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I have a matrix given as a function of is indices and another free parameter as follows

A[x_, i_, j_] = x*i*j

Now I would like to evaluate this for a given vector of indices $i, j$, e.g.

iVec = {1,2}
jVec = {1,2}

such that the matrix 

AMat[x_] = {{x,2*x},{2*x,4*x}}

is returned. We may assume that the matrix is square, i.e. iVec and jVec have the same length and the first argument of the function A is supposed to correspond to the row index of the resulting matrix.

Is there any way to do this? Ideally I would like to write something like

AMat[x_] = A[x, iVec, jVec]

but that will not work because Mathematica doesn't know which vector to treat as a transpose.

EDIT: The input should be A[x_, i_, j_] without modifying it's functional form. The reason is that I compute that automatically.

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    $\begingroup$ Perhaps A[x_,i_,j_]:=x*Outer[Times,i,j]; A[x,{1,2},{1,2}] ? $\endgroup$
    – Bill
    Jul 4, 2018 at 15:05
  • $\begingroup$ @Bill great minds... see MarcoB's answer and my comment to it. $\endgroup$
    – Wolpertinger
    Jul 4, 2018 at 15:26

1 Answer 1

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ClearAll[AMat]
AMat[x_, iVec_, jVec_] := x Outer[Times, iVec, jVec]

AMat[x, {1, 2}, {1, 2}]
(* Out: {{x, 2 x}, {2 x, 4 x}} *)

In response to the edit in the OP, in order to use the existing definition of A:

(* Existing definition of A *)
ClearAll[A]
A[x_, i_, j_] = x*i*j

(* New definition of AMat *)
ClearAll[AMat]
AMat[x_, iVec_, jVec_] := Table[A[x, i, j], {i, iVec}, {j, jVec}]

So now: 

AMat[x, {1, 2}, {1, 2}]
(* Out: {{x, 2 x}, {2 x, 4 x}} *)
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  • $\begingroup$ thanks for your answer! This works of course, but I originally wanted a way that does not modify A[x_, i_, j_]. The background is that A in terms of the indices is computed automatically. So I guess what is really missing is a way to automatically get from the form of A[x_, i_, j_] in the question to the second line in your answer. My question was probably unclear in that respect, I'll clarify it. +1 for the solution anyway! $\endgroup$
    – Wolpertinger
    Jul 4, 2018 at 15:25
  • $\begingroup$ @Wolpertinger See if the edit works for you $\endgroup$
    – MarcoB
    Jul 4, 2018 at 15:32
  • $\begingroup$ brilliant! Exactly what I wanted! $\endgroup$
    – Wolpertinger
    Jul 4, 2018 at 15:36
  • $\begingroup$ @Wolpertinger Great! Glad it worked. $\endgroup$
    – MarcoB
    Jul 4, 2018 at 15:37

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