Be $\mathbb{K}$ a field, and $\alpha_0, \ldots , \alpha_n \in \mathbb{K}$ are distinct elements. Show the application $$\mathbb{K}[x]_{\leq n} \to \mathbb{K}^{n+1}$$ $$p \mapsto (p(\alpha_0), \ldots, p(\alpha_n))$$ is an isomorphism.
Attempts
I proved the application is linear. Now I want to prove the application is injective, so I thought to look at the kernel, proving it's trivial.
The kernel contains all the polynomial of degree $n$ with the distinct roots, but I actually would prove it does contain only the trivial null vector, maybe proving the polynomials of degree $n$ with $n$ roots do not form a vector subspace... Calling $f$ the application:
$$\ker (f) = \left\{ p \in \mathbb{K}[x]_{\leq n};\ \begin{pmatrix} p(\alpha_0) \\ \ldots \\ p(\alpha_n) \end{pmatrix} = \begin{pmatrix} 0 \\ \ldots \\ 0 \end{pmatrix} \in \mathbb{K}^{n+1} \right\}$$
So $\alpha_0, \ldots, \alpha_n$ are roots of the polynomial, that is $(x - \alpha_0)$, $(x - \alpha_1), \ldots $ divide $p$.
So if $p$ has $n$ roots, it won't be a polynomial of degree less than $n$ hence the kernel does contain either the polynomials of degree $n$ or the null polynomial.
If the kernel contains the poly of degree $n$ with $n$ roots (all with the same algebraic multiplicity) then all these polynomials are of the form $\beta (x - \alpha_0)(x-\alpha_1) \ldots (x-\alpha_n)$ hence all identical except for a multiplicative constant, whence $\ker(f) = span\{ (x-\alpha_0)\ldots (x-\alpha_n)\}$
I don't actually know how to proceed. We haven't dealt a lot with proofs, and being able to prove something is still quite hard for me...
I know that I also have to prove it's surjective which will imply the inverse does exist. Then I should prove both $f$ and $f^{-1}$ are continuous. I don't really know where to start.
Any help? Thank you!
If $p$ is in the kernel then we know that it has $n+1$ distinct roots $\alpha_0,...,\alpha_n$. This means that $p \neq 0$ would immediately lead to a contradiction since a non zero polynomial of degree $n$ has at most $n < n+1$ distinct roots.
This means that you must have $p \in \ker f \implies p = 0$ and the map is injective.
Moreover since $K[x]_{\leq n}$ and $K^{n+1}$ have the same dimension $n+1$ it follows that injective linear maps are always isomorphisms (this follows from the rank nullity theorem). Indeed we have
$$ rank(f) + \dim \ker f = n+1$$
so $rank(f) = n+1$. This means that the image of $f$ is an $n+1$ dimensional subsapce of $K^{n+1}$. That is,
$$ Im(f) = K^{n+1}$$
This proves that the map is a linear isomorphism.
In your post you also mention continuity of $f$ and $f^{-1}$ this is of no interest for proving that we have isomorphism of vector spaces. However if you want to prove this too then it follows easily since linear maps are always continuous (and inverses of linear maps are also linear)