How can I determine the eigenvalue of a given eigenvector?

Question

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?

Answer 1

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 $\lambda =3$

Answer 2

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 $\lambda$.

Answer 3

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 $\vec v$.