Mathematica Asked by Ben Pepper on April 12, 2021
I am trying to obtain the eigenvectors of a unitary matrix $M(k)$ which depends on a parameter k.
This matrix $M(k)$ has dimension 6, and while for general matrices of dimension 6 it’s not possible to write down algebraic expressions for their eigenvalues, because the characteristic polynomial will also be of order 6, for the case of $M(k)$ it’s characteristic polynomial is such that it is possible to write its eigenvalues in algebraic form.
When I use the Eigenvectors[] function in Mathematica, it gives me eigenvectors which
So basically I don’t trust how Eigenvectors[] is working for this matrix. I would like to use another way of calculating the eigenvectors.
I have tried computing the eigenvectors $v$ of a matrix $M(k)$ of dimension 6 in Mathematica by using Solve[] on this equation
$(M(k) – aI)v = 0$
where a is an eigenvalue of $M(k)$.
Why does Mathematica only give me the trivial solution $v=0$?
I checked that the determinant of $(M(k) – aI)$ is zero, and so if I denote an eigenvector $v$ as $v = (v_1,v_2,v_3,v_4,v_5,v_6)$ then there should be a solution where $v_2, v_3, v_4, v_5,$ and $v_6$ are written solely in terms of $v_1$, but this solution doesn’t appear.
I tried doing the exact same procedure as above for a different matrix of dimension 2 and IT DID GIVE ME the non-trivial solutions, so I’m not sure why it can’t do it for $M(k)$.
This is the matrix
$ frac{1}{sqrt{2}}begin{bmatrix} 0 & 0 & 1 & i & 0 & 0 0 & 0 & 0 & 0 & ie^{-ik} & e^{-ik} 0 & 0 & 0 & 0 & e^{ifrac{2pi}{3}} & ie^{ifrac{2pi}{3}} ie^{ifrac{2pi}{3}} & e^{ifrac{2pi}{3}} & 0 & 0 & 0 & 0 e^{ik}e^{-ifrac{2pi}{3}}& ie^{ik}e^{-ifrac{2pi}{3}} & 0 & 0 & 0 & 0 0 & 0 & ie^{-ifrac{2pi}{3}} & e^{-ifrac{2pi}{3}} & 0 & 0 end{bmatrix}$
{{0,0,1/Sqrt[2],I/Sqrt[2],0,0},
{0,0,0,0,(I E^(-I k))/Sqrt[2],E^(-I k)/Sqrt[2]},
{0,0,0,0,E^((2 I π)/3)/Sqrt[2],(I E^((2 I π)/3))/Sqrt[2]},
{(I E^((2 I π)/3))/Sqrt[2],E^((2 I π)/3)/Sqrt[2],0,0,0,0},
{E^(I k-(2 I π)/3)/Sqrt[2],(I E^(I k-(2 I π)/3))/Sqrt[2],0,0,0,0},
{0,0,(I E^(-((2 I π)/3)))/Sqrt[2],E^(-((2 I π)/3))/Sqrt[2],0,0}}
I need to find the eigenvectors of this matrix
The 6×6 matrix $m$ looks like three 2×2 matrices, so re-ordering the matrix may be helpful. We can find the first eigenvector, in terms of the Root
expressions, this way
μ = Eigenvalues[m];
rowOrder = {4, 5, 1, 6, 2, 3};
xfrm = IdentityMatrix[6][[rowOrder]];
s = xfrm.m;
v = First@NullSpace[s - μ[[1]] xfrm];
m.v - μ[[1]] v // Simplify
(* {0, 0, 0, 0, 0, 0} *)
So, $v$ is an eigenvector for the first eigenvalue, $mu_1$. The transformation that re-ordered the rows preserved the null space. If someone can find a transformation that preserves the eigenvalues and re-orders the matrix to have a banded structure, Mathematica may (or may not) be able to find the eigenvalues that use radicals instead of Root
.
Answered by LouisB on April 12, 2021
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