User:Alexander Chervov/sandbox2

Reformulation via symmetric tensors
The identity can be reformulated as follows.
 * $$ \sum_{l=0,...,\infty} Trace_{Sym^l (V)} ((TA)^{sym(l)}) = \frac{1}{\det (I_m - TA)},$$

here V is n-dimensional complex vector space with basis xk and TA can be considered as linear operator V->V, SymlV is l-th symmetric tensor power, (TA)sym(l) is induced operator $$Sym^l (V)\to Sym^l (V)$$.

Indeed, this reformulation is almost by definition of symmetric tensor power SymlV, which can be identified with the space of polynoms of degree l in xk, k=1,...n. The expression
 * $$\prod_{i=1}^m \bigl(a_{i1}x_1 + \dots + a_{im}x_m \bigl)^{k_i}.$$

Is precisely the action of A on the Syml(V). Multiplication on $$t_1^{k_1}...t_n^{k_n}$$ corresponds to the action of T on the Syml(V).

Thus one concludes that
 * $$ G(k_1,\dots,k_m) t_1^{k_1}\cdots t_m^{k_m} $$

is exactly the diagonal element of the matrix $$(TA)^{sym(l)}$$ corresponding to basic element $$x_1^{k_1}...x_n^{k_n} \in Sym^l(V) $$, with $$l=\sum_{i=1,...,n} k_i$$. And hence the sum of all diagonal elements gives the trace - as it is written in the formula above.

Proof
Since theorem holds true for an arbitrary matrix A, and hence TA=M is an arbitrary matrix, one can reformulate the theorem without splitting M to T and A
 * $$ \sum_{l=0,...,\infty} u^l Trace_{Sym^l (V)} ((M)^{sym(l)}) = \frac{1}{\det (I_m - uM)},$$

with u a formal variable, and thus equality makes sense as a formal power series equality.

Such reformulation allows to give a simple proof. Indeed, the trace and the determinant are invariant under conjugation, thus one may prove the theorem in any basis. Diagonalizable matrices are dense among all complex matrices and hence, it is enough to prove only for such matrices. Thus, taking eigenbasis for M and denoting eigenvalues as l1,...,ln. The right hand side can be rewritten as
 * $$\begin{align}

\frac{1}{(1-l_1 u)...(1-l_n u)} &=(1+ul_1+u^2l_1^2+...)...(1+ul_n+u^2l_n^2+...) =\\ & = \sum_{p=0,...,\infty} u^p \sum_{k_1+...+k_n=p} l_1^{k_1}l_2^{k_2}...l_n^{k_n}.\\ \end{align}$$

In the last expression one can recognize the left hand side of the theorem written in eigenbasis, this proves the theorem.