# Linear Algebra II - Isomorphism

Show that $T$ is an isomorphism by defining $T^{-1}$ explicitly.
$T: P_n \rightarrow P_n$ is given by $T[p(x)] = p(x+1)$

Not completely sure how to approach this. Would you begin with determining if $T$ is linear by the vector addition and scalar multiplication?

• Well, you are given a linear transformation, so you don't need to check if it is linear. Do you know how to prove an isomorphism by definition? – Itay4 Apr 3 '17 at 16:13
• what would $T^{-1}$ be? is it always true that $T^{-1}(T(p)) = p$? – gt6989b Apr 3 '17 at 16:19
• By definition, I am aware that T is an isomorphism if it is linear (which we can conclude) and that it is either one-to-one or onto. But how would you begin by checking if it is one-to-one or onto? Would using one-to-one, the zero vector, be the easiest method? – Katte Joy Apr 3 '17 at 16:27

Approach 1

Define $T': P_n \rightarrow P_n$ by $T'(P(x)) = P(x-1)$. We now prove that $T'$ is the inverse of $T$, and therefore $T$ will be an isomorphism (as it has an inverse and is therefore bijective)

Take $P \in P_n$.

$T(T'(P(x))) = T(P(x-1)) = P(x)$

$T'(T(P(x))) = T'(P(x+1)) = P(x)$

Hence, $\forall P \in P_n: T \circ T' = 1_{P_n} = T' \circ T$. By definition, $T' = T^{-1}$

Approach 2

We prove directly that $T$ is bijective.

Let $P \in P_n$

Then $T(P(x-1)) = P(x)$. Hence, $T$ is surjective. Because $P_n$ has finite dimension and the mapping is an endomorphism, it follows that $T$ is bijective, and therefore an isomorphism.