Question

Find the Eigenvector of $\begin{bmatrix} 1 & cos\theta\\ cos \theta & 1 \end{bmatrix}$

Answer 1

Let $A = \begin{bmatrix}
1 & cos\theta\\
cos\theta & 1
\end{bmatrix}$ and the eigenvalues be represented by $\lambda$,then eigenvalues can be determined with the help of below property :

$det(A-\lambda I = 0)$

We get $2$ values of $\lambda$ as the matrix size is $2$x$2$.

$\lambda_1 = 1 + cos\theta\\\lambda_2 =  1 - cos\theta$

Now, let $\textbf{x} = \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ denotes the eigenvector, then by the property of eigenvector,

$\begin{array}\\
& A\textbf{x} &= \lambda \textbf{x} \\
\implies&(A - \lambda I)\textbf{x} &= 0
\end{array}$

On solving for $\lambda_1$, we get

$\begin{bmatrix}
-cos\theta & cos\theta\\
cos\theta & -cos\theta
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} = 0
$

which simplifies to $x_1 = x_2$.

As an eigenvector merely refers to the ratio between the two (or more) quantities, we say that for $\lambda_1, \textbf{x} = \begin{bmatrix}
1 \\1
\end{bmatrix} $.

We usually write eigenvector in normalized form, i.e. we divide each element of the eigenvector by $\sqrt{x_1^2+x_2^2\ldots x_n^2}$ for an eigenvector of size $n\text{x}1$.

So, the normalized eigenvector is $\begin{bmatrix}
\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}
\end{bmatrix}$.

Similarly, the eigenvector for the other eigenvalue can also be determined.

For more details, please see this.