Deriving a matrix from a given matrix equation

Questions
1

Dear Wolfram Forum,

I am trying to derive a specific transformation matrix given a matrix equation involving matrix multiplication and flattening operations. The equation is as follows:

A . (X // Flatten) == {Subscript[a, 0], Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]}

Where X is a 2x2 matrix defined as:

X = {{Subscript[a, 0] + Subscript[a, 3], Subscript[a, 1] - I Subscript[a, 2]}, 
     {Subscript[a, 1] + I Subscript[a, 2], Subscript[a, 0] - Subscript[a, 3]}}

And the result of the matrix multiplication A . (X // Flatten) simplifies to:

{Subscript[a, 0], Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]}

I have been able to empirically find a matrix A that satisfies the equation, which is:

{{1/2, 0, 0, 1/2}, {0, 1/2, 1/2, 0}, 1/I {0, -1/2, 1/2, 0}, {1/2, 0, 0, -1/2}}

However, I am seeking a systematic approach or methodology to determine A. I suspect this is a familiar transformation within certain fields, and thus I am reaching out to see if anyone can shed light on the official technique or formula to derive this matrix.

Any guidance on how to approach this problem, related literature, or relevant functions in Mathematica would be extremely helpful.

Thank you for your time and help.

Best regards,
Zhao

2

This

LogicalExpand[
  Reduce[
    {{a,b,c,d},{e,f,g,h},{i,j,k,l},{m,n,o,p}}.{a0+a3,a1-I a2,a1+I a2,a0-a3}==
    {a0,a1,a2,a3},
    {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p}
]]

quickly gives you five different solutions.

Formatted differently gives the same solutions

A={{a,b,c,d},{e,f,g,h},{i,j,k,l},{m,n,o,p}};
X={{a0+a3,a1-I a2},{a1+I a2,a0-a3}};
LogicalExpand[Reduce[A.(X//Flatten)=={a0, a1, a2, a3},Flatten[A]]]

But I cannot understand

I want to get the solutions that express a _ i (i=0,1,2,3) as the functions of elements of X _ ij, e.g

when your original problem was to find the 4x4 matrix A and you showed an example 4x4 solution of coefficients.

3

Thank you, but I want to get the solutions that express a _ i (i=0,1,2,3) as the functions of elements of X _ ij, e.g.:

a_0 = (X_11 + X22 ) /2
a_1 = (X_12 + X21 ) /2
a_2 = (X_21 + X12 ) / (2 I)
a_3 = (X_11 - X22 ) /2

Regards,
Zhao

4

This is my attempt:

X = {{Subscript[a, 0] + Subscript[a, 3], 
    Subscript[a, 1] - I Subscript[a, 2]}, {Subscript[a, 1] + 
     I Subscript[a, 2], Subscript[a, 0] - Subscript[a, 3]}};
coeffs = {Subscript[a, 0], Subscript[a, 1], Subscript[a, 2], 
   Subscript[a, 3]};
matrixA = Array[b, {4, 4}];
matrixXgeneric = Array[x, {2, 2}];
solA = Solve[Eliminate[{matrixA . (X // Flatten) == coeffs,
    X == matrixXgeneric},
   coeffs],
  Flatten[matrixA]]
matrixSol = matrixA /. solA[[1]] /. b[__] :> 1
matrixSol . (matrixXgeneric // Flatten) == coeffs /. 
  Thread[Flatten[matrixXgeneric] -> Flatten[X]] // Simplify
5

By running your code, I obtained the following results:

In[50]:= 
X = {{Subscript[a, 0] + Subscript[a, 3], 
    Subscript[a, 1] - I Subscript[a, 2]}, {Subscript[a, 1] + 
     I Subscript[a, 2], Subscript[a, 0] - Subscript[a, 3]}};
coeffs = {Subscript[a, 0], Subscript[a, 1], Subscript[a, 2], 
   Subscript[a, 3]};
matrixA = Array[b, {4, 4}];
matrixXgeneric = Array[x, {2, 2}];
solA = Solve[
  Eliminate[{matrixA . (X // Flatten) == coeffs, X == matrixXgeneric},
    coeffs], Flatten[matrixA]]
matrixSol = matrixA /. solA[[1]] /. b[__] :> 1
matrixSol . (matrixXgeneric // Flatten) == coeffs /. 
  Thread[Flatten[matrixXgeneric] -> Flatten[X]] // Simplify


During evaluation of In[50]:= Solve::svars: Equations may not give solutions for all "solve" variables.

Out[54]= {{b[1, 4] -> (
   x[1, 1] - 2 b[1, 1] x[1, 1] - 2 b[1, 2] x[1, 2] - 
    2 b[1, 3] x[2, 1] + x[2, 2])/(2 x[2, 2]), 
  b[2, 4] -> (-2 b[2, 1] x[1, 1] + x[1, 2] - 2 b[2, 2] x[1, 2] + 
    x[2, 1] - 2 b[2, 3] x[2, 1])/(2 x[2, 2]), 
  b[3, 4] -> (-2 b[3, 1] x[1, 1] + I x[1, 2] - 2 b[3, 2] x[1, 2] - 
    I x[2, 1] - 2 b[3, 3] x[2, 1])/(2 x[2, 2]), 
  b[4, 4] -> (
   x[1, 1] - 2 b[4, 1] x[1, 1] - 2 b[4, 2] x[1, 2] - 
    2 b[4, 3] x[2, 1] - x[2, 2])/(2 x[2, 2])}}

Out[55]= {{1, 1, 1, (-x[1, 1] - 2 x[1, 2] - 2 x[2, 1] + x[2, 2])/(
  2 x[2, 2])}, {1, 1, 1, (-2 x[1, 1] - x[1, 2] - x[2, 1])/(
  2 x[2, 2])}, {1, 1, 
  1, (-2 x[1, 1] - (2 - I) x[1, 2] - (2 + I) x[2, 1])/(
  2 x[2, 2])}, {1, 1, 
  1, (-x[1, 1] - 2 x[1, 2] - 2 x[2, 1] - x[2, 2])/(2 x[2, 2])}}

Out[56]= True

But I still cannot see how a _ i (i=0,1,2,3) are expressed via X _ ij by your results.
But I still can’t see how your result represents a _ i (i=0,1,2,3) by X _ ij (i=1,2, j = 1,2).

Regards,
Zhao

6

I thought you wanted A as a function of X. However, take another look at

matrixSol . (matrixXgeneric // Flatten) == coeffs
7

Wonderful, thank you very much for your tip. It does match up perfectly with the results I provided earlier:

In[11]:= 
matrixSol . (matrixXgeneric // Flatten) == coeffs // Simplify

Out[11]= {1/2 (x[1, 1] + x[2, 2]), 1/2 (x[1, 2] + x[2, 1]), 
  1/2 I (x[1, 2] - x[2, 1]), 1/2 (x[1, 1] - x[2, 2])} == {Subscript[a,
   0], Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]}

The above result is the same as my previous representation. Is this just a coincidence?

Regards,
Zhao

8

It does not seem to be a coincidence. There are infinitely many matrices A that solve your problem, but they all give the same relation between X and the a_i:

Simplify[(matrixA /. solA[[1]]) . (matrixXgeneric // Flatten)]
9

Wonderful confirmation and validation. This further prompts me to consider whether the issue here is related to the following discussion:

Selection_110.png
Selection_110.png1305Ă—513 73.5 KB

Regards,
Zhao