Find the characteristic polynomial and minimal polynomial of the Transformation

Question

Let $V=\{\,f(x) ∈ \Bbb R[x] : \deg f(x) ≤ n\,\}$ and T be the linear endomorphism on $V$ given by $$T(f(x)) = f(x + 1) + f(x − 1).$$

Find the minimal polynomial and the characteristic polynomial of the transformation $T$.

So essentially what I am seeing is that the function $f(x)$ is finite dimensional by being less than $n$

So think I should construct a matrix of the transformation. Such that $$\det(xI − T) =\det(xI − T_1)\det(xI − T_2)\dots\det(xI − T_n)$$ which would be the characteristic polynomial

and the minimal polynomial would be: $$\def\lcm{\operatorname{lcm}} f(x) = \lcm(f_1(x), f_2(x), \ldots , f_n(x))$$ ($\lcm$ is lowest common multiple here)

where $f_1(x), f_2(x), . . . , f_n(x)$ are minimal polynomials of $T_1, T_2, \ldots , T_n$ respectively.

How do I construct this matrix? Is my line of thought even correct for this question?

Answer 1

[I personally find the habit of designating polynomials by $f(x)$ horrible for pedagogical reasons, as it incites to confusing polynomials and polynomial functions; I will just denote polynomials as $P\in \Bbb R[X]$. The operation of substituting a value $a$ (maybe another polynomial) for $X$ now needs explicit notation; I will write $P[X:=a]$ for what would have been dissimulated as $f(a)$ in the notation of the question.]

Your operation is $T: V\to V$ defined by $T(P)=P[X:=X+1]+P[X:=X-1]$. To find the matrix of $T$ on the basis $\def\E{\mathcal E}\E=[1,X,X^2,\ldots,X^n]$ of $V$, it is maybe easiest to see $T$ as the sum of the two simpler (shift) operators $T_+:P\mapsto P[X:=X+1]$ and $T_-:P\mapsto P[X:=X-1]$, whose matrices can be found directly from the binomial theorem. In detail, $$X^k[X:=X+a]=(X+a)^k = \sum_{i=0}^k\binom kiX^ia^{k-i}$$ which shows that $T_+$ and $T_-$ have upper triangular matrices with respect to the basis$~\E$ (no higher degree terms that $X^k$ occur in $T_+(X^k)$ for instance), and binomial coefficients as entries (but with an alternating sign in case of $T_-$). Concretely for $n=5$ one gets $$\def\Mat{\operatorname{Mat}} \Mat_\E(T_+)= \pmatrix{1&1&1&1&1&1\\0&1&2&3&4&5\\0&0&1&3&6&10\\ 0&0&0&1&4&10\\0&0&0&0&1&5\\0&0&0&0&0&1} \hbox{ and } \Mat_\E(T_-)= \pmatrix{1&-1&1&-1&1&-1\\0&1&-2&3&-4&5\\0&0&1&-3&6&-10\\ 0&0&0&1&-4&10\\0&0&0&0&1&-5\\0&0&0&0&0&1} $$ and together this combines to give $$\def\Mat{\operatorname{Mat}} \Mat_\E(T)=\Mat_\E(T_+)+\Mat_\E(T_-)= \pmatrix{2&0&2&0&2&0\\0&2&0&6&0&10\\0&0&2&0&12&0\\ 0&0&0&2&0&20\\0&0&0&0&2&0\\0&0&0&0&0&2}. $$ This is upper triangular with entries$~2$ on the main diagonal, so the characteristic polynomial of $T$ is $\chi_T=(X-2)^{n+1}$.

For the minimal polynomial$~\mu_T$, I'll let you reflect a bit for yourself, but here is a hint. Clearly $T-2I$ is nilpotent, so $\mu_T$ will be a power of $X-2$. But the fact that $T$ stabilises the subspaces of even and odd polynomials (i.e., those with only even respectively only odd degree terms) means that in general a power lower than $n+1$ will suffice.