I need to solve the system of linear equations $S_i = \sum_ {r = 1}^\lambda x_ {r, i} E_r$, where $S_i$ is a $\lambda$ - dimensional vector, $i$ runs from 1 to $n$, $r$ runs from 1 to $\lambda$, and $x$ is a $\lambda\times n$ matrix.

I have no idea how to typeset that in Mathematica when I don' t know the values of $n$ and $\lambda$. The $S_i$ come from datasets of different sizes, and I build the $E_r$ from said datasets.

Thank you very much in advance.

Edit - Maybe this will help explain beter: I want to avoid having to type

Solve[S[[i,1]]==x[[1,i]]E[[1,1]] + x[[2,i]]E[[1,2]] + ...
   && S[[i,2]]== ...
   && ...

for $\lambda$ rows, each one with $\lambda$ elements. There must be a way to generalise this and typeset in a much simpler way which will then work for any matrix I feed into the code regardless of its dimensions.


  • 1
    $\begingroup$ What about: eq= Table[S[[i]] == Sum[x[[r, i]] E[[r]], {r, \[Lambda]}], {i, n}] $\endgroup$ Oct 13, 2021 at 11:00
  • $\begingroup$ @DanielHuber, and then just solve eq for x? $\endgroup$
    – LNah
    Oct 13, 2021 at 22:40
  • $\begingroup$ Yes, eq gives the equations for x $\endgroup$ Oct 14, 2021 at 7:27

1 Answer 1



(* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *)

You cannot use E as a variable name since it is used for the exponential constant. Use [DoubleStruckCapitalE] instead.

$Assumptions = Element[i | λ | n, Integers] && 
  i > 0 && λ > 0 && n > 0 && 
  Element[\[DoubleStruckCapitalE] | S, Vectors[λ]] && 
  Element[x, Matrices[{λ, n}]];

Format[\[DoubleStruckCapitalE][r_]] := 
  Subscript[Style[\[DoubleStruckCapitalE], Italic], r];

Format[S[i_]] := Subscript[S, i];

Format[x[r_, i_]] := Subscript[x, r, i]

(eqn = S[i] == 
    Sum[HoldForm[x[r, i]*\[DoubleStruckCapitalE][r]], {r, 
      1, λ}]) // TraditionalForm

enter image description here

Or in StandardForm

eqn // ReleaseHold

enter image description here

  • $\begingroup$ Thanks for pointing out that E cannot be used as a variable name. I just used it here as a placeholder, but I'll keep that in mind in the future. From your answer it is still unclear to me (or maybe I was unclear in my original question) how to solve for x. So far what I've tried either throws out errors or gives me a solution of 'x', which is most unhelpful. $\endgroup$
    – LNah
    Oct 13, 2021 at 22:52

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