Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

I have a matrix

A = {{5, 4, 2}, {4, 5, 2}, {2, 2, 2}}

The eigenvalues are 1, 1, 10 and I need to find the eigenvectors.

Solving for 1 would result in two eigenvectors. Using WolframAlpha I get (-1, 0, 2) and (-1, 1, 0)

This comes from x3 = -2 (x1 + x2)

For instance I could have:

x1 = 1, x2 = -1, x3 = 0


x1 = -1, x2 = 1, x3=0


x1 = 0, x2 = -1, x3 = 2 


x1=-1, x2=0, x3=2

How to properly order those values?

share|improve this question

marked as duplicate by Mr.Wizard Sep 2 '14 at 9:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I believe this was already answered here: (1831). Therefore I am closing this question. If anyone feels that I am in error please tell me. – Mr.Wizard Sep 2 '14 at 9:03
up vote 5 down vote accepted

I'm not quite sure if I understand your problem correctly.

Anyway, use Eigensystem to find Eigenvalues and Eigenvectors at the same time.

Here we go

A = {{5, 4, 2}, {4, 5, 2}, {2, 2, 2}};


(* {{10, 1, 1}, {{2, 2, 1}, {-1, 0, 2}, {-1, 1, 0}}} *)

Hence the Eigenvalues are {10,1,1} and the respective Eigenvectors appear ordered lexicogaphically in the list {{2, 2, 1}, {-1, 0, 2}, {-1, 1, 0}}.

Hope this helps.

Best regards, Wolfgang

share|improve this answer
Using Eigensystem will give a unique ordering, but that avoids the underlying question. The set of eigenvectors for a matrix is not unique. – Steven Bell Sep 1 '14 at 20:42

Remember that by definition, an eigenvalue is only unique up to a scalar: Ax=λx, but you can multiply x by any scalar b and still have an eigenvector. This is the difference between the first two eigenvectors you've proposed ({1, -1, 0} and {-1, 1, 0}).

The second part is that when you have a repeated eigenvalue, there may be a space of independent eigenvectors. Here, any other linear combination of the eigenvectors you've proposed (such as {-1, -1, 4}) is a valid eigenvector. The important requirement for the second eigenvector is that it be linearly independent from the first. It's not a matter of picking the eigenvector that is right, it's a matter of picking one that is convenient. Often, we'll select it to be orthogonal.

Eigenvectors are ordered based on the values of their corresponding eigenvalues. I am unfamiliar with any convention for ordering eigenvectors corresponding to repeated eigenvalues.

share|improve this answer

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