# Eigenvector (if v1 and v2 are equal)

• Jan 9th 2009, 08:04 AM
JrShohin
Eigenvector (if v1 and v2 are equal)
One eigenvalue is 2.
A= |-1 3| |2 0|

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

x=y

I don't know the answer :(
• Jan 9th 2009, 08:28 AM
running-gag
Hi
Could you explain better what you are looking for ?
• Jan 9th 2009, 08:31 AM
JrShohin
Quote:

Originally Posted by running-gag
Hi
Could you explain better what you are looking for ?

I should find eigenvectors of A matrix.

A= |-1 3| |2 0|
• Jan 9th 2009, 08:42 AM
running-gag
OK so your beginning is correct
With eigenvalue = 2, you find x=y
Therefore (1,1) is an eigenvector
• Jan 9th 2009, 09:04 AM
running-gag
Quote:

Originally Posted by JrShohin
One eigenvalue is 2.
A= |-1 3| |2 0|

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

x=y

I don't know the answer :(

Suppose you have an eigenvector X for an eigenvalue k
Then A.X = kX
And for every p real
A.(pX) = p A.X = p kX = k (pX)
Therefore pX is also en eigenvector

This is why when you try to solve your system you find 2 equivalent equations
• Jan 9th 2009, 10:26 AM
HallsofIvy
In fact, the set of all vectors v such that $Ax= \lambda x$, for $\lambda$ and eigenvalue of a A is a subspace. That's why you cannot solve for just a single pair of values for x and y. You arrived at y= x so <x, y>= <x, x>= x<1, 1> or all mutiples of <1, 1> are eigenvectors corresponding to eigenvalue 2. That is the "eigenspace" for eigenvalue 2.