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I am trying to perform a product of a large number of terms of the form MaxtrixExp[...], for example, something like this

L = Simplify[Normal[Series[
 MatrixExp[z a1[1][R] Y[1]].MatrixExp[z a[2][R] Y[2]].MatrixExp[
   z a[3][R] Y[3]].MatrixExp[z a[4][R] Y[4]].MatrixExp[
   z a[5][R] Y[5]].MatrixExp[z a[6][R] Y[6]].MatrixExp[
   z a[7][R] Y[7]].MatrixExp[z a[8][R] Y[8]].MatrixExp[
   z a[9][R] Y[9]].MatrixExp[z a[10][R] Y[10]].MatrixExp[
   z a[11][R] Y[11]].MatrixExp[z a[12][R] Y[12]].MatrixExp[
   z a[13][R] Y[13]].MatrixExp[z a[14][R] Y[14]].MatrixExp[
   z a[15][R] Y[15]].MatrixExp[z a[16][R] Y[16]].MatrixExp[
   z a[17][R] Y[17]].MatrixExp[z a[18][R] Y[18]].MatrixExp[
   z a[19][R] Y[19]].MatrixExp[z a[20][R] Y[20]].MatrixExp[
   z a[21][R] Y[21]].MatrixExp[z a[22][R] Y[22]].MatrixExp[
   z a[23][R] Y[23]].MatrixExp[z a[24][R] Y[24]].MatrixExp[
   z a[25][R] Y[25]].MatrixExp[z a[26][R] Y[26]].MatrixExp[
   z a[27][R] Y[27]].MatrixExp[z a[28][R] Y[28]].MatrixExp[
   z a[29][R] Y[29]].MatrixExp[z a[30][R] Y[30]].MatrixExp[
   z a[31][R] Y[31]].MatrixExp[z a[32][R] Y[32]].MatrixExp[
   z a[33][R] Y[33]].MatrixExp[z a[34][R] Y[34]].MatrixExp[
   z a[35][R] Y[35]].MatrixExp[z a[36][R] Y[36]].MatrixExp[
   z a[37][R] Y[37]].MatrixExp[z a[38][R] Y[38]].MatrixExp[
   z a[39][R] Y[39]].MatrixExp[z a[40][R] Y[40]].MatrixExp[
   z a[41][R] Y[41]].MatrixExp[z a[42][R] Y[42]], {z, 0, 2}]]];

where Y[m] are 13x13 matrices generated by

x = 6; nv=7; Table[e[a, b]=Table[KroneckerDelta[a, c] KroneckerDelta[b, d], {c,6+ nv}, {d, 6+ nv}], {a, 6 + nv}, {b, 6 + nv}]; Table[Y[i] = e[1, x + i] + e[x + i, 1], {i, nv}];  Table[Y[i + nv] = e[2, x + i] + e[x + i, 2], {i, nv}]; Table[Y[i + 2 nv] = e[3, x + i] + e[x + i, 3], {i, nv}];Table[Y[i + 3 nv] = e[4, x + i] + e[x + i, 4], {i, nv}]; Table[Y[i + 4 nv] = e[5, x + i] + e[x + i, 5], {i,  nv}]; Table[Y[i + 5 nv] = e[6, x + i] + e[x + i, 6], {i, nv}];

a[n][R] are just some unknowns. I am trying to find a way to rewrite the above expression for L using Product, i.e.

L2=Simplify[Normal[Series[Product[MatrixExp[z a[i][R] Y[i]], {i, 1, 42}], {z, 0, 2}]]];

but this does not yield the correct result for L as given in the first long expression.

I also need to form a product of the inverse of L, i.e. something like

Linverse=Simplify[Normal[Series[MatrixExp[-z a[42] Y[42]].Matrix[-z a[41] Y[41]] ... MatrixExp[-z a[1] Y[1]],{z,0,2}]]];

so that

L.Linverse=IdentityMatrix[13]

There must be a simple way to rewrite a product for L and (Linverse), and I'd be grateful if anyone could point this out to me. Thanks.

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1 Answer 1

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As a start, change Product to Dot@@Table as

L2 = Simplify[Normal[Series[Dot @@ Table[MatrixExp[z a[i][R] Y[i]], {i, 1, 42}], {z, 0, 2}]]]

I'll continue working on the inverse.

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  • $\begingroup$ Thanks for your answer. Using the same Dot@@Table command like yours, I construct the Linverse to be L2i = Simplify[ Normal[Series[ Dot @@ Table[ MatrixExp[-z a[(-1) (i - 42)][R] Y[(-1) (i - 42)]], {i, 0, 41}], {z, 0, 2}]]]; This indeed yields the desired result when contracted with L (a 13x13 identity matrix) $\endgroup$
    – user195583
    Nov 22, 2019 at 5:42

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