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Solving polynomial with matrix coefficients

Paul Ryan

Posted 10 years ago

I have a polynomial B in u that I want to solve, where u is an operator. It's of degree three so there will be three operators that solve it, and I want the solutions given as matricies. My current code is L[u_] = Array[L, {3, 3}] + {{u, 0, 0}, {0, u, 0}, {0, 0, u}} t[\[Epsilon]] = {{1, \[Epsilon]1, \[Epsilon]2}, {-\[Epsilon]1, 1, \[Epsilon]3}, {-\[Epsilon]2, -\[Epsilon]3, 1}} T[u_] = L[u].t[\[Epsilon]] The polynomial for B is B = T[u][[2, 3]] (T[u][[1, 2]] T[u - I][[2, 3]] - T[u][[1, 3]] T[u - I][[2, 2]]) - T[u][[1, 3]] (T[u][[1, 3]] T[u - I][[2, 1]] - T[u][[1, 1]] T[u - I][[2, 3]]) The L[i, j] are also matricies given by L[1, 1] = I/3 {{2, 0, 0}, {0, -1, 0}, {0, 0, -1}} L[2, 2] = I/3 {{-1, 0, 0}, {0, 2, 0}, {0, 0, -1}} L[3, 3] = I/3 {{-1, 0, 0}, {0, -1, 0}, {0, 0, 2}} L[1, 2] = I {{0, 0, 0}, {1, 0, 0}, {0, 0, 0}} L[1, 3] = I {{0, 0, 0}, {0, 0, 0}, {1, 0, 0}} L[2, 1] = I {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}} L[2, 3] = I {{0, 0, 0}, {0, 0, 0}, {0, 1, 0}} L[3, 1] = I {{0, 0, 1}, {0, 0, 0}, {0, 0, 0}} L[3, 2] = I {{0, 0, 0}, {0, 0, 1}, {0, 0, 0}} I tried don't know how to tell it that u is a 3x3 matrix and how to find that matrix. Up until now I was writing u= x {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}} and B resulted in me having a 3x3 matrix with a polynomial of degree 3 on each element of the diagonal which isn't what I want. Any help appreciated, thanks.

I have a polynomial B in u that I want to solve, where u is an operator. It's of degree three so there will be three operators that solve it, and I want the solutions given as matricies.

My current code is

L[u_] = Array[L, {3, 3}] + {{u, 0, 0}, {0, u, 0}, {0, 0, u}}

t[\[Epsilon]] = {{1, \[Epsilon]1, \[Epsilon]2}, {-\[Epsilon]1, 
   1, \[Epsilon]3}, {-\[Epsilon]2, -\[Epsilon]3, 1}}

T[u_] = L[u].t[\[Epsilon]]

The polynomial for B is

B = T[u][[2, 3]] (T[u][[1, 2]] T[u - I][[2, 3]] - 
     T[u][[1, 3]] T[u - I][[2, 2]]) - 
  T[u][[1, 3]] (T[u][[1, 3]] T[u - I][[2, 1]] - 
     T[u][[1, 1]] T[u - I][[2, 3]])

The L[i, j] are also matricies given by

L[1, 1] = I/3 {{2, 0, 0}, {0, -1, 0}, {0, 0, -1}}
L[2, 2] = I/3 {{-1, 0, 0}, {0, 2, 0}, {0, 0, -1}}
L[3, 3] = I/3 {{-1, 0, 0}, {0, -1, 0}, {0, 0, 2}}
L[1, 2] = I {{0, 0, 0}, {1, 0, 0}, {0, 0, 0}}
L[1, 3] = I {{0, 0, 0}, {0, 0, 0}, {1, 0, 0}}
L[2, 1] = I {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}}
L[2, 3] = I {{0, 0, 0}, {0, 0, 0}, {0, 1, 0}}
L[3, 1] = I {{0, 0, 1}, {0, 0, 0}, {0, 0, 0}}
L[3, 2] = I {{0, 0, 0}, {0, 0, 1}, {0, 0, 0}}

I tried don't know how to tell it that u is a 3x3 matrix and how to find that matrix. Up until now I was writing u= x {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}} and B resulted in me having a 3x3 matrix with a polynomial of degree 3 on each element of the diagonal which isn't what I want.

Any help appreciated, thanks.

POSTED BY: Paul Ryan

12 Replies

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Anton Antonov

Anton Antonov, Accendo Data LLC

Posted 10 years ago

You might consider using the function `MatrixFunction`.

