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

[Dick]

where the vectors associated with the eigenvalue c are (T(a vector from this space)=c*a vector from this space).

In other words, the set of all transformations such that T(X)=UX=cX, right? What conclusions have we reached about such transformations?

 

[evilpostingmong]

In other words, the set of all transformations such that T(X)=UX=cX, right? What conclusions have we reached about such transformations?

when T acts on X(vi) from an arbitrary basis and sends it to an eigenspace,
it can map to c*any vector in the basis of that eigenspace.

 

[Dick]

when T acts on X(vi) from an arbitrary basis and sends it to an eigenspace,
it can map to c*any vector in the basis of that eigenspace.

That is amazing uninformative and not very clear or correct. Can you be more specific? Don't you see what the n*d dimensionality thing might have to do with your problem?? There are two spaces here. The n dimensional vector space and the d dimensional subspace span{e1...ed}. What does X have to do with them?

 

[evilpostingmong]

That is amazing uninformative and not very clear or correct. Can you be more specific? Don't you see what the n*d dimensionality thing might have to do with your problem?? There are two spaces here. The n dimensional vector space and the d dimensional subspace span{e1...ed}. What does X have to do with them?

X can send v1 to any element in the subspace. Is this a start?

 

[Dick]

X can send v1 to any element in the subspace. Is this a start?

X sends ANY v to an element of span{e1...ed}, that would be a start. Doesn't that mean X must map R^n to R^d? I.e. X is an arbitrary mapping from an n dimensional space to a d dimensional space, isn't it?

 

[evilpostingmong]

X sends ANY v to an element span{e1...ed}, that would be a start. Doesn't that mean X must map R^n to R^d?

Yes since R^d is a subspace of R^n so all vectors in R^d are in R^n.

 

[Dick]

No... X MAPS R^n to R^d. That doesn't necessarily mean R^d is in R^n. At all. But in this case I'm going to pass that since span(e1...ed) IS a subspace of span(v1...vn). So what is the conclusion?

 

[evilpostingmong]

No... X MAPS R^n to R^d. That doesn't necessarily mean R^d is in R^n. At all. But in this case I'm going to pass that since span(e1...ed) IS a subspace of span(v1...vn). So what is the conclusion?

Let's call n 3 and d 3. So the basis is <v1, v2, v3>
and the basis for the subspace is <e1, e2, e3>. v1 can be mapped
by X to e1 or e2 or e3 v2 can be mapped by X to e1 or e2 or e3
v3 can be mapped by X to e1 or e2 or e3 so that makes 9
possiblilities, or 9 dimensions or d*n=9

 

[Dick]

Lets call n 3 and d 3. So the basis is <v1, v2, v3>
and the basis for the subspace is <e1, e2, e3>. v1 can be mapped
by X to e1 or e2 or e3 v2 can be mapped by X to e1 or e2 or e3
v3 can be mapped by X to e1 or e2 or e3 so that makmpes 9
possiblilities, or 9 dimensions or d*n=9

Now that is starting to show some promise, yes. Those aren't the only ways to do it, but they are a basis for all ways to do it. For example you could also map v1 to e1+e2. But that is just the sum of maps that take v1->e1 and v1->e2. You already deleted it but if n=3 and d=2 the dimension of the basis is 6, yes.

 

[evilpostingmong]

But I am stuck on the matrix descriptions, that's where it gets problematic.
We should focus on this part. Let's use the bases <v1, v2, v3> and <e1, e2>

 

[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.