Syntax for higher rank tensor multiplication

Question

I am trying to do some matrix multiplication in Mathematica but I just cannot figure out the correct syntax for my problem.

I want to write the following: $$ A\pmatrix{a & b\\c & d}+B\pmatrix{e & f\\g & h}=\pmatrix{Aa+Be & Ab+Bf\\Ac+Bg & Ad+Bh} $$ as $$ \pmatrix{\pmatrix{a & b\\c & d}\\\pmatrix{e & f\\g & h}}\pmatrix{A\\B}=\pmatrix{Aa+Be & Ab+Bf\\Ac+Bg & Ad+Bh} $$

So far I found out that this generates

{{{{a, b}, {c, d}}}, {{{g, h}, {i, j}}}};

MatrixForm[%]

the first tensor. But when I multiply this by {e,f} the result is incorrect. This

{{{{a, b}, {c, d}}}, {{{e, f}, {g, h}}}}.{A, B};

MatrixForm[%]

gives me this as a result: $$ \pmatrix{\pmatrix{Aa & Bb\\Ac & Bd}\\\pmatrix{Ae & Bf\\Ag & Bh}} $$

I think that my problem has an easy fix... probably just some brackets that need to be added, but I just didn't manage to figure it out.

Thanks

Philipp

Answer 1

You almost had it, with just two things going wrong. First, you had a higher-rank tensor than you wanted, which can be solved by peeling off a layer of curly braces, and the two $ 2 \times 2 $ matrices are effectively transposed, suggesting we should reverse the order of multiplication. You want:

{A, B}.{{{a, b}, {c, d}}, {{e, f}, {g, h}}} 

which gives

$$ \left( \begin{array}{cc} a A+B e & A b+B f \\ A c+B g & A d+B h \\ \end{array} \right) $$

as desired.

Answer 2

You definitely want to look up the functions Inner and Outer. Outer gives you a generalized outer product, and it is extremely useful. Inner gives you a generalized inner product. See this question and the answers below it for how to think about these functions.

It can be tricky to use and figure out the syntax, so here's a start. For your problem, let's define the matrices and vector as

mat1 = Array[a, {2, 2}];
mat2 = Array[b, {2, 2}];
vec = {A, B};

Then, the direct translation of what you're doing is

(combo1 = A Array[a, {2, 2}] + B Array[b, {2, 2}]) // MatrixForm

resulting in

Using Inner:

combo2 = Inner[Times, {Array[a, {2, 2}], Array[b, {2, 2}]}, {A, B}, Plus, 1];

You can verify that they are the same:

combo1 === combo2
(* True *)

Note that the last (optional) argument to Inner is necessary here. From the documentation:

Inner[f, Subscript[list, 1], Subscript[list, 2], g, n] contracts index n of the first tensor with the first index of the second tensor.

I guess in this case, n == 1 is not the default.

---

If you would like to make a simple function that does this without all of the extra typing, do this:

dot[twoTensor_ /; Dimensions@twoTensor === {2, 2, 2}, oneTensor_ /; Dimensions@oneTensor == {2}] := Inner[Times, twoTensor, oneTensor, Plus, 1]

---

You can also take direct advantage of Dot using replacement rules:

Clear[aMat, bMat]
Dot[{aMat, bMat}, {A, B}] /. {aMat -> Array[a, {2, 2}], bMat -> Array[b, {2, 2}]}