7
$\begingroup$

For example, I know that matrices $$A=\left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ -6 & -1 & 4 & -5 \\ 12 & 0 & -5 & 5 \\ 12 & 0 & -6 & 5 \\ \end{array} \right)$$ and $$B=\left( \begin{array}{cccc} 0 & 0 & 0 & 35 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 \\ \end{array} \right)$$ are similar. Is there any efficient way to find $P$ such that $A=PBP^{-1}$? Let's say a code SimTransform[A_,B_]:= ... that returns $P$ when $A$ and $B$ are similar.

$\endgroup$
2
  • 2
    $\begingroup$ P={{p1,p2,p3,p4},{p5,p6,p7,p8},{p9,p10,p11,p12},{p13,p14,p15,p16}}; Q=P/.ToRules[Reduce[A==P.B.Inverse[P],Flatten[P]]] returns {{p1,p2,7*p1,7*p2}, {p5,p6,-12*p1-5*p5,-12*p2-5*p6}, {-p1+p2-p5,7*p1-p2-p6,5*p1+7*p2+5*p5,49*p1+5*p2+5*p6}, {(-10*p1+4*p2-5*p5-p6)/5,8*p1-2*p2+p5-p6,-2*p1+8*p2+5*p5+p6,44*p1-2*p2-5*p5+5*p6}} if I haven't made any mistakes translating latex back into Mathematica. And Simplify[A==Q.B.Inverse[Q]] returns True $\endgroup$
    – Bill
    Oct 16, 2022 at 22:40
  • 1
    $\begingroup$ I'd go with linear algebra:amat = {{1, 1, 1, 0}, {-6, -1, 4, -5}, {12, 0, -5, 5}, {12, 0, -6, 5}}; bmat = {{0, 0, 0, 35}, {1, 0, 0, 0}, {0, 1, 0, 2}, {0, 0, 1, 0}}; simmat = Array[p, {4, 4}]; simmat /. Solve[amat . simmat == simmat . bmat, Flatten[simmat]] $\endgroup$ Oct 17, 2022 at 14:11

4 Answers 4

8
$\begingroup$
A = {{1, 1, 1, 0}, {-6, -1, 4, -5}, {12, 0, -5, 5}, {12, 0, -6, 5}};
B = {{0, 0, 0, 35}, {1, 0, 0, 0}, {0, 1, 0, 2}, {0, 0, 1, 0}};
p = Partition[Table[Unique["x"], 16], 4];
pvalue = Partition[Flatten[Values[FindInstance[p.B.Inverse[p]==A,Catenate[p]]]],4]

{{0, -1, 0, -7}, {-1, 1, 5, 7}, {0, 0, -12, 0}, {0, 0, -12, 12}}

Check:

pvalue.B.Inverse[pvalue] == A

True

$\endgroup$
5
$\begingroup$

This works if it so happens that Eigensystem returns the eigenvalues of $A$ and $B$ in the same order:

SimTransform[A_,B_]:=Module[{valsA,vecsA,valsB,vecsB},
  {valsA,vecsA}=Eigensystem[A];
  {valsB,vecsB}=Eigensystem[B];
  If[valsA===valsB,
    Transpose[vecsA].Inverse[Transpose[vecsB]],
    $Failed]];

Example:

A={{1,1,1,0},{-6,-1,4,-5},{12,0,-5,5},{12,0,-6,5}};
B={{0,0,0,35},{1,0,0,0},{0,1,0,2},{0,0,1,0}};

P=SimTransform[A,B]//Simplify
(* {{1/72,1/72,7/72,7/72},
    {-1/36,1/18,-1/36,-4/9},
    {1/36,1/36,1/36,37/36},
    {0,0,0,1}} *)

A-P.B.Inverse[P]//Simplify
(* gives zero *)
$\endgroup$
1
  • 2
    $\begingroup$ Your method only work if A and B are diagonalizable $\endgroup$
    – yode
    Oct 17, 2022 at 4:57
4
$\begingroup$
Clear["Global`*"]

SimTransform[A_?MatrixQ, B_?MatrixQ] := Module[
   {P = Array[p, {Length@A, Length@A}]},
   P /. ToRules@Reduce[A == P . B . Inverse[P], Flatten@P]] /;
  And @@ {Dimensions@A === Dimensions@B, Tr[A] == Tr[B],
    Eigenvalues@A == Eigenvalues@B}

A = {{1, 1, 1, 0}, {-6, -1, 4, -5}, {12, 0, -5, 5}, {12, 0, -6, 5}};

B = {{0, 0, 0, 35}, {1, 0, 0, 0}, {0, 1, 0, 2}, {0, 0, 1, 0}};

Format[p[arg__]] := Subscript[p, Row[{arg}]]

P = SimTransform[A, B]

enter image description here

Verifying,

A == P . B . Inverse[P] // Simplify

(* True *)
$\endgroup$
4
$\begingroup$

Your can use Jordan decomposition.

A = {{1, 1, 1, 0}, {-6, -1, 4, -5}, {12, 0, -5, 5}, {12, 0, -6, 5}};
B = {{0, 0, 0, 35}, {1, 0, 0, 0}, {0, 1, 0, 2}, {0, 0, 1, 0}};
{aTrasf, aJF} = JordanDecomposition[A];
{bTrasf, bJF} = JordanDecomposition[B];
If[aJF === bJF, T = Simplify[aTrasf . Inverse[bTrasf]]];

If Jordan forms are the same, then T is the similarity transform. I think they can differ by constant factor (at least). Didn't check that case

Simplify[T . B . Inverse[T]]===A

(* True *)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.