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I want to prove that

Any polynomial in $n$ variables over $\mathbb{R}$, is a linear combination of powers of "linear polynomials in $x_i$'s".

I've tried to do this by induction on $n$

Also note that it is enough to write every term of a certain degree (say $k$) can be written in above mentioned way.

Case 1: $n=2$

For $k$, consider $k+1$ polynomials $(x_1+ix_2)^k\text{ where }1\leq i\leq k+1$. Then since the Vander-monde determinant of the coefficient matrix is non-zero, we get that any term of the form $x_1^{k-m}x_2^m\text{ where }0\leq m\leq k$ can be written in prescribed way. Hence we are done in this way.

Now , assuming the case for $n=l$, I have to prove that the process continues to hold in $n=l+1$. I'm stuck here. Give me a hint.

P.S. PLEASE DO NOT POST A FULL SOLUTION. A HINT WILL SERVE THE PURPOSE!

EDIT: As people are not getting the question, let's give you an example

take $n=2$

then $xy=\frac{1}{2}((x+y)^2-x^2-y^2)$. Identities about $xy^2$, someone mentioned in the comment. So the question is how to write a polynomial of the form $x_1^{k_1}x_2^{k_2}\dots x_n^{k_n}=\sum (\sum a_{i(j)}x_{i(j)})^{l_j}$

user300
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  • Are you sure about this? I ask because I don't see how $xy^2$ can be written as such a linear combination. – Alex M. Mar 22 '17 at 19:47
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    $xy^2 = \frac12(x + y)^3 - \frac43(x + 2y)^3 + \frac76(x + 3y)^3 - \frac13(x + 4y)^3 = \frac16(x - y)^3 - \frac13x^3 + \frac16(x + y)^3$. – Smylic Mar 22 '17 at 19:59
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    Related: http://math.stackexchange.com/questions/137912/can-e-n-always-be-written-as-a-linear-combination-of-n-th-powers-of-linear-p – user26857 Mar 22 '17 at 23:22
  • @AlexM.: It's true because:

    you can take the coefficient vectors as vector in space $\mathbb{R}^{k+1}$, then you can consider $x_1^{k-m},x_2^m$ as the vector consisting of $0,1$'s with $1$ at $k-m+1$ th position and $0$ elsewhere and then we use linear independence.

     – user300 Mar 23 '17 at 07:09
  • @Smylic It's not about how to write! I wanted to ask about proceeding for induction step! – user300 Mar 23 '17 at 07:20

2 Answers2

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HINT: It is enough to prove for the monomial $x_1 x_2 \ldots x_n$.

For $n=3$ we have the identity

$$6 x_1 x_2 x_3 = (x_1 + x_2 + x_3)^ 3 - (x_1 + x_2)^3 - (x_1 + x_3)^3 - (x_2 + x_3)^ 3 + x_1^3 + x_2 ^3 + x_3^3$$

user26857
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orangeskid
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This may be a bit late, but here's a hint.

For $n>2$, let $A:=\mathbb{R}\left[x_1,\dots, x_{n-1}\right]$ and think of $\mathbb{R}\left[x_1,\dots, x_n\right]$ as $A\left[x_n\right]$. We claim that by an induction on $n$, $A\left[x_n\right]$ is spanned by monomials of the form $a^kx_n^m$ where $a=\sum_{i=1}^{n-1}c_ix_i$ with $c_i\in \mathbb{R}$. Now apply the $n=2$ case to $\mathbb{R}\left[a,x_n\right]$.

Daps
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