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

[Dick]

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.

 

[evilpostingmong]

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! We can end this thread.

 

[Dick]

Sure, you are welcome! Let's move on.