Show the following matrix recursion is symmetric

Question

I need to show the following matrix equation is symmetric and I'm not sure where to start:

$A_i=\sum_{j_1=1}^{i-2}2(i-j_1-1){i-2 \choose j_1-1}A_{j_1}\Big(\sum_{j_2=1}^{i-j_1-1}{i-j_1-2 \choose j_2-1}A_{j_2}A_{i-j_1-j_2} \Big)$

for $i\geq3$, where each $A_j\in\mathbb{R}^{n\times n}$ is symmetric for all $j<i$. I actually have a much larger general form for this equation, but I wanted to tackle that myself once I got started. Obviously, I would proceed by induction, but the base case $i=3$ is trivial and I'm stuck on the inductive step. My first guess was to look at some notes on noncommutative symmetric polynomials. Any help/references would be greatly appreciated.

Answer 1

Hint: You can prove the statement by mathematical induction. The expression for $A_i$ can be rewritten as $$ A_i = \sum_{j+k+\ell=i}\frac{2(i-2)!}{(j-1)!(k-1)!(\ell-1)!}A_jA_kA_\ell. $$ Split the sum into four parts, each for one of the following cases:

1. $j=k=\ell$,
2. $j,k,\ell$ are all distinct,
3. $j=\ell\neq k$,
4. $j=k$ or $\ell=k$, but not both.

In each case, prove that the partial sum constitutes a symmetric matrix.