One eigenvalue is 2.

A= |-1 3| |2 0|

-3x+3y=0

2x-2y=0

x=y

I don't know the answer :(

Printable View

- Jan 9th 2009, 07:04 AMJrShohinEigenvector (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, 07:28 AMrunning-gag
Hi

Could you explain better what you are looking for ? - Jan 9th 2009, 07:31 AMJrShohin
- Jan 9th 2009, 07:42 AMrunning-gag
OK so your beginning is correct

With eigenvalue = 2, you find x=y

Therefore (1,1) is an eigenvector - Jan 9th 2009, 08:04 AMrunning-gag
- Jan 9th 2009, 09:26 AMHallsofIvy
In fact, the set of all vectors v such that $\displaystyle Ax= \lambda x$, for $\displaystyle \lambda$ 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.