Now let $V$ be the vector space of real polynomials in $x$ of degree at most $d$ where $d > 0$. Which of the following linear mappings of V into itself are diagonalizable?
$T_3\colon f(x) \mapsto f(x + 1)$
$ T_4\colon f(x) \mapsto f(−x).$
Yes. I get the matrix for the mapping: for $T_3$ and at most degree 2, the matrix is
$\begin{bmatrix}1 & 1 &1\\0 & 1 &2\\ 0& 0& 1\end{bmatrix}$
But how should I generalize it?
It is not strictly necessary to find the matrices of these transformations, but doing so is a valid approach. Here's an approach that doesn't require matrices:
Hint: Convince yourself that $1$ is the only possible eigenvalue of $T_3$ (that is: if $f$ is a polynomial $f(x+1) = \lambda f(x)$ for some $\lambda \neq 1$, then we must have $f = 0$). If a diagonalizable transformation has $1$ as its only eigenvalue, which transformation must it be?
Alternatively, it suffices to note that the matrix of the transformation will always be upper triangular (but never diagonal) with $1$s on the diagonal.
Hint: Any polynomial can be broken up into an even part and odd part. That is, any $f$ can be written as $f(x) = f_1(x) + f_2(x)$, where $$ f_1(-x) = f_1(x), \qquad f_2(-x) = -f_2(x) $$ Recall that a transformation is diagonalizable if and only if every vector can be written as a linear combination of eigenvectors.
Alternatively, note that the matrix of the transformation is diagonal.