Finding the Eigenvector for eigenvalue 0 I tried finding the eigenvectors of the following transformation: $T:\mathbb R_2[X]\rightarrow \mathbb R_2[X] $, $T(f(x))=f'(x)$. I found the matrix of this transformation to be $$\begin{matrix} 0&0&0 \\ 2&0&0 \\ 0&1&0  \end{matrix}$$
Thus I found the eigenvalue to be 0. But how do I find the eigenvector? I found it to be 0, which isn't right... 
Thanks in advance! 
 A: You have a eigenvalue $\lambda_1 = 0$, but it has algebraic multiplicity equal to three.
I will show you how to find the first eigenvector, but only provide guiding hints for the other two.
We can find the fist the eigenvector as:
$$Av_1 = 0$$
This is the same as finding the nullspace of $A$, so we get:
$$v_1 = (0,0,1)$$
Unfortunately, this only produces a single linearly independent eigenvector as the space spanned only gives a geometric multiplicity of one. This means we need to find two generalized eigenvectors.
We try:
$$Av_2 = v_1$$
This gives us another linearly independent eigenvector (I will let you fill in the details), but again only one.
Next, we try:
$$Av_3 = v_2$$
This produces a third linearly independent eigenvector.
Sometimes, we resort to using chains to find the generalized eigenvectors.
A: The eigenvectors are the constant polynomials. You can see that either from your matrix, or directly by noticing that the derivative of a constant polynomial is always $0$.
A: Let $A$ be the matrix of the given linear transformation. Let $v = (a,b,c)^t$ note that you can consider elements of $\mathbb{R}_2[X]$ in this form.
Find $a,b,c$ such that $Av = 0$. 
