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I am doing the following problem for practice to understand linear transformations.

Consider the linear transformation $T:P_2 \rightarrow P_2$ defined by: $$ T(p(x)) = p(x+1) $$ Where $P_2$ is the vector space of polynomials of at most degree 2. Determine whether T is injective, surjective, or bijective.

The problem I am having trouble on how to start this. Just from using my intuition I would say that the $T$ is bijective partially because it is going from one vector space to the same vector space. But I am unsure of how to show this. If anyone can help me I would greatly appreciate it.

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  • $\begingroup$ 'Just from using my intuition I would say that $T$ is bijective partially because it is going from one vector space to the same vector space.' So, using your intuition, the map $\mathbb{R}\to\mathbb{R}$, $x\to0$ is also bijective? $\endgroup$ – B. Pasternak Oct 23 '17 at 11:18
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Hint:

Here's a start: to check whether $T$ is injective, assume that $T(p) = T(q)$. What's that tell you about $p$ and $q$ an their relationship to each other? If this, through your knowledge of algebra, implies that $p = q$, then $T$ is injective.

[This technique is really quite general: it's what you should first do whenever you try to prove something's injective, unless you know quite a lot more (e.g., you know it's a matrix transformation with full rank, ...)]

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Injectivity: Show that $T(p)(x)=0$ for all $x$ $\Leftrightarrow p(x)=0$ for all x.

Surjectivity: Show that for every $q\in P_2$ there is $p\in P_2$ such that $T(p)(x)=q(x)$ for every $x$.

Note: Since $P_2$ is finite-dimensional, it is enough to show one of these.

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