Apply Outer to list of matrices and vectors

Question

I have a list, M, of square $ n \times n $ matrices, {M1, M2, M3, ...}, and a list, V, of $ n \times 1 $ vectors, {v1, v2, v3, ...}, and the corresponding transposes of those vectors {r1, r2, r3, ...}. I'm trying to form the matrix {{r1.M1.v1, r2.M1.v2, r3.M1.v3, ...}, {{r1.M2.v1, r2.M2.v2, r3.M2.v3, ...}, ...}. Note that M and V are not necessarily the same length (i.e. there may only be 5 matrices, but 100 vectors).

It seemed like a method using Outer would work, such as: Outer[Transpose[#2].#1.#2 &, M, V], which should then just be a 2-dimensional matrix of scalars.

However, I think that this runs into a problem because the lists M and V are themselves technically lists of lists (M being a list of matrices, V being a list of vectors), and so the outer gets distributed into the sublists, rather than doing the calculation I want and it ends up being a high dimensional object. I've tried playing around with various flattening schemes, but haven't quite figured it out - any assistance on how to implement this functionality (is Outer even the correct functional tool to use)?

Answer 1

I believe what you want is

Outer[#2.#1.#2 &, M, V, 1]

First of all, you don't need to Transpose vectors in Mathematica. Dot knows what to do automatically. Second, you need to use the optional extra argument for Outer that tells it the level at which the "object" that you are feeding to the function lives. Otherwise, Outer seeks the lowest level by default. Since your "objects" are at level 1 in both lists (the elements of the lists which are either matrices or vectors), use the level-spec 1 as the extra argument to Outer.

Answer 2

Just use Table

SeedRandom@12;
m = Table[RandomInteger[5, {2, 2}], 3];
MatrixForm /@ m

m=$\{\left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \\ \end{array} \right),\left( \begin{array}{cc} 5 & 4 \\ 1 & 4 \\ \end{array} \right),\left( \begin{array}{cc} 5 & 3 \\ 0 & 5 \\ \end{array} \right)\}$

v = Table[RandomInteger[5, 2], 3]

v={{3, 4}, {1, 1}, {2, 5}}

Table[v[[j]].m[[i]].v[[j]], {i, Length@m}, {j, Length@m}]

{{21, 2, 14}, {169, 14, 170}, {161, 13, 175}}

Second Solution

#2.#1.#2 & @@@ Tuples[{m, v}] // Partition[#, 3] &