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

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.

• Where did you get stuck? Were you able to prove this for $m=1$? For $k=1$?
– Pedro
May 14, 2021 at 20:54
• 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 $k$. May 14, 2021 at 21:01

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