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I have a question in FindRoot of matrix operation equations.

It's my first time use Matrix operation in FindRoot, i have write a simple example, it shows FindSolve can solve Matrix equations.

a = {{1, 2}, {3, 4}}; b = {{1, 3}, {2, 2}};

{c} /. FindRoot[
  c == a.b - IdentityMatrix[2],
  {c, IdentityMatrix[2]}
  ]

Very fast and simple!

Can I write a new expression in FindRoot like

a = {{1, 2}, {3, 4}}; b = {{1, 3}, {2, 2}};
d = {{1, 1}, {3, 3}};
{c} /. FindRoot[
  var == d.b,
  c == a.var - IdentityMatrix[2],
  {c, IdentityMatrix[2]}
  ]

and return c and var, well, var is some function result I want update in FindRoot and used to the equation I want to solve like c == a.var. var isn't need to solve.

Above code have issues. Here is my idea:

a = {{1, 2}, {3, 4}}; b = {{1, 3}, {2, 2}};
d = {{1, 1}, {3, 3}};
{var = d.b, c /. FindRoot[
   c == a.var - IdentityMatrix[2],
   {c, IdentityMatrix[2]}
   ]}

Can I achieve same result in a more compact form？Putting expressions (eg. var) that don't need to solve into FindRoot will cause errors.

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  • $\begingroup$ By the way, can I display the Math symbol more comfortable when I paste the code from notebook to this question? ` [CapitalOmega]` is not easy to reading. $\endgroup$
    – lumw
    Commented Nov 30, 2021 at 15:36
  • $\begingroup$ It might be helpful if you show the matrixequation you try to solve ! Your first example seems to be unclear $\endgroup$
    – Ulrich Neumann
    Commented Nov 30, 2021 at 16:02
  • 1
    $\begingroup$ q is not defined when the definition of qE0 refers to its parts. For your simple example, Solve works: c /. Solve[{c == a . b - IdentityMatrix[2], c \[Element] Matrices[{2, 2}]}, c][[1]] $\endgroup$
    – Bob Hanlon
    Commented Nov 30, 2021 at 16:25
  • 1
    $\begingroup$ See Additional useful buttons for our M.SE editor $\endgroup$
    – Bob Hanlon
    Commented Nov 30, 2021 at 16:28
  • 1
    $\begingroup$ Why not a.d.b - IdentityMatrix[2] directively? There no equation in your examples indeed. $\endgroup$
    – cvgmt
    Commented Dec 1, 2021 at 0:16

1 Answer 1

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$\begingroup$ 
a = {{1, 2}, {3, 4}}; b = {{1, 3}, {2, 2}};
d = {{1, 1}, {3, 3}};

With[{var = d . b},
 c /. Solve[{c == a . var - IdentityMatrix[2],
     c ∈ Matrices[2]}, c][[1]]]

(* {{20, 35}, {45, 74}} *)

With[{var = d . b},
 c /. FindRoot[c == a . var - IdentityMatrix[2],
   {c, IdentityMatrix[2]}]]

(* {{20., 35.}, {45., 74.}} *)
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