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Transformation assigment

  1. Jan 25, 2004 #1
    Hi! I'm in need of some help regarding an assigment. I need to understand the concepts in the problem, and they are not explained in the literature we are using. This is the problem:

    Given a linear transformation

    [tex]\mathcal{A} = \mathbf{R}^n\rightarrow\mathbf{R}^m[/tex]

    show that it can be divided/separated as [tex]\mathcal{A}=i\circ\mathcal{B}\circ{p}[/tex], where [tex]p[/tex] is the projection on the complement to the kernel, [tex]\lbrace \vec{v}; \mathcal{A}\vec{v} = \vec{0}\rbrace[/tex] (two elements in the complement are considered identical if they differ with one element in the kernel, ie we take the ratio???), the transformation [tex]\mathcal{B}[/tex] is an invertible transformation ([tex](\mathcal{B}\vec{v})^{-1} = \vec{v}[/tex]?), from the complement to the kernel to the image under [tex]\mathcal{A}[/tex], and [tex]i[/tex] is the inclusion of the image in [tex]R^m[/tex].

    Ok, so the kernel is the set of vectors with the property [tex]\mathcal{A}\vec{v} = \vec{0}[/tex]. The complement should be [tex]\lbrace \vec{v}; \vec{v}\in\mathbf{R}^n \rbrace\backslash\lbrace\vec{v};\mathcal{A}\vec{v}=\vec{0}\rbrace[/tex]. Right? What about the projection on this set? How do you project a vector on a set? On every individual element, in which case the result is a set?

    And the what is the inclusion [tex]i[/tex]? How should I approach this problem?

    Thanks on advance,
    Nille
     
  2. jcsd
  3. Jan 25, 2004 #2

    HallsofIvy

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    I suppose it would cranky of me to suggest that you "approach" this problem by going back and learning some definitions. You seem to have only a very general idea of what these things are. In particular you "question":[tex](\mathcal{B}\vec{v})^{-1} = \vec{v}[/tex]?" doesn't even make sense (but at least is was a question) [tex]\mathcal{b}\vec{v}[/tex] is a vector. Since multiplication of vectors is not defined neither is [tex]\vec{v}^{-1}[/tex] for any vector. What you meant to write was [tex](\mathcal{B}^{-1}(\mathcal{B}(\vec{v}))= \vec{v}[/tex] although the condition that v be in the compliment of the kernel of [tex]\mathcal{B}[/tex] can be written much more simply: [tex]\mathcal{B}(\vec{v})\neq\vec{0}[/tex].
    You write the complement of the kernel of [tex]\mathcal{B}[/tex] as
    [tex]\lbrace \vec{v}; \vec{v}\in\mathbf{R}^n \rbrace\backslash\lbrace\vec{v};\mathcal{A}\vec{v}=\vec{0}\rbrace[/tex] which is correct but it is much more simply [tex]\lbrace\vec{v};\mathcal{B}\vec{v}\neq\vec{0}\rbrace[/tex].
    Actually, I suspect you are misreading the problem and that by "complement" they mean not just the "set complement" but the "orthogonal complement". The kernel of any linear transformation is a subspace but the set-complement of that set is not.

    Consider this simple R2 example, representing each vector as (x,y): Define [tex]\mathcal{B}[/tex] by [tex]\mathcal{B}[/tex](x,y)= (x- y, x- y). The kernel is all vectors of the form (x,x) (the line y= x in the xy-plane). The "set-complement" of that is the set of all (x,y) such that x [tex]\neq [/tex]y but that is clearly not a subspace (the only subspaces of R2 are straight lines through the origin). The "orthogonal complement" of the subspace {(x,y)|y=x} is the subspace {(x,y)|y= -x}.

