Differentiation maps and diagonalization Let $P$ be the vector space spanned by functions of the form $p(t) e^{\lambda t}$ where $p(t)$ is a polynomial (in $t$) of degree less than 2, and $\lambda \in \{1,2,3\}$. The map $\mathcal{D}: P \rightarrow P$ is given by $\mathcal{D} (f(t)) = f(t) + f’(t) $.
Is this map diagonalizable?
I have trouble with understanding how a map (given by a function in a vector space) looks like, because it is not defined by some matrix $D$ for example. In general, I would proceed to see if there is some basis of Eigenvectors, and if such a basis exists, the map must be diagonalizable. However, I don’t see how this connects to differentiation maps.
 A: It is not diagonizable.
Your question is not very precise concerning $P$, but I think the correct interpretation is this:
Let $V$ be the real vector space of all functions $f : \mathbb R \to \mathbb R$ and $P$ be the subspace spanned by all functions of the form $p(t) e^{\lambda t}$ where $p(t)$ is a polynomial of degree $\le 1$ and $\lambda \in \{1,2,3\}$.
Let us first prove the following lemma:
Let $f(t) = \sum_{\lambda= 1}^n p_\lambda(t)e^{\lambda t}$ with polynomials $p_\lambda$. Then $f = 0$ if and only if all $n$ polynomials $p_\lambda = 0$.
The "if" part is trivial. The "only if" part is proved by induction on $n$. For  $n=1$ we have $p_1(t)e^t = 0$, thus $p_1(t) = \dfrac{0}{e^t} = 0$. Assume it is true for some $n \ge 1$. Now let $\sum_{\lambda= 1}^{n+1} p_\lambda(t)e^{\lambda t} = 0$. Then $p_{n+1}(t) = - \sum_{\lambda= 1}^n \dfrac{p_\lambda(t)}{e^{(n+1-\lambda)t}}$, thus $\lvert p_{n+1}(t) \rvert \le  \sum_{\lambda= 1}^n \dfrac{\lvert p_\lambda(t) \rvert}{e^{(n+1-\lambda)t}}$. As $t \to \infty$, the right hand side of the inequality goes to $0$ because $e^{(n+1-\lambda)t} \ge e^t$ for $t \ge 0$ and $\dfrac{\lvert q(t) \rvert}{e^t} \to 0$ for every polynomial $q$. Hence $p_{n+1}(t) \to 0$ as $t \to \infty$ which is possible only if $p_{n+1} = 0$. Now the induction hypothesis shows that $p_\lambda = 0$ for $\lambda = 1,\ldots, n$.
This allows us to show that $P$ has a basis $\mathfrak B$ consisting of the six functions $f_{0,\lambda}(t) = e^{\lambda t}$, $f_{1\lambda}(t) = t e^{\lambda t}$, $\lambda \in \{1,2,3\}$. In fact, each $p(t) e^{\lambda t}$, where $p(t) = at +b$, has the form $af_{1,\lambda}+bf_{0,\lambda}$, thus the six functions span $P$. To show that the $f_{i,\lambda}$ are linearly independent, let $\sum_{i = 0,\lambda =1}^{1,3} c_{i,\lambda} f_{i,\lambda} = 0$ with $c_{i,\lambda} \in \mathbb R$. Then $\sum_{\lambda=1}^3 (c_{1,\lambda}t + c_{0,\lambda})e^{\lambda t} = 0$. The above lemma says that $c_{1,\lambda}t + c_{0,\lambda} = 0$, therefore all $c_{i,\lambda} = 0$.
Using the basis $\mathfrak B$, we could compute the matrix representation of $\mathcal D$, find its characteristic polynomial and determine eigenvalues and eigenvectors. We shall not do that. Instead we directly compute eigenvalues and eigenvectors. So let us find all $f  \in P$, $f \ne 0$, and all $\alpha \in \mathbb R$ with
$$\mathcal D(f) = \alpha f .$$
Write $f = \sum_{\lambda =1}^3 p_\lambda(t) e^{\lambda t}$, where the $p_\lambda(t) = a_\lambda t + b_\lambda$ are polynomials of degree $\le 1$. Then
$$\mathcal D(f) = \sum_{\lambda =1}^3 p'_\lambda(t) e^{\lambda t} + \sum_{\lambda =1}^3 \lambda p_\lambda(t) e^{\lambda t} = \alpha \sum_{\lambda =1}^3 p_\lambda(t) e^{\lambda t}$$
which implies
$$\sum_{\lambda =1}^3 (p'_\lambda(t) + (\lambda-\alpha)p_\lambda(t))e^{\lambda t} = 0 .$$
Therefore
$$p'_\lambda(t) + (\lambda-\alpha)p_\lambda(t) = (\lambda-\alpha)a_\lambda t +  a_\lambda + (\lambda-\alpha)b_\lambda = 0$$
which means
$$(\lambda-\alpha)a_\lambda = 0, a_\lambda + (\lambda-\alpha)b_\lambda = 0 .$$
If $a_\lambda \ne 0$, we must have $\lambda-\alpha = 0$ which implies $a_\lambda + (\lambda-\alpha)b_\lambda = a_\lambda \ne 0$, a contradiction. Therefore $a_\lambda = 0$, i.e. $p_\lambda(t) = b_\lambda$, and $(\lambda-\alpha)b_\lambda = (\lambda-\alpha)p_\lambda = 0$. Since we want $f \ne 0$, we cannot have all $p_\lambda = 0$. This implies that $\lambda-\alpha = 0$ for some $\lambda = \lambda_0$, and we conclude that $\lambda-\alpha \ne 0$ for $\lambda \ne\lambda_0$ which implies $p_\lambda = 0$ for $\lambda \ne\lambda_0$.
Summarizing we see that $\mathcal D$ has exactly three eigenvalues $\alpha = \lambda = 1, 2, 3$ with corresponding eigenvectors $b_\lambda f_{0,\lambda}$. Therefore the eigenspaces of the three eigenvalues are one-dimensional and $P$ (whose dimension is $6$) does not decompose into the direct sum of its eigenspaces. This means that $\mathcal D$ is not diagonalizable.
