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The following example is puzzling me.

I was given the following matrix:

$$\begin{matrix} 0 & 0 & 0 \\ 1/2 & 1 & 0 \\ 1/2 & 0 & 1 \\ \end {matrix}$$

I used an Eigenvector calculator available on the net. I was naive enough to plug my matrix in without thinking first. The calculator gave me the following eigenvalues: $1$, $1$, $0$. I've checked the characteristic equation. It was OK. The eigenvectors given by the calculator were the following column vectors: $$\begin{matrix} 0 & 0 & -2 \\ 0 & 1 & \ \ 1 \ \ \\ \ 1 \ & 0 & 1 \\ \end {matrix}$$

I've checked these eigenvectors; they worked well with the corresponding eigenvalues.

Then I found many other eigen-looking vectors that worked with the first and the second eigenvalues. Here is an example: $$\begin{matrix} 0 \\ 0.4 \\ 0.6 \\ \end {matrix}$$

Only then realized I that all vectors whose first (uppermost) component is $0$ are eigen-looking vectors of the matrix at stake -- simply because of the arrangement of the zeros and the two one's in the matrix. So the specific case was explained well.

Is there any general explanation regarding "unsolicited" eigenvalues? What could be the reason that the calculator computed exactly the above mentioned vectors?

zoli
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    Nothing is wrong: any linear combination of eigenvectors having the same eigenvalue $\lambda$ is another eigenvector with eigenvalue $\lambda$ (or is the zero vector): the solutions $v$ to $Av = \lambda{v}$ for fixed $\lambda$ are a vector subspace. Any vector $(0,a,b)$ is an eigenvector with eigenvalue $1$ and it is $a(0,1,0) + b(0,0,1)$, which is a linear combination of the original eigenvectors you found with eigenvalue $1$. No problem. You found a basis for the subspace of solutions to $Av = v$ (here $\lambda = 1$) and any other element of that subspace is a linear combination of them. – KCd Jan 12 '15 at 09:41

1 Answers1

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The naive approach would be to look at how the calculator works: it could build a linear system: for the eigenvalue $1$ it would write merely as $$x_1=0.$$

Then, in order to find the eigenvectors the calculator would suppose that $x_2=1$, and take $x_1=0$, put $x_3=0$ and conclude that the system of equations is satisfied by a non-zero vector.

After that, the calculator would make an opposite hypothesis: $x_2=0$, which would result in, necessarily, $x_3\ne 0$, hence the system is satisfied by the vector $(0,0,1)$, which would give another eigenvector.

Finally, the calculator finds the rank of the system, compares with the number of vectors found, nd outputs the result.

TZakrevskiy
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