Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the Matrix $\pmatrix{E&t\\t&-E}$

The eigenvalues are $\lambda_1=\sqrt{E^2+t^2}$ , $\lambda_2=-\sqrt{E^2+t^2}$.

Consider the first (+ve) eigenvalue.

To find the eigenvectors, we write:

$$\pmatrix{E-\lambda_1 &t\\t&-E-\lambda_1} \pmatrix{a\\b} =\pmatrix{0\\0}$$

The problem is that there are two different ways to proceed:

$$(1)\qquad(E-\lambda_1)a + tb = 0 \Longrightarrow b=\frac{-E+\lambda_1}{t}$$

$$(2)\qquad ta + (-E-\lambda_1)b = 0 \Longrightarrow a=\frac{E+\lambda_1}{t}$$

I know the general solutions is any multiple integer of the eigenvector, but what we have above is NOT! that's something different, what is above is different eigenvectors for the same eigenvalue (not simple integer multiples...)

So what's wrong? Which one of 1) or 2) is right?

share|improve this question
1  
Welcome to MSE! It really helps readability to format your questions using MathJax (see FAQ - top right). Is E the exponential or just a variable? Regards –  Amzoti Apr 1 '13 at 0:30
    
For some basic information about writing math at this site see e.g. here, here, here and here. I have tried to improve the readability of your question by introducing Tex. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. –  user1551 Apr 1 '13 at 0:40
    
E is a variable, not 2.7. Thanks all for modifying the math using Tex !! Yes, I should use it next time. Thanks. –  student1 Apr 1 '13 at 19:53
add comment

3 Answers

up vote 0 down vote accepted

An eigenvalue doesn't give an eigenvector, but an eigenspace, formed of infinite eigenvectors.

This is because of the linearity of those applications, $v$ is an eigenvector if $f(v)=\lambda v$, but then any multiple of $v$, $\mu v$, will also be an eigenvector: $f(\mu v)=\mu f(v)=\lambda\mu v$.

So in the case you were writing ($2$ dimensional spaces), the possibility is that the dimension of the eigenspace is $1$ or $2$. Basically, the eigenspace is given by the equation of that same operation you have written $(A-\lambda)v=0$. So it might be the case that the dimension of the eigenspace is not $1$ but $2$ and therefore the same eigenvalue $\lambda$ will give you different independent eigenvectors. This will be one of the keys to understand diagonalization of linear functions.

share|improve this answer
    
So you say that both the e.vector in 1) and in 2) are valid? and are independent? Remember that the e.value is NOT degenerate... –  student1 Apr 1 '13 at 20:09
    
They're both valid but they don't have to be independent, there are cases in which both equations end up being the same. I wrote "it might be the case". For example the isomorphism $\pmatrix{0&1\\1&0}$ Has e.values $\pm 1$, and each of them gives you two equivalent equations when you make $(A-\lambda)v=0$, $a=b$ for $\lambda=1$ and $a=-b$ for $\lambda=-1$. –  MyUserIsThis Apr 1 '13 at 21:58
    
OK, got it. So for a given eigenvalue I have infinite number of e. vectors, some of them are just integer multiples but some are not. And they "may" be orthogonal. –  student1 Apr 1 '13 at 22:38
    
@ali8 Yes, they may be orthogonal, as in the example I wrote above. Have in mind that eigenvectors are vector that belong to subspaces that don't change under that isomorphism, for example in the matrix in my other comment, you are just swapping the two basis vectors, that is a reflexion along the line $x=y$, so we could say just by looking and without calculating, that the eigenspaces will be lines $x=y$, which remains the same: vector $(1,1)$, and line $x=-y$ which is reflected but the vectors that belonged to that space still belong to it after the transformation is applied: vector $(1,-1)$ –  MyUserIsThis Apr 1 '13 at 23:00
add comment

It is no problem that you have different eigenvector to an eigenvalue even you may be have eigenvector space with $3$ dimension for a eigenvalue!at this problem also it is not problem!
Only it must satisfy this relation : dimension ( eigenvector space of $\lambda_1$ )+dimension (eigenvector space of $\lambda_2$ )$\le 2$ (because our matrix is $2 \times 2$) attention if $=2$ then matrix is diagonalizable.

share|improve this answer
add comment

Assuming that the value of $a$ in (1) is $1$ and the value of $b$ in (2) is $1$ (which you don't say explicitly), the vectors you got are indeed multiples (note that they do not have to necessarily be integer multiples as you say) of each other. Hint: what happens if you multiply your $b$ from (1) with your $a$ from (2)?

share|improve this answer
    
I don't see what kind of "multiplicity" is between the e.vector in 1) and that in 2), if it is not a simple integer multiplicity. And, for your hint, multiplying both of them returns $\frac{-E^{2} + L1^{2}}{t^{2}}$ –  student1 Apr 1 '13 at 20:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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