# Find eigenvalue of matrix given eigenvector

I have the following matrix:

$P= \begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix}$

And its eigenvector is :

$v= \begin{bmatrix} 4 \\ 3 \end{bmatrix}$

I would like to find its eigenvalue. This is my attempt, however I am doing something wrong:

$\begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$

Then...

$6x -3y = 4$

$2x + y = 3$

Elimination...

$x = 13/12$

$y = 10/12$

$\begin{bmatrix} 13/12 \\ 10/12 \end{bmatrix} = \lambda \begin{bmatrix} 4 \\ 3 \end{bmatrix}$

For the final part, I can't figure out a value for $\lambda$ that satisfies both 4 and 3, which means I am doing something wrong. Can someone point me in the right direction?

• Are you sure you did not mean that the eigenvalues are $4$ and $3$? I think you are confusing the two eigenvalues and eigenvectors. – Amzoti Jan 20 '15 at 1:58
• no, the question I'm doing states that [4,3] is the eigenvector of the matrix – Bolboa Jan 20 '15 at 1:59
• If you find the characteristic polynomial of $P$ you get $x^2-7x+12$. Amzoti is right. – Eleven-Eleven Jan 20 '15 at 2:00
• @Balboa: your book has an error! It is easy to verify what I am stating. – Amzoti Jan 20 '15 at 2:01
• For what it is worth, the eigenvectors are $[1,1]$and $[-3,2]$ for $3,4$ respectively – Eleven-Eleven Jan 20 '15 at 2:05

## 1 Answer

You need to write $$\begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \end{bmatrix} = \lambda \begin{bmatrix} 4 \\ 3 \end{bmatrix}$$

and then find $\lambda$. (By the way, when I do this in my head, I find that no such lambda exists, i.e., that this is NOT an eigenvector of the given matrix...but maybe my in-the-head arithmetic isn't so reliable.)

Given Amzoti's comment, you need to write $$\begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = a\begin{bmatrix} x \\ y \end{bmatrix}$$ where $a$ is either $4$ or $3$, and find the associated eigenvector. If $v$ is an eigenvector, so is $2v$, so you'll have to pick either $x$ or $y$ and then solve for the other.

• my scratch work agrees, $[4,3]^T$ is not an eigenvector. – JMoravitz Jan 20 '15 at 1:59