=1) = \frac{1}{3}$
$f(p(x)=x) = 0$
And the matrix of $f$ is equal to it's jordan normal form
$J = \begin{pmatrix} \frac{1}{3} & 0 \\\ 0 & 0 \end{pmatrix}$
and is diagonalizable.