helloo!

find the eigenvalues and eigen vectors of

A= |1 2|

|5 4|

i dont think its long to work out, but ive just started this topic and cannot work it out :o

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- July 10th 2007, 11:17 AMcharlotte_usaeigenvalues and eigen vectors!
helloo!

find the eigenvalues and eigen vectors of

A= |1 2|

|5 4|

i dont think its long to work out, but ive just started this topic and cannot work it out :o - July 10th 2007, 11:45 AMCaptainBlack
- July 10th 2007, 11:46 AMThePerfectHacker
The Eigenvalues are solutions to:

Thus,

Now solve. - July 10th 2007, 02:33 PMgalactus
A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.

Where k is the eigenvalue.

They're tied together. Lambda is mostly used, but I will use k.

You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.

In the first matrix, we have

Let y=t, then we have x=-t and y=t

In the second one, we have

Let y=t, then

Therefore, the eigenvectors are

and

- July 11th 2007, 04:54 AMcollege_clare
http://www.mathhelpforum.com/math-he...a221c356-1.gif

http://www.mathhelpforum.com/math-he...7971eda5-1.gif

i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...:confused: - July 11th 2007, 05:47 AMtopsquark
Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation ):

So we have the simultaneous equations

and

Solving this for a and b gives a = a and b = -a. So the eigenvector is

(This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)

We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:

I'll leave it to you to work out the rest.

-Dan - July 11th 2007, 06:27 AMcharlotte_usa
- July 11th 2007, 07:25 AMtopsquark