POSTED BY: Anton Antonov

Paul Ryan

Posted 10 years ago

I've tried using it, I put m = {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}} and set f[u] = MatrixFunction[B[#] &, u] (note I fixed the definition of B in the first post and wrote in as B[u] and then used the usual solve function for f[u]=m, and got some crazy answer it took about two minutes to compute. Earlier up I did write u as u = Array[x, {3, 3}], so I'm not sure if that's causing issues or not...

POSTED BY: Paul Ryan

Anton Antonov

Anton Antonov, Accendo Data LLC

Posted 10 years ago

Can you attach a notebook with your code to the discussion? It seems that your matrix equation can be unfolded into a set of equations using `Thread` and giving the equations and the corresponding variables to `Reduce`.

POSTED BY: Anton Antonov

Paul Ryan

Posted 10 years ago

Sure. Attached. Attachments:

POSTED BY: Paul Ryan

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

This is not very clear to me. Are you trying to do this? sol = u /. Solve[Flatten[B[u]] == Flatten[m], Flatten[u]]; That gives many solution matrices expressed in terms of the epsilons.

POSTED BY: Daniel Lichtblau

Paul Ryan

Posted 10 years ago

It might be what I'm trying to do... I want to get matricies u that solve B[u]=0. I put your code into mine but it didn't work - may I see your code please?

POSTED BY: Paul Ryan

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

"My" code is just your code. I didn't change anything that I recall. L[u_] = Array[L, {3, 3}] + {{u, 0, 0}, {0, u, 0}, {0, 0, u}}; t[\[Epsilon]] = {{1, \[Epsilon]1, \[Epsilon]2}, {-\[Epsilon]1, 1, \[Epsilon]3}, {-\[Epsilon]2, -\[Epsilon]3, 1}}; T[u_] = L[u].t[\[Epsilon]]; B[u_] := T[u][[2, 3]] (T[u][[1, 2]] T[u - v][[2, 3]] - T[u][[1, 3]] T[u - v][[2, 2]]) - T[u][[1, 3]] (T[u][[1, 3]] T[u - v][[2, 1]] - T[u][[1, 1]] T[u - v][[2, 3]]); L[1, 1] = I/3 {{2, 0, 0}, {0, -1, 0}, {0, 0, -1}}; L[2, 2] = I/3 {{-1, 0, 0}, {0, 2, 0}, {0, 0, -1}}; L[3, 3] = I/3 {{-1, 0, 0}, {0, -1, 0}, {0, 0, 2}}; L[1, 2] = I {{0, 0, 0}, {1, 0, 0}, {0, 0, 0}}; L[1, 3] = I {{0, 0, 0}, {0, 0, 0}, {1, 0, 0}}; L[2, 1] = I {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}}; L[2, 3] = I {{0, 0, 0}, {0, 0, 0}, {0, 1, 0}}; L[3, 1] = I {{0, 0, 1}, {0, 0, 0}, {0, 0, 0}}; L[3, 2] = I {{0, 0, 0}, {0, 0, 1}, {0, 0, 0}}; u = Array[x, {3, 3}]; v = I {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}; sol = u /. Solve[Flatten[B[u]] == 0, Flatten[u]]; Length[sol] (* Out[18]= 432 ) Here is one of them. sol[[1]] ( Out[19]= {{-((2I)/3), 0, 0}, {0, I/3, 0}, {-((I(2\[Epsilon]1\[Epsilon]2 - \[Epsilon]2^2\[Epsilon]3 - \[Epsilon]3^3 - I\[Epsilon]3Sqrt[4\[Epsilon]1^2 - \[Epsilon]2^4 - 2\[Epsilon]2^2\[Epsilon]3^2 - \[Epsilon]3^4]))/ (2(\[Epsilon]1\[Epsilon]2^2 + \[Epsilon]1\[Epsilon]3^2))), 0, I/3}} )

"My" code is just your code. I didn't change anything that I recall.