    Given any vector (x,y) we can write as a sum of vectors in these two subspaces: (x,y)= ((x+y)/2,(x+y)/2)+ ((x-y)/2,(y-x)/2).
    The first of these, ((x+y)/2,(x+y)/2), is in the kernel of [tex]\mathcal{B}[/tex]:[tex]\mathcal{B}[/tex](((x+y)/2,(x+y)/2))= (0,0).
    The second, ((x-y)/2,(y-x)/2), is in the orthogonal complement of [tex]\mathcal{B}[/tex]. The "projection" from R2 to this set is the linear transformation that maps (x,y) onto ((x-y)/2,(y-x)/2).
    The "inclusion" map from a subspace to the entire space just maps [tex]\vec{v}[/tex] to itself. The difference between it and the "identity" map is that the identity map maps the entire space to itself. The inclusion map is defined on any subspace. In the example above, since [tex]\mathcal{B}[/tex] is from R2to R2, the inclusion map is the identity map.
    Of course, if we restrict [tex]\mathcal{B}[/tex] to the orthogonal complement of the kernel, it maps only [tex]\vec{0}[/tex] to [tex]\vec{0}[/tex] and so is invertible.

    A more general example: [tex]\mathcal{B}[/tex], From R2 to R1 takes (x,y) to x- 2y. This is singular since it maps many different pairs into the same number. In particular, the kernel is all (x,y) such that x-2y= 0 or y= (1/2)x, a straight line. The orthogonal complement of that is the line y=-2x, which is, of course, orthogonal (perpendicular) to the line y= (1/2)x.

    Suppose (x0,y0) is any point in R2. Drop a perpendicular to y= -2x: it is the line y= (1/2)(x-x0)+ y0 and that crosses y= -2x where -2x= (1/2)(x- x0)+ y0 or -4x= x- x0+ 2y0. -3x= -x0+ 2y0 or x= [tex]\frac{x_0-2y_0}{3}[/tex]. Since this is on the line y= -2x, the y coordinate is then y= [tex]\frac{-2x_0+ 4y_0}{3}[/tex]. The projection operator maps (x,y) onto ([tex]\frac{x-2y}{3},\frac{-2x- 4y}{3}[/tex]).

    Notice that the set of all such numbers is a one-dimensional subspace of R2- they all lie on the line y= -2x. Now apply [tex]\mathcal{B}[/tex]. It maps all vectors of the form ([tex]\frac{x-2y}{3},\frac{-2x- 4y}{3}[/tex]) into [tex]\frac{x-2y}{3}-\frac{-2x+ 4y}{3}= \frac{3x-6y}{3}[/tex]= x-2y (Of course, [tex]\matcal{B}[/tex]is that function. (And it is non-singular on this set- it maps only (0,0) to 0.) Finally, the "inclusion" operator maps that into itself.
     
    Last edited: Jan 27, 2004
  4. Jan 29, 2004 #3
    Hi!
    Thanks for the ambitious response! However, I have failed to understand your explanation. I therefor post some more questions:

    How can I define the projection on the complement to the kernel of [tex]\mathcal{A}[/tex]?

    If [tex]i:\mathbb{R}^n \rightarrow \mathbb{R}^m[/tex], what is the purpose of [tex]i[/tex] if the result of [tex]\mathcal{B}[/tex] is a vector in [tex]\mathbb{R}^m[/tex]? If I understand it correctly, [tex]i[/tex] only maps a vector in [tex]\mathbb{R}^n[/tex] to a vector in [tex]\mathbb{R}^m[/tex]. For instance, if you were to map a vector in [tex]\mathbb{R}^2[/tex] to [tex]\mathbb{R}^3[/tex], the result would be the same vector with z=0, ie a line on the xy-plane. Is this correct?

    What does it mean that "two elements in the complement is considered identical if they differ with one element in the kernel, ie we take the ratio"?

    The problem is that the course I'm reading doesn't introduce these concepts at all. It only covers elementary linear algebra. Where could I find some more information on this? I have tried a bunch of web pages...

