# 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 ] ? Apr 26 '19 at 12:01
• This hasn't worked but thanks. It returns an empty bracket Apr 26 '19 at 12:11
• Root objects are algebraic numbers. Apr 26 '19 at 12:41
• Why do you want a solution in radical terms? Root objects are much easier to work with. One thing you might find useful about them is their well-defined ordering. Apr 26 '19 at 14:20
• Eigenvectors[M[k]] // ToRadicals gives explicit expressions without Root objects. Jan 13 at 12:17

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.