Dimension of Eigenspace of A and A^T

  • Thread starter potassium_mn04
  • Start date
  • Tags
    Dimension
In summary, the dimension of the eigenspace of A is the number of linearly independent eigenvectors associated with A, which can be calculated by finding the number of distinct eigenvalues of A. To find the eigenvalues and eigenvectors of A, you can use the characteristic polynomial method or the diagonalization method. The dimension of the eigenspace of A can tell us about the diagonalizability of A and the multiplicity of its eigenvalues. The dimension of the eigenspace of A and A^T are equal, as they share the same eigenvalues and eigenvectors. The dimension of the eigenspace of A cannot be greater than the size of A, as the maximum number of linearly independent vectors
  • #1
potassium_mn04
1
0
Homework Statement
Prove that the dimension of the eigenspace of A and dimension of eigenspace of A^T are equal
Relevant Equations
dim(E_A)=dim(E_(At))
I know that the rank of A and A^T are equal, and that the statement follows from there, but I have no idea how to prove it.
 
Physics news on Phys.org
  • #2
The eigenspace of ##A## corresponding to an eigenvalue ##\lambda## is the nullspace of ##\lambda I - A##. So, the dimension of that eigenspace is the nullity of ##\lambda I - A##. Are you familiar with the rank-nullity theorem? (If not, then look it up: Your book may call it differently.) You can apply that theorem here.
 

1. What is the dimension of the eigenspace of A?

The dimension of the eigenspace of A is equal to the number of distinct eigenvalues of A. This means that each distinct eigenvalue corresponds to a unique eigenspace, and the dimension of each eigenspace is equal to the number of linearly independent eigenvectors associated with that eigenvalue.

2. How is the dimension of the eigenspace of A related to the rank of A?

The dimension of the eigenspace of A is equal to the rank of A if and only if A is a square matrix with distinct eigenvalues. In other words, if A has n distinct eigenvalues, then the dimension of the eigenspace of A will be equal to n, which is also the rank of A.

3. Can the dimension of the eigenspace of A be greater than the dimension of A?

Yes, it is possible for the dimension of the eigenspace of A to be greater than the dimension of A. This can occur when A has repeated eigenvalues, which means that there are fewer distinct eigenvalues than the dimension of A. In this case, the eigenspace of each repeated eigenvalue will have a dimension greater than 1, resulting in a total dimension of eigenspace that is greater than the dimension of A.

4. How does the dimension of the eigenspace of A relate to the nullity of A?

The dimension of the eigenspace of A is equal to the nullity of A if and only if A is a symmetric matrix. This means that for a symmetric matrix, the eigenspace associated with each eigenvalue will be orthogonal to each other, resulting in a total dimension of eigenspace that is equal to the nullity of A.

5. Is the dimension of the eigenspace of A always less than or equal to the dimension of A?

No, the dimension of the eigenspace of A can be greater than the dimension of A in certain cases, as mentioned in question 3. However, in general, the dimension of the eigenspace of A will be less than or equal to the dimension of A, as the maximum number of linearly independent eigenvectors of A is equal to the dimension of A.

Similar threads

Replies
1
Views
344
  • Introductory Physics Homework Help
Replies
11
Views
645
  • Introductory Physics Homework Help
Replies
4
Views
633
  • Calculus and Beyond Homework Help
Replies
2
Views
381
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
526
  • Differential Equations
Replies
9
Views
287
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
628
  • Introductory Physics Homework Help
Replies
2
Views
89
Back
Top