Jordan normal form for a linear transformation

Question

The following is a question from an old exam:

Let $P_n$ be the vector space of polynomials of degree less than or equal to $n$ with complex coefficients. Define $L:P_n \to P_n$ by $$L(p(x))=p(x+1)-2p(x)+p(x-1)$$ for all $p \in P_n$. Show that $L$ is nilpotent, and determine the Jordan normal form of $L$.

My attempt:

Showing that it is nilpotent is easy (the operator lowers the degree of the polynimal by two). Since it is nilpotent, the only eigenvalue is zero. But I don't know how to determine the Jordan normalform of L. I have looked at what happens with the basis vectors, and L obvisouly sends them lower degree polynimal, which makes L upper triangular. But, I don't know how to proceed from there.

Answer 1

As you’ve noted, $L$ reduces the degree of the polynomial by two, so its kernel consists of polynomials of degree less than two. As the only eigenvalue is $0$, for $n\ge2$ this implies that there will be two blocks in the JNF. The minimal polynomial of $L$ is $x^k$, with $k=\lfloor n/2\rfloor+1$ (verify this!). This gives you the size of the largest Jordan block.

Answer 2

I'll handle the case where $n$ is even, but the case where $n$ is odd is similar.

You have correctly found that for any $p \in P_n$, $\deg(L(p)) = \deg(p) - 2$. We also find that $\ker(L) = \operatorname{span}(\{1,x\}) = P_1 \subset P_n$. We can deduce that $$ \dim \ker(L^k) = \begin{cases} 2k & k \leq n/2\\ n & k > n/2 \end{cases} $$ Thus, we find that $$ \dim \ker L^k - \dim \ker L^{k-1} = \begin{cases} 2 & k \leq n/2\\ 1 & k = n/2\\ 0 & \text{otherwise} \end{cases} $$ Now, note that $\dim \ker L^k - \dim \ker L^{k-1}$ is the number of Jordan blocks (associated with $\lambda = 0$) whose size is at least $k$ (where we define $L^0 = I$).

We deduce that the Jordan form of $L$ is $J_{n/2} \oplus J_{1+n/2}$, where $J_k$ denotes the Jordan block associated with $\lambda = 0$ of size $k$.

Answer 3

We can also reason as follows.

Let $f:p(x)\mapsto p(x+1/2)-p(x-1/2)$; then $L=f\circ f$. Moreover $dim(\ker(f^k))-dim(\ker(f^{k-1}))=1$ or $0$ and the Jordan form of $f$ is $J_{n+1}$.

Finally, the Jordan form of $L$ is the Jordan form of ${J_{n+1}}^2$, that is, the one described by Omnomnomnom .