# Thread: Need some clarification on eigenvalues/eigenvectors

1. ## Need some clarification on eigenvalues/eigenvectors

i have a quick question about eigenvalue/eigenvectors

so say i have a matrix

-3 0
2 -1

with eigen values -1 and -3 (i'm pretty sure they are right)

so for the first eigenvalue (-1) i get a matrix

-2 0
2 0

and for the second (-3)

0 0
2 2

well i not realy sure how to get the eigen vector from these?...wondering if someone could help me...this would really help, thanks

2. Originally Posted by action259
i have a quick question about eigenvalue/eigenvectors

so say i have a matrix

-3 0
2 -1

with eigen values -1 and -3 (i'm pretty sure they are right)

so for the first eigenvalue (-1) i get a matrix

-2 0
2 0

and for the second (-3)

0 0
2 2

well i not realy sure how to get the eigen vector from these?...wondering if someone could help me...this would really help, thanks
You have the eigenvalues right, but what is this matrix associated with the eigenvalue? I've never heard of it. Anyway, to get the eigenvectors:

$\lambda = -1$:

Let the eigenvector be $\left ( \begin{array}{c} a \\ b \end{array} \right )$

So:
$\left ( \begin{array}{cc} -3 & 0 \\ 2 & -1 \end{array} \right ) \left ( \begin{array}{c} a \\ b \end{array} \right ) = (-1) \left ( \begin{array}{c} a \\ b \end{array} \right )$

$\left ( \begin{array}{c} -3a \\ 2a - b \end{array} \right ) = \left ( \begin{array}{c} -a \\ -b \end{array} \right )$

Thus $-3a = -a$ or a = 0, and $2a - b = -b$ which says that $0 - b = -b$, leaving b undetermined. So the eigenvector for $\lambda = -1$ is of the form
$\left ( \begin{array}{c} 0 \\ b \end{array} \right )$
where b is undetermined.

You can use the same method to get the eigenvector for $\lambda = -3$.

-Dan

3. The characteristic polynomial is your friend; you'll use it a lot. That is,

Ax = (lambda)x --> (A - (lamdba)(I))x = 0, where I is the identity matrix.