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KerF and ImF of linear copying

  1. Feb 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Find just one base of the set of pictures and one base of the kernel.

    [tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^3\ ,f(x,y,z)=(x+2y-z,y+z,x+y-2z)[/tex]

    2. Relevant equations

    [tex]Im F=\{f(v)| v\in V\}[/tex]

    [tex]ker F=\{v\in V | F(v)=0\}[/tex]

    3. The attempt at a solution

    Actually, I don't know the principle of solving this tasks. My text book is rubbish. It doesn't say anything about this problem, or how to solve it. Thanks for your time, and help.
     
  2. jcsd
  3. Feb 22, 2008 #2

    HallsofIvy

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    I find it very difficult to believe your book does not have anything about the definitions of "image" (what you are calling a "picture"- translation problem obviously. "Picture" is a very specific word, "image" has a much more general meaning that can include "picture".) or "kernel" (exactly the right word)- and that's all you should need. You should not expect your textbook to tell you how to solve every problem it has- that defeats the purpose of "education"- which is not just memorizing formulas or how to solve specific problems.

    f(x,y,z)= (x+ 2y- z, y+ z, x+ y- 2z) has kernel given by x+ 2y- z= 0, y+ z= 0, x+y- 2z= 0 (that's directly from the definition of "kernel"). Solve those equations for x, y, and z to see what value give <0, 0, 0>. Adding the first two equations, x+ 3y= 0. Adding twice the second equation to the third, x+ 3y= 0. Those two equations are the same and are satisfied by any (x, y, z) such that x= -3y and, from the second equation, z= -y. In particular if we take y= 1, then x= -3, z= 1 so <-3, 1, -1> is a vector in this one dimensional subspace. Since that has dimension 1, you should know from the "dimension theorem" (you may know it by another name), that the image must have dimension 3- = 2.

    To find the image, look at <x+ 2y- z, y+ z, x+y- 2z> and think about the values of those components for different values of the three variables, x, y, and z. As I have suggested before, let x= 1, y= z= 0, and you get <1, 0, 1>. Let x= z= 0, y= 1 and you get <2, 1, 1>. Let x= y= 0, z= 1 and you get <-1, 1, -2>. But notice that <-1, 1, -2>= <1, 0, 1>- 2<2, 1, 1>! (Which makes nonsense of what I said about this choice of "1" and "0" for each variable guarentee independent vectors!). Since the third vector is a linear combination of the first two, {<1, 0, 1>, <2, 1, 1>} is a basis for the image of the linear transformation- whidh is, as we checked before, dimension 2.
     
  4. Feb 22, 2008 #3
    I'll show you sometimes my book, you'll see. There are just equations, and not any examples. Ok, I understand all this. But I have few questions. For example,
    How you know that the vector is one dimensional subspace, so you don't get 2-nd series of values for x,y or y,z or x,z?
     
  5. Feb 22, 2008 #4

    HallsofIvy

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    For the kernel, solving the equations defining the kernel, I arrived at the two independent equations x+ 3y= 0 and y+ z= 0. That told me that I could choose any one of x, y, or z however I wished and solve for the other 2. I have "one degree of freedom" and so the subspace is one dimensional. Of course, if all three equations had been independent, I could have solved for the unique solution: x= y= z= 0: in that case the linear transformation would be "one-to-one".

    It is possible, say the linear transformation were T<x,y,z>= <x- y+ z, 2x- 2y+ 2z, -x+ y- z>, that I would find that all 3 equations, x- y+ z= 0, 2x- 2y+ 2z= 0, -x+ y- z= 0 were just multiples of one another: they all reduce to x- y+ z= 0. In general, one equation allows us to solve for one of the variables and reduce the dimension by one. Since there is only one independent equation here, the dimension of the kernel is 3- 1= 2.

    More precisely, I can solve for y= x+ z so I am free to choose any values I want for x or z and then solve for y: If x= 1 and z= 0 (I'm a simple soul- I like simple numbers) y= 1 so we have <1, 1, 0>. If x= 0 and z= 1, y= 1 so <0, 1, 1>. A basis for the kernel is {<1, 1, 0>, <0, 1, 1>}. Any vector in that space can be written as a<1, 1, 0>+ b<0, 1, 1>= <a, a+b, b>. I'll leave it to you to see that the linear transformation applied to <a, a+ b, b> gives <0, 0, 0>.

