I apologize in advance for the mess. This is my attempt.
The matrix in question is
\begin{align} \textbf{A} = \frac{1}{100}\cdot\begin{pmatrix}
92 & 0 & -144 \\
0 & 100 & 0 \\
-144 & 0 & 8
\end{pmatrix}.\end{align}
We have to find the normalised eigenvector $v = \begin{pmatrix} v_{1} \\ v_{2} \\ v_{3} \end{pmatrix}$ for the eigenvalue $\lambda = -1$, such that $\lvert v \rvert = 1 $.
I solved the equation $$(\textbf{A} – \lambda \textbf{I})\cdot v = 0$$
i.e. $$\begin{bmatrix} 192 & 0 & -144 \\ 0 & 200 & 0 \\ -144 & 0 & 108 \end{bmatrix}\cdot\begin{bmatrix}v_{1} \\ v_{2} \\ v_{3}\end{bmatrix} = 0$$
I used Gaussian elimination and found that $$v = \begin{bmatrix}\frac{3}{4}v_{3} \\ 0 \\ v_{3} \end{bmatrix}.$$
Using the fact that $\lvert v \rvert = 1$, I found that $$ 1 = \Big(\frac{3}{4}v_{3}\Big)^{2} + \Big(v_{3}\Big)^{2} \implies v_{3} = \pm \frac{4}{5}.$$
I tried plugging both $\begin{bmatrix}\frac{3}{5} \\ 0 \\ \frac{4}{5} \end{bmatrix}$ and $\begin{bmatrix} -\frac{3}{5} \\ 0 \\ -\frac{4}{5} \end{bmatrix}$ in the original equation and they both work. So I'm wondering if it's possible that matrix $\textbf{A}$ has two eigenvectors for the eigenvalue $\lambda = -1$, or is the sign irrelevant? If so, which one should I consider?
Any help or suggestions is appreciated.
Best Answer
Yes, there can be more than one eigenvector associated with a certain eigenvalue; we call the number of eigenvalues of a eigenvalue its geometric multiplicity $m_g(\lambda)$.
The eigenvectors need to be linearly independent however for them to count as two (or more) different ones. In your case, one is a scalar multiple of $-1$ of the other and thus not linearly independent so they are they 'same' eigenvector, i.e. there is only one for the eigenvalue $\lambda = -1$.
In general, the number of eigenvectors $m_g(\lambda)$ corresponding to a eigenvalue is the dimension of the eigenspace $E_\lambda$ which is spanned by the eigenvectors $\xi_1, \xi_2, \ldots$. If you get two eigenvectors which are linearly dependent, the dimension of $E_\lambda$ does not change so we do not consider it as a different one.
As an example of a matrix with 2 eigenvectors belonging to one eigenvalue, consider the matrix $$A = \begin{pmatrix} 2 &0\\ 0&2 \end{pmatrix}.$$ Since $A$ is upper triangular, its eigenvalues $\lambda$ are given by the elements on the diagonal which are both $2$. If we calculate the eigenvectors corresponding to $2$ we get two linearly independent ones (check this!) \begin{align*} \left\{\begin{pmatrix} 1\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \end{pmatrix}\right\}. \end{align*}