    Thanks again!
    Nille
     
  5. Jan 31, 2004 #4

    HallsofIvy

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    In what sense do you mean this? In the general sense, the projection onto the (orthogonal) complement of the kernel of [tex]\mathcal{A}[/tex] is just the usual definition of projection onto a subset.
    How one defines the projection for a specific [tex]\mathcal{A}[/tex] depends on [tex]\mathcal{A}[/tex], of course.

    For example suppose [tex]\mathcal{A}[/tex] is from [tex]\mathbb{R}^3[/tex] to [tex]\mathbb{R}^2[/tex] defined by [tex]\mathcal{A}[/tex](x,y,z)= (x-y,z-y). The kernel consists of all (x,y,z) such that (x-y,z-y)= 0 or x=y, z= y. That can be represented by the straight line through (0,0,0) and (1,1,1) or the parametric equations x= t, y= t, z= t. The "orthogonal complement" of the kernel can be represented by the plane, containing (0,0,0) perpendicular to that line and has equation x+ y+ z= 0. Given any point (x0,y0,z0) we can find its projection onto that plane (subspace) by taking the line through that point perpendicular to the plane: its parametric equations are x= x0+ t, y= y0+t, z= z0+ t. The point of intersection of that line with the plane must satisfy x+ y+ z= x0+ t+ y0+t+z0+ t= x0+ y0+ z0+ 3t= 0 or t= (x0+ y0+ z0)/3.
    Since x= x0+(x0+ y0+ z0)/3 from the equation for the line,
    x= (4/3)x0+(1/3)y0+ (1/3)z0
    Similarly,
    y= (1/3)x0+(4/3)y0+ (1/3)z0
    z= (1/3)x0+(1/3)y0+ (4/3)z0

    Since this is a linear transformation, we can represent it as a matrix:
    [4/3 1/3 1/3]
    [1/3 4/3 1/3]
    [1/3 1/3 4/3]

    Actually, this is used in to different ways. Is m> n or m< n?
    (Obviously, if m= n, [tex]i[/tex] is just the identity function).
    If [tex]i:\mathbb{R}^n \rightarrow \mathbb{R}^m[/tex] with n> m, then this is a projection mapping: (x,y,z)-> (x,y,0) as you say. More often we use [tex]i:\mathbb{R}^n \rightarrow \mathbb{R}^m[/tex] with m> n so that (x,y)-> (x,y,0) to identify [tex]mathbb{R}^2[/tex] with a subspace of [tex]mathbb{R}^3[/tex]. Typically, if p is a vector on some subspace N, of vector space M, we use [tex]i:M\rightarrow N[/tex] to indicate that we were previously thinking of p as a vector in N and are now thinking of it as a vector in M (of which N is a subspace).

    I have no idea what that means and I wonder if you copied it correctly (for one thing, if it really said "two elements is" it is terribly bad English!). Two elements in the orthogonal complement of the kernel are "considered identical" only if they are identical! Two elements in the whole space will give the same projection onto the complement of the kernel (and can be "considered identical" in that sense) if and only if their difference is a member of the kernel. As for "ie we take the ratio", I have no idea what that could mean since "ratio" is not defined for vectors. They seem to be confusing "difference" with "ratio".
     
  6. Jan 31, 2004 #5
    Hi, and thanks again!
    The problem with this assignment is that nothing is defined. The question is asked so that the answer should apply to some arbitary linear transformation [tex]\mathcal{A}:\mathbb{R}^n\rightarrow\mathbb{R}^m[/tex]. This is where the problem appears - I don't know how to prove it. Since [tex]\mathcal{A}[/tex] isn't defined, the kernel isn't defined (obviously) and thus the complement to the kernel isn't defined. How can I define projection onto the complement of a kernel of an arbitary transformation [tex]\mathcal{A}:\set{R}^n\rightarrow\set{R}^m[/tex]?

