Mathematica, solving non linear system of equations with lot of equations and variables

Question

I need to find a square matrix A satisfying the equation

A.L.A = -17/18A -2(A.L.L + L.A.L + (L.L).A) + 3(A.L + L.A) -4L.L.L + 8L.L - 44/9L + 8/9*(ID)

,where L is a diagonal matrix L = {{2/3,0,0,0},{0,5/12,0,0},{0,0,11/12,0},{0,0,0,2/3}}.

I can find the answers in the case that A is of dimension 2 and 3, but there is a problem with dimension 4 and above.

Actually, the matrix A has to satisfy the equation A.A = A too, but with a suitable matrix L only the equation above equation is enough.

This is my code ;

A = Table[a[i,j],{i,1,4},{j,1,4}]

B = A.L.A

ID = IdentityMatrix[4]

M = -17/18A -2(A.L.L + L.A.L + (L.L).A) + 3(A.L + L.A) -4L.L.L + 8L.L - 44/9L + 8/9*(ID)

diff = (B - M)//ExpandAll//Flatten ( so I get 16 non linear system of equations here )

A1 = A/.Solve[diff == 0][[1]]

After running this code for quite sometime, the error come up with there is not enough memory to compute.

In this case there are 16 equations and 16 variables. Some of the entries are parameters but I just do not know which one until I get the result.

I am not sure if there is anyway to solve this problem. I need the answer to be rational(probably integers) which is possible theoretically.

Could this problem be solved by matrix equation or any other method? I see one problem for this is there are too many equations and variables.

Answer 1

This evaluates fairly quickly and with modest memory for a problem this size.

L = {{2/3, 0, 0, 0}, {0, 5/12, 0, 0}, {0, 0, 11/12, 0}, {0, 0, 0, 2/3}};
A = {{a, b, c, d}, {e, f, g, h}, {i, j, k, l}, {m, n, o, p}};
Reduce[{A.L.A == -17/18 A - 2 (A.L.L + L.A.L + (L.L).A) + 3 (A.L + L.A) - 
   4 L.L.L + 8 L.L - 44/9 L + 8/9*IdentityMatrix[4]},
   {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p}, Backsubstitution->True
]

Then you just have to sort through the 143 potential solutions that it returns. You might be able to Select from those that satisfy your A.A==A. You can also use ToRules on the result returned from Reduce to put this into a form similar to that returned from Solve, but check this carefully to make certain it is doing what you expect.

Check this very carefully to make certain I haven't made any mistakes.