L[u_] = Array[L, {3, 3}] + {{u, 0, 0}, {0, u, 0}, {0, 0, u}};
t[\[Epsilon]] = {{1, \[Epsilon]1, \[Epsilon]2}, {-\[Epsilon]1, 
    1, \[Epsilon]3}, {-\[Epsilon]2, -\[Epsilon]3, 1}};
T[u_] = L[u].t[\[Epsilon]];
B[u_] := T[u][[2, 3]] (T[u][[1, 2]] T[u - v][[2, 3]] - 
      T[u][[1, 3]] T[u - v][[2, 2]]) - 
   T[u][[1, 3]] (T[u][[1, 3]] T[u - v][[2, 1]] - 
      T[u][[1, 1]] T[u - v][[2, 3]]);
L[1, 1] = I/3 {{2, 0, 0}, {0, -1, 0}, {0, 0, -1}};
L[2, 2] = I/3 {{-1, 0, 0}, {0, 2, 0}, {0, 0, -1}};
L[3, 3] = I/3 {{-1, 0, 0}, {0, -1, 0}, {0, 0, 2}};
L[1, 2] = I {{0, 0, 0}, {1, 0, 0}, {0, 0, 0}};
L[1, 3] = I {{0, 0, 0}, {0, 0, 0}, {1, 0, 0}};
L[2, 1] = I {{0, 1, 0}, {0, 0, 0}, {0, 0, 0}};
L[2, 3] = I {{0, 0, 0}, {0, 0, 0}, {0, 1, 0}};
L[3, 1] = I {{0, 0, 1}, {0, 0, 0}, {0, 0, 0}};
L[3, 2] = I {{0, 0, 0}, {0, 0, 1}, {0, 0, 0}};
u = Array[x, {3, 3}];
v = I {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};

sol = u /. Solve[Flatten[B[u]] == 0, Flatten[u]];
Length[sol]

(* Out[18]= 432 *)

Here is one of them.

sol[[1]]

(* Out[19]= {{-((2*I)/3), 0, 0}, {0, I/3, 0}, 
{-((I*(2*\[Epsilon]1*\[Epsilon]2 - \[Epsilon]2^2*\[Epsilon]3 - \[Epsilon]3^3 - 
         I*\[Epsilon]3*Sqrt[4*\[Epsilon]1^2 - \[Epsilon]2^4 - 
            2*\[Epsilon]2^2*\[Epsilon]3^2 - \[Epsilon]3^4]))/
          (2*(\[Epsilon]1*\[Epsilon]2^2 + \[Epsilon]1*\[Epsilon]3^2))), 0, I/3}} *)

POSTED BY: Daniel Lichtblau

Paul Ryan

Posted 10 years ago

When I change from u=Array[x,{3,3}] to u={{-((2I)/3), 0, 0}, {0, I/3, 0}, {-((I(2\[Epsilon]1\[Epsilon]2 - \[Epsilon]2^2\[Epsilon]3 - \[Epsilon]3^3 - I\[Epsilon]3Sqrt[4\[Epsilon]1^2 - \[Epsilon]2^4 - 2\[Epsilon]2^2\[Epsilon]3^2 - \[Epsilon]3^4]))/ (2(\[Epsilon]1\[Epsilon]2^2 + \[Epsilon]1*\[Epsilon]3^2))), 0, I/3}} B[u] doesn't equal zero though?

When I change from

u=Array[x,{3,3}]

u={{-((2*I)/3), 0, 0}, {0, I/3, 0}, 
{-((I*(2*\[Epsilon]1*\[Epsilon]2 - \[Epsilon]2^2*\[Epsilon]3 - \[Epsilon]3^3 - 
         I*\[Epsilon]3*Sqrt[4*\[Epsilon]1^2 - \[Epsilon]2^4 - 
            2*\[Epsilon]2^2*\[Epsilon]3^2 - \[Epsilon]3^4]))/
          (2*(\[Epsilon]1*\[Epsilon]2^2 + \[Epsilon]1*\[Epsilon]3^2))), 0, I/3}}

B[u] doesn't equal zero though?

POSTED BY: Paul Ryan

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

Properly substituted, you should get zero. Together[B[u] /. Thread[Flatten[u] -> Flatten[sol[[1]]]]] (* Out[20]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}} *)

POSTED BY: Daniel Lichtblau

Paul Ryan

Posted 10 years ago

I see. Why is it not sufficient to just put it in normally? Sorry for not following - I've never seen the Flatten stuff before.

POSTED BY: Paul Ryan

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

I guess I made it too complicated. You are correct that a direct substitution can be done. Here is what I should have shown: B[sol[[1]]] // Together (* Out[24]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}} *)

POSTED BY: Daniel Lichtblau

Paul Ryan

Posted 10 years ago

Ah okay. Thank you very much! Why can't the solution be substituted into B[u] normally though?

POSTED BY: Paul Ryan

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