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.
- 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?
Best Answer
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 $\lambda =3$