The Remez alternation linear system has a unique solution

Define the linear map $T:{\mathcal{P}}_n(I)\times \mathbb{R}\to {\mathbb{R}}^{n+2}$ by $T(p,\lambda {)}_i=p(x_i)+(-1{)}^i\lambda$. Choose a polynomial representative $P$ of $p$ using [F3] and [F4]. If $T(p,\lambda )=0$ and $\lambda =0$, then $P$ vanishes at all $n+2$ distinct nodes, so [F5] forces $P=0$ and hence $p=0$. If $\lambda \mathrm{\ne }0$, then $P(x_i)=-(-1{)}^i\lambda$ are nonzero alternating values, so $P$ is nonzero and [F1] rules this out. Thus $\ker T=\{(0,0)\}$.

The restrictions of $1,x,\dots ,x^n$ span ${\mathcal{P}}_n(I)$ by [F3] and [F4]. If a linear combination of them restricts to zero on $I$, its representing polynomial vanishes at $x_0,\dots ,x_n$. These are $n+1$ distinct roots, so [F5] makes that polynomial zero; the coefficient representation then makes every coefficient zero. Thus the restrictions are linearly independent and form a basis, so $\dim {\mathcal{P}}_n(I)=n+1$. Therefore $(1,0),(x,0),\dots ,(x^n,0),(0,1)$ is a basis of the domain, while the standard coordinate vectors form a basis of the codomain; both dimensions are $n+2$. By [F2] and step 1.1, $T$ has image dimension $n+2$ and is therefore bijective. Its equation $T(p,\lambda {)}_i=f(x_i)$ has exactly one solution.