Eigenvalues of the differentiation operator I have a linear operator $T_1$ which acts on the vector space of polynomials in this way:
$$T_1(p(x))=p'(x).$$
How can I find its eigenvalues and how can I know whether it is diagonalizable or not?
 A: Take the derivative of $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ (with $a_n\neq 0$), and set it equal to $\lambda a_nx^n+\cdots+\lambda a_0$.  Look particularly at the equality of the coefficients of $x^n$ to determine what $\lambda$ must be.  Once you know what the eigenvalues are, consider which possible diagonalized linear transformations have that eigenvalue set, and whether such linear transformations can be similar to differentiation.
A: Differentiating lowers the degree, so the only case where you get out a scalar multiple of what you put in is when you differentiate a constant.
A: In the finite dimensional case, finding the eigenvalues can be done by considering the matrix of the operator, computing the characteristic polynomial, and finding the roots. This is not possible in the infinite dimensional case (as occurs in the case of the vector space of all polynomials with coefficients in $F$), because there is no matrix for the operator and no characteristic polynomial.
Instead, you have to go back to the definitions. An eigenvalue of $T$ is a scalar $\lambda$ for which there exists a nonzero vector $\mathbf{x}$ with $T(\mathbf{x}) = \lambda\mathbf{x}$.
Suppose $\mathbf{x}$ is an eigenvector. What can we say about $\mathbf{x}$ and $\lambda$? As usual in this kind of cases, we write down what everything means, and see what this entails/implies. Often, we can gain enough information to figure out who $\mathbf{x}$ and $\lambda$ have to be.
Let's write $\mathbf{x} = a_nx^n + \cdots + a_0$, with $a_n\neq 0$ (we know at least one coefficient has to be nonzero for $\mathbf{x}$ to be nonzero, a precondition for being an eigenvector; and so we may as well write it going up to just the degree we need; so we are going to write $x^2+0x+1$, but not $0x^4+0x^3+x^2+0x+1$, in order to make our life easier). 
Then the equation  $T(\mathbf{x}) = \lambda\mathbf{x}$ becomes
$$ na_nx^{n-1}+\cdots + a_1 = \lambda a_nx^n + \lambda a_{n-1}x^{n-1}+\cdots + \lambda a_0.$$
This gives you a system of equations
\begin{align*}
\lambda a_n &= 0\\
\lambda a_{n-1} - na_n &= 0\\
\lambda a_{n-2} - (n-1)a_{n-1} &= 0\\
&\vdots\\
\lambda a_1 - 2a_2 &= 0\\
\lambda a_0 - a_1 &=0.
\end{align*}
Since we are assuming $a_n\neq 0$, it should be an easy matter to determine all eigenvalues, and all corresponding eigenvectors from this.
Now, a linear transformation is diagonalizable if and only if there is a basis for the vector space that consists entirely of eigenvectors. Since you know by now what all the eigenvectors of $T$ are, you can figure whether you can find a linearly independent set of eigenvectors that spans the vector space of all polynomials. If you can, then $T$ is diagonalizable. If you cannot, then $T$ is not diagonalizable.
