Finding the Jordan form of $T(p(t))=p(t+1)$

Question

Let $T:P_n(\mathbb{R})\to P_n(\mathbb{R})$ be a linear operator defined by $T(p(t))=p(t+1)$.

(a) Find the Jordan form for $T$.

(b) if $n=4$, find a base $B$ of $T$ such that $[T]_B$ is in the jordan form.

My attempt:

Consider the standard base for $P_n(\mathbb{R})$ which is $A=\{1,x,x^2,...,x^n\}$, with this base, we have $T(1)=1, T(x)=x+1, T(x^2)=x^2+2x+1,...$ and so $[T]_A=\begin{bmatrix} 1 & 1 & 1 & ... \\ 0 & 1 & 2 & ... \\ 0 & 0 & 1 & ... \\ 0 & 0 & 0 & ... \\ \end{bmatrix}$

So I thought about finding the Jordan for using induction if $n=1$, it's trivial.

If $n=2$ we have $[T]_A=\begin{bmatrix} 1 & 1 \\ & 1 \\ \end{bmatrix}$and the jordan form is $\begin{bmatrix} 1 & 1 \\ & 1 \\ \end{bmatrix}$.

If $n=3$ then the jordam form is $\begin{bmatrix} 1 & 1 & \\ & 1 & 1 \\ & & 1 \\ \end{bmatrix}$ so on, I can say then that the jordan form is $\begin{bmatrix} 1 & 1 & & & & \\ & 1 & 1 & & & \\ & & 1 & 1 & & \\ & & & 1 & 1 & \\ & & & & 1 & \\ & & & & & ... \\ \end{bmatrix}$?

(b) If $n=4$ then $[T]_A=\begin{bmatrix} 1 & 1 &1 & 1 \\ & 1 & 2 & 3\\ & & 1 & 3 \\ & & & 1 \\ \end{bmatrix}$ and the jordan form is $J=\begin{bmatrix} 1 & 1 & & \\ & 1 & 1 & \\ & & 1 & 1 \\ & & & 1 \\ \end{bmatrix}$, in this case we have that $P=\begin{bmatrix} 6 & 6 & 1 & \\ & 6 & 3 & \\ & & 3 & \\ & & & 1 \\ \end{bmatrix}$ is such that $PJP^{-1}=[T]_A$ then, we have that $B=\begin{bmatrix} 6 & 6 & 1 & \\ & 6 & 3 & \\ & & 3 & \\ & & & 1 \\ \end{bmatrix}$ is the base that solves the problem.

Is this correct? Thanks

Answer 1

Regarding part a: you haven't proved that your answer is correct. One way to prove that $$ J = \begin{bmatrix} 1 & 1 & & & & \\ & 1 & 1 & & & \\ & & 1 & 1 & & \\ & & & 1 & 1 & \\ & & & & 1 & \\ & & & & & \ddots \\ \end{bmatrix} $$ is the correct Jordan form is to show that $T$ (which is an operator over an $(n+1)$-dimensional space) satisfies $(T-I)^n \neq 0$ and $(T-I)^{n+1} = 0$ (where $I$ denotes the identity operator). Alternatively, it also suffices to show that $T$ has $1$ as its only eigenvalue and that the dimension of the kernel of $T-I$ is $1$.

Regarding part (b): your answer doesn't make sense; $B$ must be a basis of the space $P_4(\Bbb R)$, so it must consist of polynomials (i.e. "vectors" from the vector space $P_4(\Bbb R)$).

An easy way to find a Jordan basis is to find a vector $p \in P_4(\Bbb R)$ for which $(T-I)^n (p) \neq 0$. For this vector, the basis $$ \{(T-I)^4(p),(T-I)^3(p), (T-I)^2(p), (T-I)(p),p\} $$ is a Jordan basis.

Answer 2

Let $n$ be the dimension of $P_n(\mathbb{R})=\operatorname{Span}_\mathbb{R}(\{1,x,\ldots,x^{n-1}\})$. I will not prove that $T$ is indeed linear.

Let us define $$ p_k(x) = {x \choose k} = \frac{x(x-1)(x-2)\cdots (x-(k-1))}{k!}. $$ For example this definition sets $p_0(x) = 1$ and $p_1(x) = x$.

I argue that $\{p_k(x)\}_{k=0}^{n-1}$ is a basis of generalised eigenvectors of $T$, and that $T(p_k) = p_k + p_{k-1}$ for $k\geq 1$ (together with $T(p_0)=p_0$). $$ T(p_k) = {x+1 \choose k} = {x \choose k} + {x \choose k-1} $$ This will remind you of Pascal's rule, and indeed it is the same thing. If you're not convinced, just expand it $$ {x+1 \choose k} = \frac{(x+1)x (x-1)\cdots (x+1-(k-1))}{k!} = \frac{(x+1)x (x-1)\cdots (x-(k-2))}{k!} $$ Write $x+1 = (x-(k-1)) + k$ and expand to get $$ \frac{(x -(k-1)) x (x-1)\cdots (x-(k-2))}{k!} + \frac{ k x (x-1)\cdots (x+1-(k-1))}{k!} = $$

$$\frac{x (x-1)\cdots (x-(k-2))(x-(k-1))}{k!} + \frac{x (x-1)\cdots (x-(k-1)-1)}{(k-1)!}= $$ $$ ={x \choose k} + {x \choose k-1}. $$