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

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Homework Help Overview

The problem involves a fixed nxn matrix U and an operator T defined on the space of nxn matrices. The task is to show that the dimension of the eigenspace associated with T is nd, given that the dimension of the eigenspace associated with U is d.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss defining a basis for the eigenspace of U and how it relates to the eigenspace of T. There are attempts to visualize the dimensions involved and how matrices can be treated as eigenvectors. Questions arise about the dimensionality of the spaces and the implications of linear independence among eigenvectors.

Discussion Status

Participants are actively engaging with the problem, exploring various interpretations and clarifying concepts related to eigenvalues and eigenspaces. Some guidance has been offered regarding the structure of the basis and the relationship between the dimensions of the spaces, but no consensus has been reached.

Contextual Notes

There is ongoing confusion regarding the implications of linear independence among eigenvectors and how they can share the same eigenvalue. Participants are also considering the dimensionality of the matrices involved and how they relate to the operator T.

  • #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.
 
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  • #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.
 

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