Isomorphism of a Polynomial Linear Transformation

Question

Show that $T: \mathcal{P}_n \to \mathcal{P}_n$ defined by $T(p(x))=p(x+ 2 )$ is bijective by showing that it is injective and surjective by definition (injective: $\vec u \neq \vec v \implies T(\vec u) \neq T(\vec v)$, surjective: $range(T) = \mathcal{P}_n$).

Answer 1

We need to show that $T$ is one-to-one and onto.

For one-to-oneness:

Let $$T(P_1(x))=T(P_2(x))$$

$$P_1(x+2)= P_2(x+2)$$ be true for all $x$

Thus it is also true for $x-2$ that is $$P_1(x-2+2)= P_2(x-2+2)$$ That is $$P_1(x) = P_2(x)$$

For onto-ness note that $$T(P(x-2)) = P(x-2+2)=P(x)$$ Thus for any P(x) there is a preimage namely $P(x-2)$

Answer 2

Notation can trip you up here.

Better to write $(Tp)(x) = p(x+2)$. It just shifts the graph of $p$ to the left.

The antidote is to shift to the right by the same amount: $(Sp)(x) = p(x-2)$.

Check $(S(Tp))(x) = (Tp)(x-2) = p((x-2)+2 = p$, and similarly $(T(Sp))(x) = p$.

Hence $S$ is the inverse, it follows that $T$ is bijective.