Firstly consider the following notations:

Given any $\alpha =(\alpha _1,\cdots, \alpha _m)\in \mathbb{N}_0^m$ we define:

  1. $\color{red}{|\alpha|}:=\sum_{i=1}^m\alpha _i$
  2. $\color{red}{\alpha !}:=\alpha _1!\alpha _2!\cdots\alpha _m!$ with $0!:=1$
  3. If $v:=(v_1,\cdots,v_m)\in \mathbb{R}^m$, then $\color{red}{v^\alpha} :=v_1^{\alpha _1}v_2^{\alpha _2}\cdots v_m^{\alpha _m}$ with $0^0:=1$.

I want to prove the following proposition:

PROPOSITION: Let $\varphi :(\mathbb{R}^m)^k\to \mathbb{R}$ be a symmetric $k$-linear form. Suppose that $(e_1,\cdots, e_m)$ is the canonical basis of $\mathbb{R}^m$. If $v\in \mathbb{R}^m$, then we have

$$\varphi (v,\cdots, v)=\sum _{|\alpha |=k}\frac{k!}{\alpha !}v^\alpha \varphi^\alpha $$

in which $\color{red}{\varphi^\alpha }:=\varphi (\underbrace{e_1,\cdots, e_1}_{\alpha _1\,\,\text{ times}}, \underbrace{e_2,\cdots, e_2}_{\alpha _2\,\,\text{ times}},\cdots, \underbrace{e_m,\cdots, e_m}_{\alpha _m\,\,\text{ times}})$ for all $(\alpha _1,\cdots,\alpha _m)\in \mathbb{N}_0^m$.

I tried to prove the above proposition using induction in $k$. But I wasn't able to finish.

Thank you for your attention!

  • $\begingroup$ Where did you get stuck? Were you able to prove this for $m=1$? For $k=1$? $\endgroup$
    – Pedro
    May 14, 2021 at 20:54
  • $\begingroup$ That proposition is a generalization of the multinomial theorem. So I tried to mimic the demonstration of that theorem. However I could not prove that that proposition is true in case $k+1$ assuming that it is true for $k$. $\endgroup$
    – rfloc
    May 14, 2021 at 21:01

1 Answer 1


If you want to prove it the same way as the multinomial theorem, then you can just expand the $\varphi(v,\ldots,v)$ in the following way: $$ \varphi(v,\ldots,v)=\varphi\left(\sum_{i=1}^{n}v_ie_i,\ldots,\sum_{i=1}^{n}v_ie_i\right)=[\text{linearity}]=\sum_{i_1,\ldots,i_k=1}^{n}v_{i_1}\ldots v_{i_k}\cdot\varphi(e_{i_1},\ldots,e_{i_k}). $$ Now, since $\varphi$ is symmetric $k$-linear form, we can collect equal terms in the previous expansion. Namely, for any $\alpha=(\alpha_1,\ldots,\alpha_m)$ with $\alpha_i\in\mathbb{N}_{0}$ and $\sum_{i=1}^{m}\alpha_{i}=k$ there will be exactly $\frac{k!}{\alpha!}$ terms with $v^{\alpha}\varphi(v)^{\alpha}$ in your notation (why?).

Can you continue now?

By the way, $\varphi(v)^{\alpha}$ is a bit weird notation since there is no dependence on $v$).

  • $\begingroup$ I had the same reasoning as you (thinking that way I knew that that identity is true). But I couldn't formalize it. I think this reasoning is informal and I wanted something more formal. $\endgroup$
    – rfloc
    May 22, 2021 at 18:08

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