How can I determine the eigenvalue of a given eigenvector? I was wondering if there was a good way to determine the eigenvalue of a particular given eigenvector. I was thinking of starting out with the equation $(A-\lambda I)\hat{x}=\hat{0}$ and solving for $\lambda$, but I am a little unsure how to do this.


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*Simple Example: Let A =$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ and eigenvector $\hat{v_{1}} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. So this would yield: $ (\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}) \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \hat{0}$. How would I solve for $\lambda$ in this case?

 A: Just use the definition of eigen vectors: $v$ is an eigen vector if $Av={\lambda}v$, (and $v~!=\hat{0}$). So if you're sure that $v$ is an eigen vector, just compute $Av$ and see how many times it is $v$, this is your lambda.
Hope it helps!
EDIT: to solve the example you provided:
$ (\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}) \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \hat{0}$
leads to the two same equations: 
$3-\lambda =0$ and
$3-\lambda =0$ which is equivalent to $\lambda =3$
A: To solve $A\vec {v_1}=\lambda \vec{v_1}$ for $\lambda, $ 
can you solve $$ \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}  \begin{bmatrix} 1 \\ 1 \end{bmatrix}=   \lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ for $\lambda?$
Note that $$ \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}  \begin{bmatrix} 1 \\ 1 \end{bmatrix}=  \begin{bmatrix} 3 \\ 3 \end{bmatrix},$$
so you must solve  $$\begin{bmatrix} 3 \\ 3 \end{bmatrix}=\lambda  \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
for $\lambda$.
A: In this particular case we may observe that
$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \tag 1$
so if in addition we have
$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \tag 2$
then
$\lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix}  = 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \tag 3$
or
$\begin{bmatrix} \lambda  \\ \lambda  \end{bmatrix}  = \begin{bmatrix} 3  \\ 3  \end{bmatrix}, \tag 4$
from which it is readily seen that
$\lambda = 3. \tag 5$
More generally, if $\vec v \ne 0$ is an eigenvector of some $n \times n$ matrix $A$, then since
$A \vec v = \lambda \vec v, \tag 6$
every entry of $A \vec v$ is the same multiple, $\lambda$, of the corresponding entry of $\vec v$; since $\vec v \ne 0$ we have
$\exists i, 1 \le i \le n, \; \vec v_i \ne 0; \tag 7$
then
$(A\vec v)_i = \lambda \vec v_i, \tag 8$
whence
$\lambda = \dfrac{(A\vec v)_i}{\vec v_i}, \tag 9$
and thus $\lambda$ is determined by the eigenvector $\vec v$.
