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Let $n \in \mathbb{N}$ for all $k \in \{1,2...,n\}$ and $P_k=x^k(1-x)^{n-k}.$ Prove $\{P_0,...,P_n\}$ is basis for $\mathbb{R}$$_{n}[x]$

I need prove two items.
i) $\{P_0,...,P_n\}$ are linearly independent.
ii) $\{P_0,...,P_n\}$ Generate the space.

For i)

Let $a_0,a_1, a_2,...,a_n\in \mathbb{R}$
If $a_0P_0+...a_nP_n=0$ Then we need to show $a_0=a_1...=a_n=0$

I think in this:
Suppose $a_0\neq 0$ Then, $P_0=\frac{a_1}{a_0}P_1-...-\frac{a_n}{a_0}P_n=x^0(1-x)^{n-0}=\frac{a_1}{a_0}x^1(1-x)^{n-1}-...-\frac{a_n}{a_0}x^n(1-x)^{n-n}=\frac{a_1}{a_0}x(1-x)^{n-1}-...-\frac{a_n}{a_0}x^n$

In other words: $(1-x)^{n}=\frac{a_1}{a_0}x(1-x)^{n-1}-...-\frac{a_n}{a_0}x^n$

I don't see the contradiction here. Can someone help me with this exercise?

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    $\begingroup$ I would look at the term of lowest degree of each polynomial. $\endgroup$ – Orest Bucicovschi Oct 12 '17 at 23:38
  • $\begingroup$ I would also formulate this by induction on $n$. $\endgroup$ – David Hill Oct 12 '17 at 23:40
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To answer the questions, you really need to utilize the linearly independence of $\{ 1, x, x^2, \cdots, x^n \}$, induction should work just fine for (i).

For part (ii), again the argument follows through an induction manner. Notice that $P_n = x^n$ and $P_{n - 1} = x^{n - 1} - x^n$, you can express $x^{n - 1} = P_{n -1} + P_n$ and the exact same follows for $x^{n-2},x^{n-2},\cdots,1$. As $\{1,x,\cdots,x^n\}$ spans $\mathbb{R}^n[x]$ and every $x^k$ where $k \in [1,n]$ can be expressed as a linear combination of $P_k$'s, $P_k$'s generate the space. Moreover, as the $P_k$'s are linearly independent, $\{P_0,P_1,\cdots,P_n\}$ form a basis.

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  • $\begingroup$ Thanks for all! $\endgroup$ – Bvss12 Oct 13 '17 at 13:23

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