I have a matrix, $S$,
$$ S = \begin{bmatrix} 1.15 & 5.76 \\ 5.76 & 29.54 \end{bmatrix}.$$
I want to find eigen value and eigen vector of $S$, $$ (S -\lambda I)a = 0. $$ Calculating the Eigen value is easy since $S$ is a matrix $2\times 2$, hence I got
$$\lambda_1 = 30.66; \\
\lambda_2= 0.03. $$
Supposed that the eigenvector of $S$, $a$, has properties $a^Ta = 1$, hence I have a relation
$$ a^Ta = \begin{bmatrix} a_{11} & a_{12} \end{bmatrix}\begin{bmatrix} a_{11} \\ a_{12} \end{bmatrix} \\ a_{11}^2 + a^2_{12} = 1. $$
Since both $a_{11}$ and $a_{12}$ are in quadratic terms, it must have two values for each $a_{11}$ and $a_{12}$.
So then, for eigenvalue $\lambda_1 = 30.66$, I will get more than one eigenvector for one eigenvalue. It is confusing me which eigenvector should I use in my calculation, I need a suggestion or solution for this problem. Thank you in advance.
Best Answer
When $a$ is an eigenvector for eigenvalue $\lambda$, any non-null vector proportional to $a$ is also an eigenvector for $\lambda$.
The equation $a^Ta=1$ is only useful if you want an eigenvector with norm 1. So you don't have to worry about that immediately. First determine $a_i$ such that $Sa_i=\lambda_i a_i$, for $i=1$ and $2$. This will give you solutions $a_i=\gamma_i b_i$, where $\gamma_i \in \mathbb{R}$ can take any non-null value.
Then, if you want $a_i$ with norm $1$, compute the norm of $b_i$ and adjust $\gamma_i$ accordingly. You'll get two solutions, which will give two opposite vectors.