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I have the following matrix given and am asked to find all its eigenvalues, eigenvectors and the geometric multiplicity of each eigenvalue:

$$A_{n\times n} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

I used the characteristic polynom to find the eigenvalues $λ_{1} = λ_{2} = 0$ and $λ_{3} = 1$. I then attempted to find the eigenvectors by solving the system $(A-λI)v=0$ for $λ_{1}=λ_{2}=0$ but once I get the system

$$ \begin{array}{ccc|c} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} $$

I am not sure how to get the eigenvectors. I know that there need to be 2, because $\dim (E_{λ_{1}}) = n- \operatorname{Rang}(A) = 2$ and MatLab tells me they are

$$ v_{1} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\quad v_{2} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} $$

but I don't quite understand why. The way I learned it you are supposed to set one of the free variable to one and all the others to zero, then calculate a solution and repeat for every free variable there is. However, here there is only one free variable as far as I can see (the one in the first column). How can I get two eigenvectors out of it and why is the $-1$ trick valid here but not in other cases?

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You have a degeneracy $\lambda_1=\lambda_2=0$. –  Jon Jan 14 '12 at 13:33
    
$A-0I=A$ has rank 2; so, the dimension of its null space is 1. (So, the dimension of the eigenspace corresponding to $\lambda=0$ is 1). –  David Mitra Jan 14 '12 at 13:35
    
Dear Robin: I think it's simpler to observe this: If $a=(0,1,0),b=(0,0,1),c=(1,0,0)$, then $Aa=a,Ab=c,Ac=0$. –  Pierre-Yves Gaillard Jan 14 '12 at 13:54
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up vote 3 down vote accepted

There aren't three eigenvalues; there are only two: $\lambda_1=0$ and $\lambda_2=1$. At this point you can not say anything about the dimension of the eigenspace corresponding to the eigenvalue $\lambda_i$, other than it at most the multiplicity of the corresponding factor in the characteristic polynomial. So here, all you can say is $\dim(E_{\lambda_1})\le 2$ and $ \dim(E_{\lambda_2})\le 1$.

To find a basis of eigenvectors for an eigenvalue $\lambda$, you examine the matrix $A-\lambda I$. A basis for the eigenspace is given by a basis for the null space of this matrix (that is, of the solutions to $(A-\lambda I)x=0$)

For $\lambda=0$, this gives the matrix $A$ again. As you observed, there is only one free variable. Thus $\dim(E_{\lambda_1})=1$. There is, up to scaling, only one eigenvector.

(Note the rank of $A$ is two, so $\dim(E_{\lambda_1}) = n -{\rm rank}(A)=3-2=1$; your error was here...)

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Thanks a lot, this clears things up! –  Robin Jan 15 '12 at 20:07
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