# Eigenvectors

• Jan 7th 2009, 02:27 AM
LooNiE
Eigenvectors
The matrix i have is |3 4| |4 -3| How do I find the corresponding eigenvectors? 2x2 matrix. 1st row 3 4, 2nd is 4 -3.
I have done the first step of the process, and found eigenvalues of +-5. Now how do I find the eigenvectors? I have aksed this question before, and I have recieved help which I already know how to do. When I substitue in the eigenvalues to find the eigenvectors, I get x1=0 and x2=0, so there must be some other way to find the eigenvector. If possible, find a final solution to the problem so I know the method works. Thanks
• Jan 7th 2009, 03:35 AM
mr fantastic
Quote:

Originally Posted by LooNiE
The matrix i have is |3 4| |4 -3| How do I find the corresponding eigenvectors? 2x2 matrix. 1st row 3 4, 2nd is 4 -3.
I have done the first step of the process, and found eigenvalues of +-5. Now how do I find the eigenvectors? I have aksed this question before, and I have recieved help which I already know how to do. When I substitue in the eigenvalues to find the eigenvectors, I get x1=0 and x2=0, so there must be some other way to find the eigenvector. If possible, find a final solution to the problem so I know the method works. Thanks

$\displaystyle \lambda = 5$: $\displaystyle \left( \begin{array}{cc} 3 & 4 \\ 4 & -3\end{array}\right)$ $\displaystyle \left( \begin{array}{c} x \\ y \end{array}\right)$ $\displaystyle = 5 \left( \begin{array}{c} x \\ y \end{array}\right)$

Therefore:

$\displaystyle 3x + 4y = 5x \Rightarrow x = 2y$ .... (1)

$\displaystyle 4x - 3y = 5y \Rightarrow x = 2y$ .... (2)

Therefore the corresponding eigenvector has the form $\displaystyle \left( \begin{array}{c} 2y \\ y \end{array}\right)$ $\displaystyle = y \left(\begin{array}{c} 2 \\ 1 \end{array}\right)$ and so the eigenvector is $\displaystyle \left( \begin{array}{c} 2 \\ 1 \end{array}\right)$.

Do the same thing for $\displaystyle \lambda = -5$.