Prove for $p(x) \in \mathbb R_n[x]$ there exist unique coefficients For all $n\in\mathbb N$ define
$$\mathbb R_n[x] =\{p(x) \in\mathbb R[x]: \deg(p(x)) \leq n\}.$$

Let $n\in\mathbb N$. Suppose polynomials $p_0(x),p_1(x),\dots,p_n(x)\in\mathbb R_n[x]$ have degrees $0,1,2,\dots,n$, respectively. Prove that for all $p(x) \in\mathbb R_n[x]$ there exist unique $a_0,\dots,a_n \in \mathbb R$ such that
  $$p(x) = a_0p_0(x)+a_1p_1(x) + \cdots+a_np_n(x).$$

I used induction, and have proved when $n = 0$ (base case). But don't know how to do the inductive step. Thanks.
 A: Existence: if $\deg p\leq n-1$ we are done by induction. If $\deg p=n$ then there exists $a_n\in\mathbb R$ such that $\deg(p-a_np_n)\le n-1$ and apply induction again. 
Uniqueness: if $$p(x) = a_0p_0(x)+a_1p_1(x) + \cdots+a_np_n(x)$$ and $$p(x) = b_0p_0(x)+b_1p_1(x) + \cdots+b_np_n(x),$$ then $$(a_0-b_0)p_0(x)+(a_1-b_1)p_1(x) + \cdots+(a_n-b_n)p_n(x)=0.$$ In the left hand side $X^n$ appears only in $p_n$, so $a_n-b_n=0$, and so on.
A: Write
\begin{align*}
a_{0}p_{0}\left(x\right) & =a_{0}c_{0,0}\\
a_{1}p_{1}\left(x\right) & =a_{1}c_{1,0}+a_{1}c_{1,1}x\\
 & \vdots\\
a_{n}p_{n}\left(x\right) & =a_{n}c_{n,0}+a_{n}c_{n,1}x+\ldots+a_{n}c_{n,n}x^{n}.
\end{align*}
Then the system we are interested in solving is
$$
\left[\begin{array}{cccc}
c_{0,0}\\
c_{1,0} & c_{1,1}\\
\vdots & \vdots & \ddots\\
c_{n,0} & c_{n,1} & \ldots & c_{n,n}
\end{array}\right] \textbf{a} = \textbf{c}
$$
where $\textbf{c}$ is the vector of coefficients of $p_0\left(x\right)$. Then the determinant of the matrix in the equation above is nonzero exactly when $c_{i,i}\neq0 \hspace{1em} \forall i$. This can be shown by using the definition of the determinant.
