Obtaining eigenvectors without using Eigenvectors

Introduction

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

1. Cannot be written algebraically, i.e. it involves expressions with roots and # when it should be possible to avoid this since the eigenvalues are algebraic.
2. Discontinuous when I don't think they should be for this matrix.

So basically I don't trust how Eigenvectors[] is working for this matrix. I would like to use another way of calculating the eigenvectors.

Question

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

$$\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 0 & 1 & i & 0 & 0\\ 0 & 0 & 0 & 0 & ie^{-ik} & e^{-ik} \\ 0 & 0 & 0 & 0 & e^{i\frac{2\pi}{3}} & ie^{i\frac{2\pi}{3}} \\ ie^{i\frac{2\pi}{3}} & e^{i\frac{2\pi}{3}} & 0 & 0 & 0 & 0\\ e^{ik}e^{-i\frac{2\pi}{3}}& ie^{ik}e^{-i\frac{2\pi}{3}} & 0 & 0 & 0 & 0 \\ 0 & 0 & ie^{-i\frac{2\pi}{3}} & e^{-i\frac{2\pi}{3}} & 0 & 0 \\ \end{bmatrix}$$

{{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

• NullSpace[ M - a I ] ? – LouisB Apr 26 '19 at 12:01
• This hasn't worked but thanks. It returns an empty bracket – Ben Pepper Apr 26 '19 at 12:11
• Root objects are algebraic numbers. – John Doty Apr 26 '19 at 12:41
• But objects like Root[#^6 + #^1 + 1 &, 2] cannot be written any simpler is my point. Eigenvectors[] gives me expressions like Root[#^6 + #^1 + 1 &, 2] which should not be there, there is no need for them. The problem is this: When I do Eigenvalues[M], mathematica returns expressions with Root that cannot be simplified because they are for polynomials of order 6. But the eigenvalues can be written perfectly well without using Root, i.e. it can be written explicitly. I know this because if I calculate CharacteristicPolynomial[] and then solve this polynomial, there are no Roots – Ben Pepper Apr 26 '19 at 13:01
• Try setting one of the elements of the eigenvector v to be equal to 1 in order to force it off of the trivial solution. – MikeY Apr 26 '19 at 13:48

The 6x6 matrix $$m$$ looks like three 2x2 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.