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Math Help - Quotient Space

  1. #1
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    Quotient Space

    Let f:X\to X be a linear transformation, and V\subset X an invariant subspace of f, i.e., f(V)\subset V.

    Prove that f induces a linear transformation \hat{f}:X/V\to X/V

    Since f is invariant, we know f(V)=\text{Im} \ f\subset V. I am struggling with quotient spaces. I know it means X mod V where \{x\in X: x+ V\}.

    I need some guidance and a good explanation of what is going on if possible.
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  2. #2
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    Re: Quotient Space

    what we would like to do is define \hat{f}(x+X) = f(x)+X.

    first we need to be sure that this is well-defined, since we are working with cosets, instead of elements.

    so suppose y is in x + X (that is, that y + X = x + X), so that y = x + v, for some vector v in X. then f(y) = f(x + v) = f(x) + f(v) = f(x) + v',

    (where v' is some element of X) since X is invariant for f, so f(y) is in f(x) + X, thus f(y) + X = f(x) + X.

    hence \hat{f}(y+X) = f(y) + X = f(x) + X = \hat{f}(x+X), so \hat{f} is indeed well-defined.

    from here. it's all down-hill, the linearity of \hat{f} is a direct consequence of the linearity of f:

    \hat{f}((\alpha u + \beta v) + X) = f(\alpha u + \beta v) + X = \alpha f(u) + \beta f(v) + X

    = (\alpha f(u) + X) + (\beta f(v) + X) = \alpha \hat{f}(u) + \beta \hat{f}(v)

    *******

    it might be helpful to see a simple concrete example.

    let V =\mathbb{R}^2, the Euclidean plane, with the usual vector operations, and suppose X = \{(x,0) \in \mathbb{R}^2\}. what do the elements of V/X look like?

    well anything in (x,y) + X has a 2nd coordinate of y, so the elements of V/X are all horizontal lines (we get one for each different real number y).

    so suppose f(x,y) = (3x+y, 2y). it should be clear X is an invariant subspace for f.

    then \hat{f} is the mapping that takes the line going through y, to the line going through 2y. in other words, \hat{f} acts "just like" the function a-->2a (of one real variable).

    the reason being, when we act "mod X", we are "shrinking" the entire x-dimension down to 0. so what f does on the first coordinate becomes irrelelvant, as far as \hat{f} is concerned.
    Last edited by Deveno; October 6th 2011 at 06:22 AM.
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  3. #3
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    Re: Quotient Space

    Quote Originally Posted by Deveno View Post
    what we would like to do is define \hat{f}(x+V) = f(x)+V.

    first we need to be sure that this is well-defined, since we are working with cosets, instead of elements.

    so suppose y is in x + V (that is, that y + V = x + V), so that y = x + v, for some vector v in V. then f(y) = f(x + v) = f(x) + f(v) = f(x) + v',

    (where v' is some element of V) since V is invariant for f, so f(y) is in f(x) + V, thus f(y) + V = f(x) + V.

    hence \hat{f}(y+V) = f(y) + V = f(x) + V = \hat{f}(x+V), so \hat{f} is indeed well-defined.

    from here. it's all down-hill, the linearity of \hat{f} is a direct consequence of the linearity of f:

    \hat{f}((\alpha u + \beta v) + V) = f(\alpha u + \beta v) + V = \alpha f(u) + \beta f(v) + V

    = (\alpha f(u) + V) + (\beta f(v) + V) = \alpha \hat{f}(u) + \beta \hat{f}(v)

    *******

    it might be helpful to see a simple concrete example.

    let V =\mathbb{R}^2, the Euclidean plane, with the usual vector operations, and suppose X = \{(x,0) \in \mathbb{R}^2\}. what do the elements of V/X look like?

    well anything in (x,y) + X has a 2nd coordinate of y, so the elements of V/X are all horizontal lines (we get one for each different real number y).

    so suppose f(x,y) = (3x+y, 2y). it should be clear X is an invariant subspace for f.

    then \hat{f} is the mapping that takes the line going through y, to the line going through 2y. in other words, \hat{f} acts "just like" the function a-->2a (of one real variable).

    the reason being, when we act "mod X", we are "shrinking" the entire x-dimension down to 0. so what f does on the first coordinate becomes irrelelvant, as far as \hat{f} is concerned.
    How did you know how to define \hat{f}\mbox{?}

    I also don't really understand your example either.
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  4. #4
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    Re: Quotient Space

    that is the usual definition of an "induced map". the map V--->V/X given by v-->v + X is "canonical", in that it does not depend on the choice of a basis for V.

    you can also easily show it is linear, let's say we call it T. thus f (via T) "induces" the map \hat{f} by setting:

    \hat{f} \circ T = T \circ f

    the reason we need X to be invariant for f, is because if x is in X, but f(x) is not in X, then f(v + X) might not be in the coset f(v) + X (although it will be in the coset f(v) + f(X), but this is not "the same" canonical map T).

    in this problem, all we have to work with is the map v-->v+X, and the map f, so our answer must only depend on those.

    as for the example, the situation is like this: linear spaces are very well-behaved, and have natural geometrical interpretations. a 1-dimensional space is a line, a 2-dimensional space is a plane, a 3-dimensional space is like the world we live in (ignoring the curvature introduced by the minkowski metric). higher-dimensional spaces are hard to "visualize" but their behavior is "the same" as lower dimensional spaces, just...more basis vectors to specify.

