Let $c_{n,j}(x) := \binom{n}{j} x^j(1-x)^{n-j} \in P_n(\mathbb{R})$
Prove that $\{c_{n,0},...,c_{n,n}\}$ is a basis for the vector space of all polynomials over $\mathbb{R}$ with degree at most n $P_n(\mathbb{R})$
I am looking for another proof idea or a simplification of mine:
Consider only the set $B_n:=\{x^0(1-x)^{n},...,x^n\}$ and essentially "forget" the binomial coefficients as $span\{\binom{n}{j} x^j(1-x)^{n-j}\} = span\{x^j(1-x)^{n-j}\}$ because $\binom{n}{j}$ is only a scalar factor.
Prove the hypothesis by induction on $n \in \mathbb{N}$. Assume the induction hypothesis is true for some n.
One has to prove the linear indepence of the set $B_{n+1}$. It follows that $B_{n+1}$ is a Basis for $P_{n+1}(\mathbb{R})$ because we have n+1 linearly independent vectors in $P_{n+1}(\mathbb{R})$.
Let $\lambda_{n+1}x^{n+1} + \lambda_{n}x^{n}(1-x)^1 +...+ \lambda_{1}x(1-x)^{n}+\lambda_{0}(1-x)^{n+1} = \\ \lambda_{n+1}x^{n+1} + (1-x)(\lambda_{n}x^{n}+...+ \lambda_{1}x(1-x)^{n-1}+\lambda_{0}(1-x)^{n}) = 0$
For specific $\lambda_{i} \in \mathbb{R}$ $\lambda_{n}x^{n}+...+ \lambda_{1}x(1-x)^{n-1}+\lambda_{0}(1-x)^{n} = \alpha_{n}x^n + ... + \alpha_{0}, \: \alpha_{i} \in \mathbb{R}$.
Then $\lambda_{n+1}x^{n+1} + (1-x)(\lambda_{n}x^{n}+...+ \lambda_{1}x(1-x)^{n-1}+\lambda_{0}(1-x)^{n}) = \\ \lambda_{n+1}x^{n+1} + (1-x)(\alpha_{n}x^n + ... + \alpha_{0}) = \\ \lambda_{n+1}x^{n+1} - \alpha_{n}x^{n+1} + \alpha_{n}x^{n} \pm... - \alpha_{0}x + \alpha_{0} = 0$
So $\alpha_{i} = \lambda_{n+1} \: \forall i=0,...,n$
$\lambda_{n+1}x^{n+1} + \lambda_{n+1} = 0$ is valid for all $x \in \mathbb{R}$ if and only if $\lambda_{n+1} = 0$. Further $\lambda_{0} = ... = \lambda_{n}= 0$ holds by induction hypothesis which implies the linear independence of the set $B_{n+1}$ q.e.d.
I do not like the abandonment of the binomial coefficients and the reduction to the canonical basis during the proof. I would be glad to see some more direct approach, if anyone has one. Thanks in advance!
In the relation $$ \lambda_{n+1}x^{n+1}+(1-x)\cdot(\dots) $$ (of degree $n+1$) we can specialize $x=1$ to get the immediate vanishing of the main coefficient. Then use the fact that $(1-x)$ is not a zero divisor in the polynomial ring, and simplify to get a linear relation of degree $n$, then proceed inductively as in the OP.
Alternatively, to use somehow the binomial coefficients, we may want to show the generation property of the given system. (Rather than showing it is independent). We can then add the elements of the system, so $1=(x+(1-x))^n$ is in the span. Then using $x=x\cdot (x+(1-x))^{n-1}$ we see that also $x$ is in the span. (And we have to adjust slightly the binomial coefficients of hte given system.) Proceed in the same way with $x^2,\dots$