How do you decompose a polynomial matrix into its matrix coefficients?

Question

Let's say I have a matrix, $\mathbf{M}$, that is polynomially dependent on a single variable, such as

M = {{15 + a^2, a + 5 a^2}, {a - 5 a^2, 2}}

and I want to find the individual matrices, $\mathbf{A}_i$, such that

$$\mathbf{M} = \mathbf{A}_0 + a \mathbf{A}_1 + a^2 \mathbf{A}_2 + \ldots$$

How do I do this? What do I use if there are multiple variables? Also, can I specify that certain symbols are not to be treated in this manner, e.g.

M = {{15 + a^2, a + 5 a^2}, {a - 5 a^2, 2 c}}

where the constant matrix would be {{15, 0}, {0, 2 c}}? Ideally, this should be applicable to vectors and tensors, also.

Answer 1

Here is a very simple way to do it:

Table[1/i! D[M, {a, i}] /. a -> 0, {i, 0, 3}]

(*
==> {{{15, 0}, {0, 2}}, {{0, 1}, {1, 0}}, {{1, 5}, {-5, 0}}, {{0, 0}, {0, 0}}}
*)

This works even if the entries are not polynomials. If they are, you can replace the arbitrary maximum 3 in the Table index by the degree of the polynomial:

Max[Exponent[M, a]]

Edit

Looking at the other solutions, this solution (due to its fundamental simplicity) is the only one that works without modification for arbitrary rank tensors and simultaneously for arbitrary functions that can be expanded in the variable a, be they polynomial or not.

Answer 2

Here's one quick way for polynomial matrices:

polyMat = {{15 + a^2, a + 5 a^2}, {a - 5 a^2, 2}};
Transpose[PadRight[CoefficientList[polyMat, a]], {2, 3, 1}]

{{{15, 0}, {0, 2}}, {{0, 1}, {1, 0}}, {{1, 5}, {-5, 0}}}

Alternatively (as Jens hints), you can do Flatten[PadRight[CoefficientList[polyMat, a]], {3}].

You can check that the matrix polynomial is reproduced with Fold[(#1 a + #2) &, 0, Reverse @ %].

Answer 3

This is similar to J. M.'s solution, but generalized to rank $n$ tensors.

Clear[decomposePolyMat]
decomposePolyMat[m_, var_] := 
     Module[{rank = ArrayDepth@m, coeffs = CoefficientList[m, var],len},
         len = Max@Map[Length, coeffs, {rank}];
         Flatten[Map[PadRight[#, len] &, coeffs, {rank}], {rank + 1}]
     ]

Answer 4

First, if you don't know the degree you can compute it:

m =  {{15 + a^5, a + 5 a^2}, {a - 5 a^2, 2 c}};
degree = Max[Flatten[Exponent[#, {a}] & /@ Flatten[m]]]

(I have increased the degree of m to verify that intermediate zero matrices are correctly output.)

And now, with degree in hand, why not use a function intended for this task?

 Coefficient[m, a, #] & /@ Range[0, degree]

It works for tensors of any rank, including scalars.