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The problem goes: or i $\in $ $[0,n-1]$, $v_{i}\in \mathbb{R}^{n}$ is defined by $v_{i} = (1^{i},2^{i},...,n^{i})$. Prove that the list $(v_{0},v_{1},...,v_{n-1})$ is a basis for $\mathbb{R}^{n}$.

Since the list has a length equal to the dimension of $\mathbb{R}^{n}$, my approach is to prove that the list is linearly independent, i.e. the linear combination $a_{0}v_{0}+a_{1}v_{1}+...+a_{n-1}v_{n-1}$ of the list is zero iff $a_{0}=a_{1}=...+a_{n-1}=0$. The backward direction is obvious but I am stuck on the forward direction. To prove that I can say since $v_{0},v_{1},...,v_{n-1}\neq 0$ as defined the proof is complete. But I doubt the validity of this method since it seems way too easy.

Can anyone give me some advice on this? Thanks a lot.

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  • $\begingroup$ Do you know about the Vandermonde matrix ? $\endgroup$ – Traklon Feb 23 '14 at 16:30
  • $\begingroup$ Sorry I know nothing about it. $\endgroup$ – verticese Feb 23 '14 at 16:31
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So let's call $V$ the matrix constructed with the $v_i$ as columns.

Suppose there exists $X = (x_0,x_1,...,x_{n-1})$ in $\mathbb{R}^n$ such that $VX = 0$.

Then : \begin{cases} x_0 + 1 x_1 + 1^2 x_2 + &\dots + 1^{n-1} x_{n-1}=0\\ x_0 + 2 x_1 + 2^2 x_2 + &\dots + 2^{n-1} x_{n-1}=0\\ & \dots \\ x_0 + n x_1 + n^2 x_2 + &\dots + n^{n-1} x_{n-1}=0 \end{cases}

Finally, let's consider the polynomial $P(Y)=\sum_{i=0}^{n-1} x_i Y^i$.

We can see that $1,2,...,n-1,n$ are roots of P so P has at least $n$ dinstincts roots. Therefore, $P=0$, so $X=0$, which gives you linear independance.

(Props to Wikipedia France for the proof).

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  • $\begingroup$ Sorry I don't get how you see the roots of P? $\endgroup$ – verticese Feb 23 '14 at 17:45
  • $\begingroup$ Well, you don't see them, but replacing Y by $1$, or $2$... or $n$ makes one of the lines written above to appear, and you know they are equal to 0, so $1$,...$n$ are roots. $\endgroup$ – Traklon Feb 23 '14 at 18:41
  • $\begingroup$ OH GREAT I got it now. Thank you very much. $\endgroup$ – verticese Feb 23 '14 at 19:09

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