Mathematica Asked on December 28, 2020
I have difficulty in finding the transformation matrix of two similar matrices.
It is known that matrix $A=left(begin{array}{ccc}
-2 & -2 & 1
2 & x & -2
0 & 0 & -2
end{array}right)$ is similar to matrix $B=left(begin{array}{lll}
2 & 1 & 0
0 & -1 & 0
0 & 0 & y
end{array}right)$.
I found the specific values of $x$ and $y$ by using the following method:
Aλ = λ*
IdentityMatrix[3] - {{-2, -2, 1}, {2, x, -2}, {0, 0, -2}};
Bλ = λ*
IdentityMatrix[3] - {{2, 1, 0}, {0, -1, 0}, {0, 0, y}};
Control`PCS`SmithForm[Aλ, λ]
Control`PCS`SmithForm[Bλ, λ]
SolveAlways[-(λ + 2) (-λ^2 -
2 λ + λ x + 2 x - 4) == (λ -
2) (λ + 1) (-(y - λ)), λ]
When I want to find an invertible matrix P such that $P^{-1}AP=B$, I have a problem:
eqs = Thread[
Flatten[Inverse[{{x11, x12, x13}, {x21, x22, x23}, {x31, x32,
x33}}].({{-2, -2, 1}, {2, x, -2}, {0, 0, -2}} /.
x -> 3).{{x11, x12, x13}, {x21, x22, x23}, {x31, x32,
x33}}] ==
Flatten[{{2, 1, 0}, {0, -1, 0}, {0, 0, y}} /. y -> -2]];
FindInstance[eqs, {x11, x12, x13, x21, x22, x23, x31, x32,
x33}, Reals]
The above code has been running, unable to return the results, what is a good way to find this matrix $P$?
Your specific example can be solve with a general $P$, see code below.
(*Data*)
A = {{-2, -2, 1}, {2, x, -2}, {0, 0, -2}};
B = {{2, 1, 0}, {0, -1, 0}, {0, 0, y}};
(*Search for x and y based on characteristic polynomial*)
n = Length@A;
Id = IdentityMatrix@n;
solxy = SolveAlways[Det[A - l*Id] == Det[B - l*Id], l]
(*Update data*)
A = A /. solxy[[1]];
B = B /. solxy[[1]];
(*Solve for general P*)
P = Array[p, {n, n}];
solP = Solve[P.B == A.P, Flatten@P];
P = P /. solP[[1]]
(*Check*)
B == [email protected] // Simplify
You can then enter some values for the free components of $P$.
Correct answer by Mauricio Fernández on December 28, 2020
The purely linear algebraic way to do this is to reduce both matrices to Jordan form:
{sa, ja} = JordanDecomposition[{{-2, -2, 1}, {2, x, -2}, {0, 0, -2}}];
{sb, jb} = JordanDecomposition[{{2, 1, 0}, {0, -1, 0}, {0, 0, y}}];
Inspecting both ja
and jb
shows that they are both diagonal, so we can proceed:
Diagonal[ja]
{-2, 1/2 (-2 + x - Sqrt[-12 + 4 x + x^2]), 1/2 (-2 + x + Sqrt[-12 + 4 x + x^2])}
Diagonal[jb]
{-1, 2, y}
A moment's consideration leads us to letting y == -2
. To try finding x
, we try equating one of ja
's unknown eigenvalues to one of jb
's:
Solve[1/2 (-2 + x - Sqrt[-12 + 4 x + x^2]) == -1, x]
{{x -> 3}}
(Exercise: look at what happens if you take all corresponding pairs of eigenvalues between ja
and jb
, and equate them.)
Thus, we can assemble the similarity transformation as:
pa = (sa /. x -> 3).Inverse[sb[[All, {3, 1, 2}]]]
{{-1/2, -13/6, -1/4}, {1, 4/3, 1/2}, {0, 0, 1}}
(Exercise: derive the expression I used for the similarity transformation)
Check:
Inverse[pa].({{-2, -2, 1}, {2, x, -2}, {0, 0, -2}} /. x -> 3).pa -
({{2, 1, 0}, {0, -1, 0}, {0, 0, y}} /. y -> -2)
{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}
Answered by J. M.'s ennui on December 28, 2020
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