Is the linear transformation $T: P_{n} \to \mathbb{R}^{n+1}$ an isomorphism? So, I need to prove that the linear transformation $T: P_{n} \to \mathbb{R}^{n+1}$, defined as $T(p(x)) = \{p(0),p(1),...,p(n)\}$ is an isomorphism, with $P_{n}$ being the polynomials with a degree equal or lesser than $n$, but I have a problem.
I tried to check if $T$ is injective, and it is since any polynomial can only have at max $n$ roots, so any polynomial that gets sent to $\underbrace{(0,0,0,...,0)}_\text{n+1 times}$ must be the zero polynomial. Then, $T$ is injective. And, since $P_{n}$ and $\mathbb{R}^{n+1}$ have the same dimension, then $T$ is an isomorphism.
Then, I tried to see what $T$ does to the base of $P_{n}$. I got a bunch of vectors and tried to see if they're linearly independent, so I got a matrix, and since the first row is all $0$, then the determinant is $0$, and since the determinant is $0$, then these vectors are not a base for $\mathbb{R}^{n+1}$. Then, since an isomorphism has to send a base to a base, $T$ is not an isomorphism.
What went wrong?
Edit: I got the vectors $(1,1,1,...,1), (0,1,2,...,n),(0,1,4,...,n^{2}),...,(0,1,2^{n},...,n^{n})$, so I did the equation $$a_{0}(1,1,1,...,1) + a_{1}(0,1,2,...,n) + ... + a_{n}(0,1,2^{n},...,n^{n}) = 0$$ and then got this:
\begin{align}
a_{0} + a_{1}(0) + a_{2}(0) +...+ a_{n}(0) &= 0\\
a_{0} + a_{1}(1) + a_{2}(1) +...+ a_{n}(1) &= 0\\
\vdots &\\
a_{0} + a_{1}(n) + a_{2}(n^{2}) +...+ a_{n}(n^{n}) &= 0\\  
\end{align}
and got the coefficient matrix:
\begin{matrix}
0 & 0 & 0 &...& 0\\
0 & a_{1} & a_{2} &...& a_{n}\\
\vdots & \vdots & \vdots & \vdots & \vdots \\
0 & a_{1}(n) & a_{2}(n^{2}) &...& a_{n}(n^{n})
\end{matrix}
 A: The first colunm is all of 1's (if you take as the first element of the basis of $P_n$ the polynomial $p_0(x)=1$, then $p_0(0)=p_0(1)=\ldots=p_0(n+1)=1$), so there no problem, or I mean, you have to prove that the determinat of the matrix
$$
\begin{pmatrix}
1 & 0 & 0 &...& 0\\
1 & 1 & 1 &...& 1\\
1 & 2 & 4 &...& 2^n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & n & n^{2} &...& n^{n}
\end{pmatrix}
$$
is not $0$ (which is true, and not so difficult to prove).
But maybe, what you are trying is not the easier way to prove that the function is surjective, you can just use the definition.
Indeed, let $(y_0,\ldots,y_{n+1})\in \mathbb{R}^{n+1}$, and consider the polynomial
$$
p(x):=\frac{y_0}{(n+1)!}\cdot \prod_{i=0,\ i\neq 0}^{n+1} (i-x)+\frac{y_1}{-(n)!}\cdot \prod_{i=0,\ i\neq 1}^{n+1} (i-x)+\ldots +\frac{y_{n+1}}{(-1)^n(n)!}\cdot \prod_{i=0,\ i\neq n+1}^{n+1} (i-x)=
\sum_{m=0}^{n+1} \left( \frac{y_m}{(-1)^m(n+1-m)!m!} \cdot \prod_{i=0,\ i\neq m}^{n+1} (i-x) \right)
$$
then it is easy to see that $p$ has degree $n$ (so it is in $P_n$) and $T(p)=(y_0,\ldots,y_{n+1})$.
Indeed, if $0\leq j\leq n$, then evaluating $p(j)$ you have that all the addends in the sum are $0$ except for the $j$-th, which is
$$
\frac{y_j}{(-1)^j(n+1-j)!j!} \cdot \prod_{i=0,\ i\neq j}^{n+1} (i-j)=\frac{y_j}{(-1)^j(n+1-j)!j!}\cdot (-j\cdot (-j+1)\cdot \ldots\cdot  -1\cdot 1\cdot \ldots \cdot (n+1-j))=y_j.
$$

Maybe you are also interested in Vandermonde matrices.
