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

[evilpostingmong]

Basically, yes. Let {v1...vn} is any basis and {e1...ed} are the eigenvectors. Now define X_{i,j} by X_{i,j}(vi)=ej and X_{i,j}(vk)=0 if k is not equal to i. Can you see how the X_{i,j} define a basis for the eigenmatrices X?

Okay so a matrix for the transformation would be this
X1,1 X2,2 X 3,3 now if d=3 then these (X1,1 X2,2 X3,3) are "used" to
map v1 to c*ei and X4,4 X5,5 X6,6 are used to map v2 to c2*ei
in a different basis of dimension d and X7,7 X8,8 and X9,9 are used to map v3
to another basis of dimension d to c3*ei. Why 3 at a time? Because
we only map to spaces of 3 dimensions, and one whos
basis is <v4, v5, v6> is not in the basis of <v1, v2, v3>

 

[Dick]

I'm not trying to write the matrix components by X_{i,j}. I mean X_{1,1} is the matrix that maps v1 to e1. X_{1,2} maps v1 to e2. X_{3,7} maps v3 to e7. Etc, etc. They are all nxn matrices. So in X_{i,j} i goes from 1 to n and j goes from 1 to d.

 

[evilpostingmong]

I'm not trying to write the matrix components by X_{i,j}. I mean X_{1,1} is the matrix that maps v1 to e1. X_{1,2} maps v1 to e2. X_{3,7} maps v3 to e7. Etc, etc. They are all nxn matrices. So in X_{i,j} i goes from 1 to n and j goes from 1 to d.

still don't get those symbols

 

[Dick]

still don't get those symbols

Then try these symbols. Define X[1,1] to be the linear tranformation that maps v1 to e1 and all other vi to 0. Define X[i,j] to be the linear transformation that maps vi to ej and all of the other v's to 0. Do you get that?

 

[evilpostingmong]

Then try these symbols. Define X[1,1] to be the linear tranformation that maps v1 to e1 and all other vi to 0. Define X[i,j] to be the linear transformation that maps vi to ej and all of the other v's to 0. Do you get that?

so X[1,2] maps from v1 to v2 and X[1,3] maps from v1 to e3
and X[1,4] in a different matrix maps from v1 to e4, but in a different basis, that's what I had
in mind since e4 is not in <e1, e2, e3>

 

[Dick]

so X[1,2] maps from v1 to v2 and X[1,3] maps from v1 to e3
and X[1,4] in a different matrix maps from v1 to e4, but in a different basis, that's what I had
in mind since e4 is not in <e1, e2, e3>

I'm sure what "e4 is not in <e1, e2, e3>" has to do with it. But do you understand what the X[i,j] are, and can you see that they are a basis for the eigenmatrices X such that UX=cX?

 

[evilpostingmong]

I'm sure what "e4 is not in <e1, e2, e3>" has to do with it. But do you understand what the X[i,j] are, and can you see that they are a basis for the eigenmatrices X such that UX=cX?

you said that X_{3,7} maps v3 to e7. So there is a v3 at row 7 of the 7th coordinate.
So it maps v3 to e7 but since e7 is not in <e1, e2, e3> there must be other bases
(that is if the dimension is 9 mapping to 3) so X maps the v3 at row 7 to
the space with the basis <e7, e8, e9>

 

[Dick]

you said that X_{3,7} maps v3 to e7. So there is a v3 at row 7 of the 7th coordinate.
So it maps v3 to e7 but since e7 is not in <e1, e2, e3> there must be other bases
(that is if the dimension is 9 mapping to 3) so X maps the v3 at row 7 to
the space with the basis <e7, e8, e9>

I don't know what you are talking about. I also said for X[i,j] that i goes from 1 to n and j goes from 1 to d. If the basis for the eigenvectors is <e1,e2,e3> there is no X[3,7] because there is no e7. Let's take another tack. The vector space is n dimensional and the eigenspace is d dimension. So X:R^n->R^d. Do you know what the dimension of the space of such transformations is?

 

[evilpostingmong]

I don't know what you are talking about. I also said for X[i,j] that i goes from 1 to n and j goes from 1 to d. If the basis for the eigenvectors is <e1,e2,e3> there is no X[3,7] because there is no e7. Let's take another tack. The vector space is n dimensional and the eigenspace is d dimension. So X:R^n->R^d. Do you know what the dimension of the space of such transformations is?

yes it is d.

 

[Dick]

yes it is d.

No it isn't. Abstractly you can think of a mapping A:R^n->R^d as nxd matrix. What's the dimension of such mappings regarded as a vector space?

 

[evilpostingmong]

uh, nxd?

 

[Dick]

uh, nxd?

Yes, n*d.

 

[evilpostingmong]

now what/

 

[Dick]

now what/

You want to show dim[Esub(c)(T)]=nd. What is Esub(c)(T)?

 

[evilpostingmong]

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

 

[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.