# Thread: How to find eigenvectors

1. ## How to find eigenvectors

$\displaystyle \begin{array}{ccc}-1&0&1\\0&-4&-3\\2&0&-2\end{array}$

I have found the eigenvalues -3 and -4 but have no idea how to find the eigenvectors. Any help? thanks.

Ok ive found a bit more but am then stuck after

2x1 + x3 = -3x1
-x2 + 3x3 = -3x2
2x1 + x3 = -3x3

And am now stuck

2. Let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle A$. Then a vector $\displaystyle v$ that satisfies $\displaystyle Av=\lambda v \Leftrightarrow (A-\lambda I)v = 0$ is an eigenvector.

Thus if you find a vector $\displaystyle v_1$, satisfying $\displaystyle (A+3I)v_1 = 0$ and a vector $\displaystyle v_2$ with $\displaystyle (A+4I)v_2= 0$, then you're done.

You think you can do that?

3. Originally Posted by Dinkydoe
Let $\displaystyle \lambda$ be an eigenvalue of $\displaystyle A$. Then a vector $\displaystyle v$ that satisfies $\displaystyle Av=\lambda v \Leftrightarrow (A-\lambda I)v = 0$ is an eigenvector.

Thus if you find a vector $\displaystyle v_1$, satisfying $\displaystyle (A+3I)v_1 = 0$ and a vector $\displaystyle v_2$ with $\displaystyle (A+4I)v_2= 0$, then you're done.

You think you can do that?
i think thats the last bit ive done and i dont know what to do after

4. $\displaystyle A+3I =$

2 0 1
0 -1 -3
2 0 1

Thus we must solve the following system:

2x+z = 0
-y - 3z = 0

This gives: $\displaystyle z = -2x, -3z = y$. Thus $\displaystyle v = (1, 6 , -2)$

is an eigenvector with $\displaystyle \lambda = -3$

Now you try the other one.

5. Originally Posted by Dinkydoe
$\displaystyle A+3I =$

2 0 1
0 -1 -3
2 0 1

Thus we must solve the following system:

2x+z = 0
-y - 3z = 0

This gives: $\displaystyle z = -2x, -3z = y$. Thus $\displaystyle v = (1, 6 , -2)$

is an eigenvector with $\displaystyle \lambda = -3$

Now you try the other one.
I understand up till you go to v from z x and y how do you do that? thanks for your help by the way its much appreciated.

6. So you understand how we found the relations:

$\displaystyle z= -2x, -3z = y$.

A vector that satisfies these relations is an eigenvector with $\displaystyle \lambda = -3$.

Now we may choose $\displaystyle x= c$ arbitrary. These relations give the values of y and z. We may choose x = c arbitrary because if v is an eigenvector, cv is an eigenvector as well.

Let x = c then by the above relations follows, $\displaystyle z=-2c, y = -6c$. Thus $\displaystyle v = (c,-2c,-6c) = c(1,-2,-6)$ is an eigenvector for any $\displaystyle c$.

The relations only show how $\displaystyle x,y,z$ relate to eachother. That's why one co-ordinate x,y or z may be freely chosen.

7. Originally Posted by Dinkydoe
So you understand how we found the relations:

$\displaystyle z= -2x, -3z = y$.

A vector that satisfies these relations is an eigenvector with $\displaystyle \lambda = -3$.

Now we may choose $\displaystyle x= c$ arbitrary. These relations give the values of y and z. We may choose x = c arbitrary because if v is an eigenvector, cv is an eigenvector as well.

Let x = c then by the above relations follows, $\displaystyle z=-2c, y = -6c$. Thus $\displaystyle v = (c,-2c,-6c) = c(1,-2,-6)$ is an eigenvector for any $\displaystyle c$.

The relations only show how $\displaystyle x,y,z$ relate to eachother. That's why one co-ordinate x,y or z may be freely chosen.
oh i see so (2 , -4, -12) is an eigenvector for -3

8. There is no eigenvector for -4

because

3 0 1
0 0 -3
2 0 2

3x + z = 0
-3z = 0
2x + 2z = 0

z = 0 and x = 0

or is 0,1,0 an eigenvector?

9. There is no eigenvector for -4
Wrong
Indeed: x=z=0 but as we see y = c can be freely choosen

So (0,c,0)= c(0,1,0) is an eigenvector for any c.

10. Originally Posted by Dinkydoe
Wrong
Indeed: x=z=0 but as we see y = c can be freely choosen

So (0,c,0)= c(0,1,0) is an eigenvector for any c.
ye i understand now thank you