Base for Eigenvalues

1. Oct 22, 2008

summer

Suppose for a given matrix A, Sarah finds the eigenvectors v1 = [1 3 4 5]' and v2 = [5 6 3 4]' form a base for eigenspace of labmda = 2. Now suppose Janie finds the eigenvectors v3 = [1 2 2 3]' and v4 = [7 8 7 6]' form a base for eigenspace of lambda = 4. Is Janie's solution compatible with Sarah's?

Okay, so I know that if v1 and v2 form a base for the eigenspace of lambda, they must be linearly independent. This same fact goes for v3 and v4. Now, my question is, to check whether or not Janie's solution is compatible with Sarah's, would I simply make sure that they are all linearly independent? If so, then they are compatible? The part that's throwing me off is that they belong to different eigenvalues.

So, can you have the same vector (or a linear combination of it), only belonging to a different eigenvalue of the same matrix? Or are the vectors unique to the eigenvalues?

Help much appreciated.

2. Oct 23, 2008

Defennder

What does "compatible" here mean?
You can check this out easily. Suppose for an eigenvalue $$\lambda_1$$ you have 2 eigenvectors v1 and v2 which are linearly independent. Then suppose we have an eigenvector v3 associated with a different $$\lambda_2$$. And suppose $$v_3 = a_1 v_1 + a_2 v_2$$.

What can you say about $$Av_3 = \lambda_2v_3$$ and $$Av_3 = A(a_1v_1 + a_2 v_2)$$?

3. Oct 23, 2008

summer

Okay, so we know that if lambda is an eigenvector of A, then A(lambda) = (lambda)x, where x is the eigenvector. So in the case above Av3 = A(a1v1 + a2v2), it would be implied that a1v1 + a2v2 would also be an eigenvector belonging to lambda1.

So, then, eigenvectors are unique to the eigenvalues. Yes?

4. Oct 23, 2008

Defennder

You can have an infinite number of linearly dependent eigenvectors for each eigenvalue but only one eigenvalue for each eigenvector. If that is what you meant, then yes you're right.

5. Oct 23, 2008

summer

Thank you! I got it now.