find $rank(T)$ when $T: P_2\to P_2:T(p(x)) = p(x+1)$

Question

An linear map $T: P_2 \to P_2$ is defined: $T(p(x)) = p(x+1)$. If $a = \operatorname{rank}T$ and $b = \operatorname{rank}(T- I)$ where $I$ denotes the identity transformation, then find $(a,b)$.

This is my problem in my textbook. In my thinking, $\operatorname{rank}T$ is equal to rank of $(p(x)) = 3$. But, I think it 's wrong, because I haven't used $p(x+1)$ yet. At b, I really don't know how to find.

Thanks :)

Answer 1

$\{ 1, x, x^2 \}$ is a basis so that $T$ has a matrix representation

$$ \left( \begin{array}{ccc} 1&1&1 \\ 0 &1 & 2 \\ 0&0& 1 \end{array}\right) .$$ This is easy. For instance $ T(x^2) = x^2 + 2x +1$. And $T-I$ has a matrix representation $$\left( \begin{array}{ccc} 0 &1&1 \\ 0 & 0 & 2 \\ 0&0& 0 \end{array} \right).$$ Accordingly $a=3$ and $b=2$.