# Solving polynomial with matrix coefficients **URL:** \ **Category:** Questions\ **Tags:** mathematica, algebra\ **Created:** [November 29, 2014, 9:35pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159 "2014-11-29T21:35:44Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [November 29, 2014, 9:35pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/1 "2014-11-29T21:35:44Z")
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. ---
**Author:** ![antononcube](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/antononcube/32/7348_2.png) [@antononcube](https://community.wolfram.com/u/antononcube)\ **Post date:** [November 29, 2014, 10:49pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/2 "2014-11-29T22:49:33Z")
You might consider using the function `MatrixFunction`. ---
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [November 30, 2014, 1:00am UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/3 "2014-11-30T01:00:55Z")
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… ---
**Author:** ![antononcube](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/antononcube/32/7348_2.png) [@antononcube](https://community.wolfram.com/u/antononcube)\ **Post date:** [November 30, 2014, 1:09am UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/4 "2014-11-30T01:09:47Z")
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`. ---
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [November 30, 2014, 1:15am UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/5 "2014-11-30T01:15:48Z")
Sure. Attached. ---
**Author:** ![danl](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/danl/32/41267_2.png) [@danl](https://community.wolfram.com/u/danl)\ **Post date:** [December 2, 2014, 2:34pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/6 "2014-12-02T14:34:19Z")
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. ---
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [December 2, 2014, 2:54pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/7 "2014-12-02T14:54:46Z")
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? ---
**Author:** ![danl](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/danl/32/41267_2.png) [@danl](https://community.wolfram.com/u/danl)\ **Post date:** [December 2, 2014, 3:42pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/8 "2014-12-02T15:42:46Z")
“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}} *) ``` ---
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [December 3, 2014, 5:29pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/9 "2014-12-03T17:29:52Z")
When I change from ``` u=Array[x,{3,3}] ``` to ``` 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? ---
**Author:** ![danl](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/danl/32/41267_2.png) [@danl](https://community.wolfram.com/u/danl)\ **Post date:** [December 3, 2014, 7:02pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/10 "2014-12-03T19:02:38Z")
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}} *) ``` ---
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [December 3, 2014, 7:28pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/11 "2014-12-03T19:28:41Z")
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. ---
**Author:** ![danl](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/danl/32/41267_2.png) [@danl](https://community.wolfram.com/u/danl)\ **Post date:** [December 3, 2014, 7:32pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/12 "2014-12-03T19:32:07Z")
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}} *) ``` ---
**Author:** ![ryanp16](https://avatars.discourse-cdn.com/v4/letter/r/d07c76/32.png) [@ryanp16](https://community.wolfram.com/u/ryanp16)\ **Post date:** [December 3, 2014, 8:46pm UTC](https://community.wolfram.com/t/solving-polynomial-with-matrix-coefficients/4159/13 "2014-12-03T20:46:20Z")
Ah okay. Thank you very much! Why can’t the solution be substituted into B[u] normally though?