How to power-series expand determinants?

Question

Say $g$ is a matrix which is given as, $g = g_0 + xg_2 + x^2 g_4 .. +x^{d/2 -1}g_{d-2}+ x^{d/2}(g_d + h_d(\log (x)))$ where $d$ is an even number and each $g_i$ is a matrix (same dimension as $g$) and $h_d$ is another matrix.

• For such a set of arbitrary matrices, how can one power-series expand $\sqrt {\det(g)}$ in $x$?

Answer 1

You can use my function MatrixD from the question det simplification:

MatrixD[expr_, x__] := With[
    {old = OptionValue[SystemOptions[], "DifferentiationOptions"->"ExcludedFunctions"]},

    Internal`WithLocalSettings[
        SetSystemOptions["DifferentiationOptions"->"ExcludedFunctions"->Join[old, {Det, Inverse, Tr}]];

        Unprotect[D];
        (* handle list derivatives *)
        D[h:((Det|Tr|Inverse)[m_]), {z_, n_Integer}] := Nest[D[#, Replace[z, _List :> {z}]]&, h, n];
        D[h:((Det|Tr|Inverse)[m_]), {z_List}] := D[h, #]& /@ z;
        D[h:((Det|Tr|Inverse)[m_]), z_, y___] := D[D[h, z], y];

        (* define derivatives for Det, Tr, and Inverse *)
        D[Det[m_], z:Except[_List]] := Det[m] Tr[Inverse[m] . D[m,z]];
        D[Tr[m_], z:Except[_List]] := Tr[D[m,z]];
        D[Inverse[m_], z:Except[_List]] := -Inverse[m] . D[m, z] . Inverse[m],

        D[expr, x],

        SetSystemOptions["DifferentiationOptions"->"ExcludedFunctions"->old];
        Clear[D];
        Protect[D]
    ]
]

For your example:

g[x_] := g0 + x g2 + x^2 g4 + x^3 g6 + x^3 hd Log[x];

Then:

coeffs = Table[
    MatrixD[Sqrt[Det[g[x]]] /. Log[x]->u, {x, n}]/n! /. {x->0, u->Log[x]},
    {n, 0, 3}
];
Print @* TeXForm /@ coeffs;

$\sqrt{\left| \operatorname{g0}\right| }$

$\frac{1}{2} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right]$

$\frac{1}{2} \left(\frac{1}{2} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).\operatorname{g2}+\operatorname{g0}^{-1}.(2 \operatorname{g4})\right]+\frac{1}{4} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right]^2\right)$

$\frac{1}{6} \left(\frac{1}{2} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[2 \left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).(2 \operatorname{g4})+\left(-\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).\operatorname{g2}.\operatorname{g0}^{-1}-\operatorname{g0}^{-1}.\operatorname{g2}.\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right)-\operatorname{g0}^{-1}.(2 \operatorname{g4}).\operatorname{g0}^{-1}\right).\operatorname{g2}+\operatorname{g0}^{-1}.(6 \operatorname{g6}+6 \operatorname{hd} \log (x))\right]+\frac{3}{4} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right] \operatorname{Tr}\left[\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).\operatorname{g2}+\operatorname{g0}^{-1}.(2 \operatorname{g4})\right]+\frac{1}{8} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right]^3\right)$

Let's evaluate the first few terms of the series using an explicit set of matrices:

SeedRandom[2];
rules = Thread[{g0, g2, g4, g6, hd} -> RandomReal[1,{5,3,3}]]

{g0 -> {{0.72224, 0.109449, 0.470703}, {0.535582, 0.583178, 0.293942}, {0.165154, 0.601258, 0.754218}}, g2 -> {{0.771123, 0.778574, 0.0236104}, {0.922757, 0.992454, 0.350409}, {0.0450047, 0.501359, 0.633756}}, g4 -> {{0.642208, 0.389875, 0.664971}, {0.843882, 0.56904, 0.398212}, {0.238652, 0.673513, 0.419507}}, g6 -> {{0.587398, 0.00833523, 0.942441}, {0.771263, 0.147503, 0.964774}, {0.898747, 0.332963, 0.204548}}, hd -> {{0.839035, 0.250388, 0.238638}, {0.616616, 0.879303, 0.404504}, {0.402517, 0.516192, 0.292009}}}

And the comparison:

N @ CoefficientList[Series[Sqrt[Det[g[x]]] /. rules, {x, 0, 3}], x]

coeffs /. rules //Expand

{0.507317, 0.663253, 0.436586, 0.0826776 + 0.572028 Log[x]}

{0.507317, 0.663253, 0.436586, 0.0826776 + 0.572028 Log[x]}

Answer 2

For any square matrix M which is the sum of two similar matrices M = A + B the determinant can be written as a sum of determinants as follows (example for two dimensions):

det(M) = det( ( A11 + B11, A12 + B12), (A21 + B21, A22 + B22) )

= det( ( A11 + 0, A12 + 0),  (A21 + B21, A22 + B22) )
+ det( ( 0 + B11, 0 + B12),  (A21 + B21, A22 + B22) )

and, expanding the lower row similarly,

= det( ( A11 + 0, A12 + 0),  (A21 + 0, A22 + 0) )
+ det( ( A11 + 0, A12 + 0),  (0 + B21, 0 + B22) )

+ det( ( 0 + B11, 0 + B12),  (A21 + 0, A22 + 0) )
+ det( ( 0 + B11, 0 + B12),  (0 + B21, 0 + B22) )

= det( ( A11, A12),  (A21, A22) )
+ det( ( A11, A12),  (B21, B22) )
+ det( ( B11, B12),  (A21, A22) )
+ det( ( B11, B12),  (B21, B22) )

Letting B = x C gives then the expansion

det(M) = 
= det( ( A11, A12),  (A21, A22) )
+ x det( ( A11, A12),  (C21, C22) )
+ x det( ( C11, C12),  (A21, A22) )
+ x^2 det( ( C11, C12),  (C21, C22) )

Here we recognise det(A) and det(C) but also determinants of matrices mixed between A and C, more exactly, with replacement of rows.

This procedure obviously generalizes to your problem. You might wish to write it down in MMA terms.

Regards, Wolfgang