    so 3-space mod a line, would be like partitioning space into a set of parallel "straws" (really skinny ones). the result is 2-dimensional, by specifying a point in a plane which cuts through all the straws, we can tell "which" straw we're at...that is, which coset of the line (straw) that goes through the origin.

    similarly, 3-space mod a plane, is like partitioning space into sheets parallel to a plane that goes through the origin. if we have a line that crosses all the "sheets", specifying a point on that line, tells us which "sheet" (that is coset of the plane going through the origin) we are in.

    in vector spaces, "modding" via a subspace is analogous to "modding mod n" in the integers. when we take an integer mod n, we are essentially declaring "all multiples of n are 0" which just leaves a cycle between 0 and n-1. when we take V/X, we are essentially setting the entire subspace X as {0}, leaving only the remaining dim(V) - dim(X) dimensions as relevant.

    in fact, what the rank-nullity theorem says is: dim(V) = dim(X) + dim(V/X). this is just the first isomorphism theorem of group theory in the context of vector spaces, and can be proved the same way.
    Last edited by Deveno; October 5th 2011 at 05:12 PM.
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  5. #5
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    Re: Quotient Space

    So \hat{f} isn't monic then, correct?

    What would be an example that shows \hat{f} isn't monic?
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    Re: Quotient Space

    it might be, it might not be. in the example i gave, \hat{f} was monic, but if f is the 0-map, then so is \hat{f}.

    even if f isn't monic, \hat{f} might be: suppose f(x,y,z) = (x,x,z) in \mathbb{R}^3.

    now f certainly isn't monic, f(1,2,3) = f(1,4,3), for example.

    but if our subspace X was all vectors of the form (0,y,0), (which is 1-dimensional, with basis {(0,1,0)},

    then certainly f(0,y,0) = (0,0,0) is in X.

    so \hat{f}((x,y,z)+X) = (x,x,z) + X = (x,0,z) + X (since (x,y,z) - (x,0,z) = (0,y,0) is in X).

    i claim \hat{f} is monic. suppose \hat{f}((x,y,z) + X) = \hat{f}((x',y',z') + X).

    then (x,0,z) + X = (x',0,z') + X, so we must have x' = x, and z' = z.

    thus (x,y,z) - (x',y',z') = (x,y,z) - (x,y',z) = (0,y-y',0) is in X, so

    (x,y,z) + X = (x',y',z') + X.
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  7. #7
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    Re: Quotient Space

    For this problem then, \hat{f} is monic.

    \hat{f}(x+V)=\hat{f}(y+V)\Rightarrow f(x)+V=f(y)+V\Rightarrow (f(x)-f(y))+V=0

    So f(x)-f(y)\in X/V

    f(x)-f(y)=0\Rightarrow f(x)=f(y)\Rightarrow x=y

    Is this correct?
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  8. #8
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    Re: Quotient Space

    NO!

    how can you deduce that x = y from f(x) = f(y)?

    IF (and that's a big if, at high decibal levels) f is monic, then \hat{f} will be, too (you can't make a map "less monic" by factoring out information).
    also f(x) + X = f(y) + X, merely implies f(x) and f(y) are in the same coset of V, that is:

    (f(x) - f(y)) + X = X, not "0" (of course 0 + X is the 0-vector (coset) in the coset space V/X, but is is unwise to denote this by simply "0", it's not the same type of vector

    as a vector in V, it's a vector composed of a SET of vectors in V (just as a residue class in the integers mod n isn't an integer, but an equivalence class of a SET of integers)).

    saying f(x) - f(y) is in X/V is meaningless....what is X/V? even if you meant f(x) - f(y) is in V/X, this is still wrong, f(x), f(y) are elements of V,

    NOT elements of V/X.

    part of your confusion is due to a bad typo in my orginal post, which i have edited.

    the cosets in V/X are cosets of X in V. i am more used to the notation U/V (which is more common), so most of the "stuff + V" expressions in my original post should have been "stuff + X". embarrassing o.o
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  9. #9
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    Re: Quotient Space

    \hat{f} isn't monic. If I define f as f:X\to X by f(x), then f is a linear transformation.

    But \hat{f} isn't monic since every element in the domain is mapped to 0+V

    Is this correct?
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  10. #10
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    Re: Quotient Space

    again, no. i don't know why you're fixated on whether or not \hat{f} is monic. sometimes it is, sometimes it isn't, it depends on f and X.

    look all \hat{f} is, is a linear transformation "as much like f as possible" on V/X. "modding by X" could take some of the domain values where multiple

    values in V get mapped by f to the same point, to a single point in V/X, but it might not.

    as far as the example you gave, how do you propose to extend f to a map defined on all of V?

    but sure, if X = V, and f is the identity on V, f(x) = x, then of course \hat{f} takes everything to the "0" of V/V,

    because V/V has just ONE coset, with all of V in it! V/V = {0+V}, so \hat{f} is the 0-map AND monic!

    but seriously, i think you are making this be far more complicated than it needs to be.

    a vector space is just an abelian group with some "extra structure" (namely, the scalar multiplication).

    and the quotient space is just the quotient group, as with any abelian group, along with an "induced" scalar multiplication:

    a(v + X) = av + X. \hat{f} is just f "ignoring what happens in X" (since f(X) is a subset of X, which all just gets dropped in the + X part of the coset).
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