    You can define the kernel as the set [tex]\mathbb{K} = \lbrace \vec{v}; \mathcal{A}\vec{v} = \vec{0} \rbrace[/tex], and the orthogonal complement to this would then be [tex]\lbrace \vec{u}; \vec{u}\cdot\mathbb{K} = 0 \rbrace[/tex]. But what does this mean, exactly? In your example, the complement to the kernel was a plane, and it is easy to project a vector on a plane. But in this case, the complement isn't defined. So how could I express a projection on some subspace [tex]\set{R}_n \subset \set{R}^n[/tex]?

    The result of the transformation [tex]p[/tex] is some vector [tex]\vec{v}_p = p\vec{v}[/tex]. The next transformation is [tex]\mathbb{B}[/tex], which is some arbitary invertable linear transformation [tex]\mathbb{B} = \left(\mathrm{ker}\ \mathcal{A}\right)^c \rightarrow \mathbb{R}^m[/tex]. If the transformation matrix (can one be definied? Is [tex]\mathrm{dim}\ \left(\mathrm{ker}\ \mathcal{A}\right)^c = \mathrm{dim}\ \mathbb{R}^m[/tex]?) is defined as [tex]\mathbf{B}[/tex], we know that [tex]\mathrm{det}\ \mathbf{B} \ne 0[/tex]. What can I conclude from this? Would the result of this be [tex]\vec{v}_p[/tex] expressed in the coordinatesystem with the base vectors [tex]\mathbf{B} = (\lbrace\vec{b}_i\rbrace_1^m)[/tex]?

    Your explanation of the inclusion was impeccable, so finally I understand that part!

    As for the spelling and grammatic errors: the original text was written in swedish, and I messed up the translation. But the bit about the ratio is a correct translation. It just doesn't make sense!

    And another excuse: the course I'm reading is an elementary one, consisting of 3 weeks of linear algebra only (the rest is multivariable analysis, or whatever it's called in english). This assignment isn't within the scope of the course.

    Thanks again (and thanks for your patience!),
    Nille
     
  7. Jan 31, 2004 #6
    I think HallsofIvy has answered this question much better than I could have answered it, but I think I know what you mean by "two elements in the complement is considered identical if they differ with one element in the kernel, ie we take the ratio"-- although I suspect the word "ratio", isn't quite the proper one. I'll try to recap a bit first and then simplify-- mine is the easy job since HallsofIvy has already provided all the details:

    Number 1 (decomposition of a space via a linear map):
    The linear transformation A between R^n and R^m induces a decomposition of R^n into the kernel of A and the image of A.
    In other words R^n = Ker A + Im A. That's kind of surprising, and a little misleading since I'm using an "equals sign", where I should be using something else (the Image of A is actually in a completely different set (namely in R^m)) however if you'll let me use '=' to mean 'isomorphic', then we'll get along just fine. We could quibble about the '+' as well, but humor me.
    Number 2 (remark on terminology):
    What you mean by the complement of the 'Ker A' is the subset of R^n that corresponds to the 'Im A'. Let's be really, really careful and call that set COMP.

    So now we say R^n = Ker A + COMP (and I'm not lying about the '=' anymore), and we're saying that for any elt of R^n there is a way to write it of the form x = c + d where c is in the kernel of A and d is in COMP, even cooler, there is a UNIQUE way to do this. so we can...
    Number 4:
    define p(x) = d... that's the projection.
    Now... B is a linear transformation from COMP to IM A
    And the i includes the Image of A (or Image of COMP if you prefer) into R^m. Now we have an honest map from R^n to R^m

    So what was up with the definition of complement?
    Think about R^n/Ker A. This defines a set of equivalence classes on the original R^n:
    for c,d in R^n we defined c~d iff c-d is in Ker A.
    Surprise number 1: This is a vector space
    Surprise number 2: This is isomorphic to Im A
    Surprise number 3: This is isomorphic to COMP.
    so... the really slick way of defining your maps is:
    A:R^n -> R^m
    Can be decomposed into the projection p:R^n -> R^n/Ker A
    B is an invertible map from R^n/Ker A -> Im A
    i is the inclusion of Im A back into R^m.
    Whew. Hope that helped.
     
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