# Thread: Prove isomorphism

1. ## Prove isomorphism

Let $\displaystyle a_0,...,a_n$be distinct real numbers.
Let the linear map $\displaystyle T: R[X]_{\leq n} \rightarrow R^{n+1}$ be defined by $\displaystyle T(q(x)) = [q(a_0),....,q(a_n)]^T$ for all $\displaystyle q(x)$ in $\displaystyle R[X]_{\leq n}$

Prove that $\displaystyle T$ is an isomorphism

For this question is it enough to say that $\displaystyle R[X]_{\leq n}$ and $\displaystyle R^{n+1}$ are both of dimension $\displaystyle n+1$and therefore isomorphic? Im not sure i can apply this to a transformation. Thanks for any help!

2. No. These two are isomorphic does not show that the map T is an isomorphism. However, since these two spaces have the same dimension, it suffices to prove that the kernel (or nullspace) of T is trivial, which is easy (but not so trivial) to prove.

3. Well its clear that the only polynomial that gives 0 for all values of x is the 0 polynomial therefore the kernal of T is trivial. Im confused as to how knowing that the kernal of T consist of only the 0 polynomial. And since the kernal of T is 0, it is therefore invertible. Which makes it isomorphic? Is this correct?

Thanks!