Showing That dim[Esub(c)(T)]=nd w/ Fixed nxn Matrix U

  • Thread starter Thread starter evilpostingmong
  • Start date Start date
  • Tags Tags
    Interesting Proof
Click For Summary
SUMMARY

The discussion centers on proving that if the dimension of the eigenspace of a fixed nxn matrix U is d, then the dimension of the operator T defined by T(A) = UA is nd. Participants clarify the definitions of eigenspaces and eigenvalues, emphasizing that the eigenspace Esub(c)(U) consists of eigenvectors corresponding to the same eigenvalue c. The conversation highlights the relationship between the dimensions of the eigenspaces and the structure of the matrices involved, ultimately confirming that dim[Esub(c)(T)] equals nd.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically eigenspaces and eigenvalues.
  • Familiarity with matrix operations and transformations, particularly with nxn matrices.
  • Knowledge of the properties of linear transformations and their representations.
  • Experience with dimensional analysis in vector spaces.
NEXT STEPS
  • Study the properties of eigenvalues and eigenvectors in linear transformations.
  • Learn about the relationship between matrix dimensions and eigenspaces in linear algebra.
  • Explore the concept of linear independence in the context of eigenspaces.
  • Investigate the implications of matrix multiplication on the dimensions of resulting matrices.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, and eigenvalue problems. This discussion is beneficial for anyone looking to deepen their understanding of eigenspaces and their applications in linear transformations.

  • #61
No, we shouldn't. If you want to do that exercise then do it. If you pick a basis where v1=e1, v2=e2..., vd=ed and the rest of the vi's are arbitrary linearly independent vectors outside of span(e1...ed) you should be able to conclude that matrix of the X's is an nxn matrix with an nxd submatrix filled arbitrary entries and the rest of the matrix filled with zeros. Making the dimension n*d. If that's not apparent to you from what we've already done, then you should probably review it.
 
Physics news on Phys.org
  • #62
Dick said:
No, we shouldn't. If you want to do that exercise then do it. If you pick a basis where v1=e1, v2=e2..., vd=ed and the rest of the vi's are arbitrary linearly independent vectors outside of span(e1...ed) you should be able to conclude that matrix of the X's is an nxn matrix with an nxd submatrix filled arbitrary entries and the rest of the matrix filled with zeros. Making the dimension n*d. If that's not apparent to you from what we've already done, then you should probably review it.

oh okay that makes sense. I appreciate your time:smile:! We can end this thread.
 
  • #63
Sure, you are welcome! Let's move on.
 

Similar threads

Linear Algebra - Kernel and range of T
  • · Replies 5 ·
Replies
5
Views
2K
Does there exist a 2x2 non-singular matrix with only one 1d eigenspace?
  • · Replies 2 ·
Replies
2
Views
2K
How do I know if my eigenvectors are right?
  • · Replies 12 ·
Replies
12
Views
3K
What Is the Transition Matrix for T in This Transformation?
  • · Replies 4 ·
Replies
4
Views
2K
Avoid unpleasant integrals in solving IVP
  • · Replies 3 ·
Replies
3
Views
2K
What information can be found in the columns of the transition matrix?
  • · Replies 3 ·
Replies
3
Views
1K
Show that T is a nonlinear transformation
  • · Replies 1 ·
Replies
1
Views
2K
[LinAlg] Set Notation, Subspaces, Bases, Ranges, & Kernels
  • · Replies 9 ·
Replies
9
Views
2K
Undergrad On a bound on the norm of a matrix with a simple pole
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Fundamental matrix of a second order 2x2 system of ODEs
  • · Replies 2 ·
Replies
2
Views
4K