Question

If I have an eigenvalue of multiplicity 2, how do I know when it will result in a single eigenvector and when it gives me two eigenvectors?

Answer 1

if you row reduce the matrix:
\[A-2I\]
and you end up with 2 free variables, there will be two eigenvectors.
If you end up with one free variable there will be one eigenvector (which is bad!)

Answer 2

I'm not sure what a free variable is, can we do an example?

Answer 3

sure, do you have a matrix in mind?

Answer 4

Let's say the matrix reduces to:

1 0
0 0

Answer 5

so you look at the columns, and one column has a one in it, and another column doesnt. This means there is one free variable (because of the column without the leading 1). So you would end up with one eigenvector.

Answer 6

that eigenvector being [0, 1] or [0, C] where C is any constant right?

Answer 7

lets say you ended up with:

1 0 3
0 1 -2
0 0 0

There are 2 pivots, 1 free variable. So one eigenvector.

1 2 3
0 0 0
0 0 0

1 pivot, 2 free variables, so 1 eigenvectors.

Answer 8

your right, the eigenvector would be (0,1)

Answer 9

can you give me a 2x2 case where it gives me 2 eigenvectors from one eigenvalue?

Answer 10

oops, my last sentence in that post with the matrices should be "...so 2 eigenvectors."

Answer 11

For a 2 x 2 case im not sure if its possible to end up with 2 eigenvectors from one eigenvalue without the matrix being totally trivial like:

2 0
0 2

let me see if i can think of something...

Answer 12

in that case, you would get [0, C] as one and [K, 0] as another eigenvector right? where C and K are constants of course.

However those two rows are not multiples of each other

Answer 13

yeah, its not possible. other than a multiple of the identity matrix like i have above. In that case, both of the eigenvalues are 2, and every vector will be an eigenvector because:
\[Ax = 2Ix = 2x\]

Answer 14

Ok sounds good, you've been a big help =)