- #1
brushman
- 113
- 1
I have the matrix:
1 1 0
0 1 0
0 0 0
First question: Is it correct that this is a 3 x 3 matrix (as opposed to a 2 x 2 matrix, since the last row and column are 0s)?
I have found the eigenvalues to be λ1 = 1, λ2 = 1, λ3 = 0. Then I have found the corresponding eigenspaces to be E1 = E2 = span (1 0 0)^T and E3 = span (0 0 1)^T.
Second question: Are λ1 and λ2 considered different eigenvalues? If so, is each considered to have its own space, even if they are just equal?
I have the following theorem (part b. on p324 LA w/ Otto Bretscher):
There exists an eigenbasis for an n x n matrix A if (and only if) the geometric multiplicities of the eigenvalues add up to n.
Final question: Is there an eigenbasis for the above matrix (which I have already found the eigenvalues and eigenvectors for)? Here are my two trains of thought and I do not know which is correct:
Train 1: the multiplicity of λ1 is 1, the multiplicity of λ2 is 1, and the multiplicity of λ3 is 1. Therefore, 1+1+1 = 3 = n. Therefore, an eigenbasis exists.
Train 2: the repeated eigenvalue (λ1 = λ2 = 1) is treated as 1 eigenvalue. Then we have the multiplicity of the repeated eigenvalue is 1 and the multiplicity of the other eigenvalue is 1. 1+1 = 2 < n = 3. Therefore, an eigenbasis does not exist.
Which is correct?
Thanks.
1 1 0
0 1 0
0 0 0
First question: Is it correct that this is a 3 x 3 matrix (as opposed to a 2 x 2 matrix, since the last row and column are 0s)?
I have found the eigenvalues to be λ1 = 1, λ2 = 1, λ3 = 0. Then I have found the corresponding eigenspaces to be E1 = E2 = span (1 0 0)^T and E3 = span (0 0 1)^T.
Second question: Are λ1 and λ2 considered different eigenvalues? If so, is each considered to have its own space, even if they are just equal?
I have the following theorem (part b. on p324 LA w/ Otto Bretscher):
There exists an eigenbasis for an n x n matrix A if (and only if) the geometric multiplicities of the eigenvalues add up to n.
Final question: Is there an eigenbasis for the above matrix (which I have already found the eigenvalues and eigenvectors for)? Here are my two trains of thought and I do not know which is correct:
Train 1: the multiplicity of λ1 is 1, the multiplicity of λ2 is 1, and the multiplicity of λ3 is 1. Therefore, 1+1+1 = 3 = n. Therefore, an eigenbasis exists.
Train 2: the repeated eigenvalue (λ1 = λ2 = 1) is treated as 1 eigenvalue. Then we have the multiplicity of the repeated eigenvalue is 1 and the multiplicity of the other eigenvalue is 1. 1+1 = 2 < n = 3. Therefore, an eigenbasis does not exist.
Which is correct?
Thanks.