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?
Best Answer
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...)