From the notes from my linear algebra course:
Given a positive integer n, the vector spaces $\mathbb{P}_n(\mathbb{R})$ and $\mathbb{R}^{n+1}$ are isomorphic.
[To show this, define] $T: \mathbb{P}_n(\mathbb{R})\to \mathbb{R}^{n+1}$ by $$ T \left( \sum_{j=0}^{n} p_j t^j \right) = \begin{pmatrix} p_0 \\ p_1 \\ \vdots \\ p_n \end{pmatrix} \in \mathbb{R}^{n+1} $$ and verify $T$ is linear and bijective.
I haven't had any good ideas on where to start by showing that $T$ is linear—the one thing that keeps coming to mind is for me to apply $T$ to each term in $\sum_{j=0}^{n} p_j t^j$, but that would obviously be circular since it assumes $T$ is linear. (Edit) Another thing I've tried is expressing the vector on the RHS, $(p_0, p_1, \ldots, p_n)^T$ as a linear combination $p_0 (1, 0, \ldots, 0)^T + p_1 (0, 1, \ldots, 0)^T + \ldots + p_n (0, 0, \ldots, 1)^T$, but I'm still not sure where I could even go using that approach; would I say that each basis vector could be expressed as a transformation, i.e. $(1, 0, \ldots, 0)^T = T\textbf{p}$, for some $\textbf{p} \in P_n(\mathbb{R})$? (End edit)
As for showing that $T$ is bijective, it seems clear that $T$ is injective since each $p \in \mathbb{P}_n(\mathbb{R})$ gets mapped to a unique $\textbf{v} \in \mathbb{R}^{n+1}$, and that $T$ is surjective since each polynomial has a representation as a vector in $\mathbb{R}^{n+1}$; however, I don't how to show that formally. What could I do here?
Let $p,q$ be two polynomials, and $\lambda$ a real.
Then, we have: $$\begin{split} T(p + \lambda q) &= T\left(\sum_{k = 0}^{n} p_k X^k + \lambda \sum_{k=0}^n q_k X^k\right)\\ &= T\left(\sum_{k = 0}^n (p_k + \lambda q_k) X^k\right)\\ &= \pmatrix{p_0 + \lambda q_0 \\ \vdots \\ p_n + \lambda q_n}\\ &= \pmatrix{p_0 \\ \vdots \\ p_n} + \lambda \pmatrix{q_0 \\ \vdots \\ q_n} = T(p) + \lambda T(q)\end{split}$$ Hence $T$ is linear.