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

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evilpostingmong said:
I'm guessing that thinking about this using matrices must've screwed me over
big time. X is the eigenvector with c as its eigenvalue so TX=UX=cX

Now that makes sense. Ok, so UX=cX. Apply that to any vector v. UX(v)=cX(v). What does that tell you about the vector X(v)?
 
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X(v) is within the set of eigenvectors that have c as the eigenvalue.
 
evilpostingmong said:
X(v) is within the set of eigenvectors that have c as the eigenvalue.

Good! We are writing the set of eigenvectors that have eigenvalue c as span(e1,...,ed), right? So X has to be a transformation that maps an n-dimensional space to the d-dimensional subspace span(e1,...,ed), agree?
 
Dick said:
Good! We are writing the set of eigenvectors that have eigenvalue c as span(e1,...,ed), right? So X has to be a transformation that maps an n-dimensional space to the d-dimensional subspace span(e1,...,ed), agree?

Ok so about those choices. X(v1) can map to c*e1 or anyone else in the span.
That means that X(v1) can map to a different uh coordinate? when it maps
to c*e1 as compared to when it maps to c*e2, since it maps to a different coordinate,
so the eigenspace must have the right amount of dimensions for X to map v1 to linear independent elements so it maps v1 to d elements.

Just to clarify, if X maps v1 to elements in span(e1, e2, e3) (eigenspace with c as the eigenvalue those elements are associated with)
X must be able to map v1 to any of the three elements in their respective coordinates or axis.
 
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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?
 
Dick said:
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>
 
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.
 
Dick said:
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
 
evilpostingmong said:
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?
 
Dick said:
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>
 
evilpostingmong said:
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?
 
Dick said:
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>
 
evilpostingmong said:
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?
 
Dick said:
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.
 
evilpostingmong said:
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 said:
now what/

You want to show dim[Esub(c)(T)]=nd. What is Esub(c)(T)?
 
where the vectors associated with the eigenvalue c are (T(a vector from this space)=c*a vector from this space).
 
evilpostingmong said:
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?
 
Dick said:
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.
 
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evilpostingmong said:
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?
 
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Dick said:
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?
 
evilpostingmong said:
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?
 
Dick said:
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.
 
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?
 
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Dick said:
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
 
evilpostingmong said:
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.
 
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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>
 

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