$T: \mathbb{C}_{2020}[x]\to \mathbb{C}_{2020}[x]$, $\sum_{i=0}^{2020}a_ix^i \mapsto \sum_{i=0}^{2020}a_i(x-1)^i$

Question

$T: \mathbb{C}_{2020}[x]\to \mathbb{C}_{2020}[x]$

$\sum_{i=0}^{2020}a_ix^i \mapsto \sum_{i=0}^{2020}a_i(x-1)^i$

I have to find the presentation matrix of $T$ in order to find Jordan form.

I tried to find it by the standard basis of $\mathbb{C}_{2020}[x]$ but I am little confused about the $\sum_{i=0}^{2020}a_i(x_i-1)^i.$

I cant present the polynomials with the basis standard.

Help please?

Answer 1

Here is how to put it into Jordan form without calculation.

Let $\lambda$ be an eigenvalue of $T$ and $f \in \Bbb C_n[x]$ be an eigenvector for $\lambda$, where $n = 2020$. That is, $T(f) = \lambda f$, i.e. $f(x - 1) = \lambda f$.

By comparing the leading coefficient, we see that $\lambda$ must be equal to $1$. Moreover, $f(x) = f(x - 1)$ can only happen when $f$ is a constant polynomial, by e.g. arguing with the sum of all roots of $f$.

Thus $T$ has only one eigenvalue $\lambda = 1$ and the space of eigenvectors for $\lambda = 1$ is one dimensional. Therefore the Jordan form has only one block, with diagonal and subdiagonal entries equal to $1$ and other entries equal to $0$.

Answer 2

Hint As mentioned in the comments, we can compute the coefficients of the matrix representation $[T]$ of $T$ with respect to the standard basis $(1,x,\ldots,x^{2020})$ of $\Bbb C_{2020}[x]$ by expanding using the Binomial Theorem: $$T(x^j) = (x - 1)^j = x^j + {j \choose 1}(-1)^1 x^{j - 1} + \cdots = x^j - j x^{j - 1} + \cdots .$$ In particular, $T(x^j)$ is the sum of $x^j$ and a linear combination of (strictly) smaller powers of $x$.