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

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 :)

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What is $P_{2}$? –  Quinn Culver Nov 9 '12 at 12:59
P2 is set of all polynomial that degree less or equal than 2 :) –  hqt Nov 9 '12 at 13:01
How did you compute "rank of $(p(x))=3$"? –  P.. Nov 9 '12 at 13:19

$\{ 1, x, x^2 \}$ 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 $b=2$.
$P_2$ is a three dimensional vector space That is to say it is finite dimensional space. And $T$ is a linear transformation. So $T$ has a matrix representation ! It is a basic fact of linear algebra. Anyway $x^2$, $x$ and 1 is corresponded to the vectors (0,0,1), (0,1,0), (1,0,0) For instance $x+1$ is corresponded to (1,1,0). And $T(x^2)=(x+1)^2$ so that $T$ sends (0,0,1) to (1, 2, 1). –  HK Lee Nov 9 '12 at 13:37