$\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}$

Question

ACHTUNG

TO FOLLOWS I USE ENSTEIN CONVENTION TO WROTE DETERMINANTS!

Definition

If $A$ is an $n\times n$ matrix we define its $k\times k$ minotr to be the determinant of $k\times k$ submatrix of $A$, e.g. $$ A^{i_1,...,i_k}_{j_1,...,j_k}:=\det\begin{bmatrix}A_{j_1}^{i_1}\ &\cdots&A_{j_k}^{i_1}\ \\\vdots&\ddots&\vdots\\A_{j_1}^{i_k}&\cdots&A_{j_k}^{i_k}\end{bmatrix}=\delta^{h_1,...,h_k}_{j_1,...,j_k}A^{i_1}_{h_1}...A^{i_k}_{h_k}=\delta^{i_1,...,i_k}_{h_1,...,h_k}A^{h_1}_{j_1}...A^{h_k}_{j_k} $$ where the $\delta$-symbol is the generalised Kronecker Delta as here defined.

So I ask to prove that $$ \frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k} $$ for any non-singular matrix. In particular using the above definition I calculate that $$ \frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\\ \frac{1}{k!}\sum_{h_1,...,h_k=1}^N\Big(\delta^{l_1,...,l_k}_{h_1,...,h_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\Big)\Big(\delta^{h_1,...,h_k}_{m_1,...,m_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}\Big)=\\ \frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{h_1,...,h_k}\delta^{h_1,...,h_k}_{m_1,...,m_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}=\\ \frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{m_1,...,m_k}A^{i_1}_{l_1}\big(A^{-1}\big)^{m_1}_{j_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_k}_{j_k} $$ but unfortunately I do not be able to conclude anything. So could someone help me, please?

Answer 1

The generalized delta is just an antisymmetrizer. You just need to use the commutation identity that you wrote at the end of the first displayed equation to slide the $A$'s through the antisymmetrizer.

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Edit with a few more details:

I said the generalized delta $$ \delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}=\sum_{\sigma\in S_k} {\rm sgn}(\sigma) \delta^{i_{\sigma(1)}}_{j_1}\cdots \delta^{i_{\sigma(k)}}_{j_k} $$ is an antisymmetrizer because if you contract it to a general tensor $T_{i_1,\ldots,i_k}$ and this way define a new tensor $$ W_{i_1,\ldots,i_k}:= \delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ T_{j_1,\ldots,j_k} $$ (I used Einstein's convention, but I don't care about upstairs/downstairs) what you get is an antisymmetric tensor, i.e., $W_{i_1,\ldots,i_k}$ changes sign if any two indices are switched.

Now the commutation, or sliding of $A$'s through the antisymmetrizer is the identity $$ \delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}A^{j_1}_{h_1}\cdots A^{j_k}_{h_k}= A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}\delta^{j_1,\ldots, j_k}_{h_1,\ldots, h_k}\ . $$

As for the proof of the wanted identity, one can start as in the OP (a bad move that needs fixing) $$ \frac{1}{k!} A^{i_1,\ldots,i_k}_{h_1,\ldots,h_k} (A^{-1})^{h_1,\ldots,h_k}_{j_1,\ldots,j_k}= \frac{1}{k!} \left( A^{i_1}_{p_1}\cdots A^{i_1}_{p_k}\delta^{p_1,\ldots, p_k}_{h_1,\ldots, h_k}\right) \left(\delta^{h_1,\ldots, h_k}_{q_1,\ldots, q_k}(A^{-1})^{q_1}_{j_1}\cdots (A^{-1})^{q_k}_{j_k}\right) $$ but the $A$'s and $A^{-1}$'s are on the wrong side of their attached generalized delta. So use the commutation twice to rewrite this as $$ \frac{1}{k!} \left(\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}A^{p_1}_{h_1}\cdots A^{p_k}_{h_k}\right) \left((A^{-1})^{h_1}_{q_1}\cdots (A^{-1})^{h_k}_{q_k}\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}\right) $$ $$ =\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k} (A^{p_1}_{h_1}(A^{-1})^{h_1}_{q_1})\cdots (A^{p_k}_{h_k}(A^{-1})^{h_k}_{q_k}) \delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k} $$ $$ =\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k} \delta^{p_1}_{q_1}\cdots\delta^{p_k}_{q_k} \delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k} =\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k} \delta^{p_1,\ldots, p_k}_{j_1,\ldots, j_k} =\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ . $$

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Edit with even more details:

For completeness, let me prove the identity about the last two contracted generalized deltas giving a single one.

By definition, and summing over permutation $\sigma,\tau$ in the symmetric group $S_k$, $$ \delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}= \sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau) \delta^{i_{\sigma(1)}}_{p_1}\cdots\delta^{i_{\sigma(k)}}_{p_k} \ \delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k} $$ But, by permuting factors, $$ \delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}= \delta^{p_1}_{j_{\tau^{-1}(1)}}\cdots\delta^{p_k}_{j_{\tau^{-1}(k)}}\ . $$ We insert this equality and do the summation over the $p$ indices and get $$ \delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}= \sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau) \delta^{i_{\sigma(1)}}_{j_{\tau^{-1}(1)}}\cdots\delta^{i_{\sigma(k)}}_{j_{\tau^{-1}(k)}} $$ $$ =\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau) \delta^{i_{\sigma(\tau(1))}}_{j_1}\cdots\delta^{i_{\sigma(\tau(k))}}_{j_k} $$ by reordering the factors again. To finish, use ${\rm sgn}(\sigma){\rm sgn}(\tau)={\rm sgn}(\sigma\tau)$ and notice that each permutation $\rho=\sigma\tau$ appears exactly $k!$ times.

Answer 2

If $A$, $B$, $C$ are $(1,1)$ tensors such that $A\cdot B = C$, that is $\sum_h A^{i}_h B^h_j = C^i_j$. If we define $A^{I}_J=A^{i_1, \ldots, i_k}_{j_1, \ldots, j_k} = A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}$ we have $\sum_H A^I_H B^H_J = C^I_J$, which we can write $(A)(B) = (AB)$ Consider now a special $(k,k)$ tensor, the generalized $\delta$, like @Abdelmalek Abdesselam: indicated, and the tensor $[A] = \Delta \cdot (A)=(A)\cdot \Delta$ (last equality needs a bit of thinking, not true for any $(k,k)$ tenson). The crucial euqality is $$\Delta \Delta = k! \Delta$$ Now we want to compare $[A]\cdot [B]$ and $[AB]$. So we write $$[A]\cdot [B]= (A)\Delta \Delta (B)= k! \Delta (A)(B)= k! \Delta (AB) = k! [AB] $$

If instead of $\Delta$ we consider $P$ ( no signs), then we have permanents, and the corresponding Binet-Cauchy for them.

Note:

We have $(\Delta \cdot (A) )^I_J= \sum \epsilon(\sigma) (A)^{\sigma(I)}_J$ while $( (A) \cdot \Delta)^I_J= \sum \epsilon(\sigma) (A)^{I}_{\sigma(J)}$

Now the tensor $(A)$ is such that $(A)^{\sigma(I)}_{\sigma(J)} = (A)^I_J$ ( commutativity in some product), and moreover, $\epsilon(\sigma) = \epsilon(\sigma^{-1})$.