    Of course, any subspace has an infinite number of possible bases. We could as easily have solved x-y+z= 0 for z= y- x. Now taking x= 1, y= 0, then z= -1 and if we take x= 0, y= 1, then z= 1. We get {<1, 0, -1> and <0, 1, 1> which is another basis for the same subspace. You can prove that by showing that <1, 0, -1> and <0, 1, 1> can be written as linear combinations of <1, 1, 0> and <0, 1, 1> and vice-versa.

    How about the image of this linear transformation? T<x, y, z= <x- y+ z, 2x- 2y+ 2z, -x+ y- z> so, taking x= 1, y= z= 0, T<1, 0, 0>= <1, 2, ->, taking x= z= 0, y= 1, T<0, 1, 0>= <-1, -2, 1> and taking x= y= 0, z= 1, T<0, 0, 1>= <1, 2, -1>. It's easy to see that those are all multiples of one another so we really have just one independent vector. We could use any one of those as the single vector in the basis. The image has dimension 1 and again we see that the dimension of the kernel plus the dimension of the image is the dimension of the whole space, 3.

    By the way do you see how my recommendation of using "x= 1, y= z= 0, then x= z= 0, y= 1, etc." just gives the standard basis: <1, 0, 0>, <0, 1, 0>, <0, 0, 1>?
     
  6. Feb 22, 2008 #5
    Yes, nice recommendation. Again thank you very much for the help. You're one of the rarely good and patient persons. Thanks for your efforts to help me, even with my problem with English language.
     
  7. Feb 23, 2008 #6
    I have one task in my book:
    [tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^2\ ,f(x,y,z)=(x-2y,x-2y)[/tex]
    Just tell me please, do I solved the problem correctly?
    First for kerF.
    x-2y=0
    x-2y=0

    x=2y
    y=1
    (2,1)
    So the base is one dimensional.
    3-1=2
    The image will be 2 dimensional.
    x=1
    y=0
    <1,1,0>
    x=-1
    y=1
    <-1,-1,0>
    x=3
    y=0
    <3,3,0>
    But <3,3,0>=3<1,1,0>
    <1,1,0>=-1<-1,-1,0>
    The dimension will be 1 here?
    What is the problem?
     
  8. Feb 23, 2008 #7

    HallsofIvy

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    When I did this about 2 hours ago, I did it in a rush and was completely wrong.

    In looking at the kernel, you ignored z! Yes, to be in the kernel, x must equal 2y and so <2, 1, 0> is a vector in the kernel. But that is true for any z. In particular, < 0, 0, 1> is in the kernel also. A basis is {<2, 1, 0>, <0, 0, 1>} (or {<2, 1, 0>, <2, 1, 1> etc.) and the kernel has dimension 2.

    Yes, it should be obvious that no matter what x, y, and z are the result of the linear transformation is <a, a> since the two components are the same. A basis for the image is {<1, 1>} and the image is 1 dimensional: 2+ 1= 3, the dimension of the domain space.
     
    Last edited: Feb 23, 2008
  9. Feb 23, 2008 #8
    Can I implement here the "dimension theorem" and how?? Is it correct, if I write for the image {1,1,0} instead of {1,1}? Can the image be {1,1,1}?
     
  10. Feb 23, 2008 #9
    You have some mistake here. It should be 3-2=1. In the last example of mine, if I use the dimension theorem, should I get [tex]dimR^2=2 or R^3=3[/tex]?
     
  11. Feb 23, 2008 #10

    HallsofIvy

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    You can't write the image as "the subspace spanned by <1, 1, 0> because the image is a subset of R2. And, by the way, the image is not "{1, 1, 1}" because that is a single vector. The image, a subspace of R2, is the space spanned by the vector <1, 1>: all multiples of that vector or all vectors of the form <a, a>

    No, there is not a mistake. The kernel is all <x, y, z> such that T<x, y, z>= <0, 0, 0>. As I showed, the all reduces to single equation x- y+ z= 0. Since R3 has dimension 3, that one equation reduce the dimension of the kernel to 3- 1= 2. We could also do this: if x= 1, y= 0, z satisfies 1- 0+ z= 0 so z= -1. <1, 0, -1> is a vector in the kernel. If x= 0, y= 1, z satisfies 0- 1+ z= 0 so z= 1. <0, 1, 1> is a vector in the kernel and {<1, 0, -1> , <0, 1, 1>} is a basis for the kernel. Again, it has dimension 2.

    The dimension of the kernel is 2 so the dimension of the image is 3- 2= 1. In fact, it is easy to see (taking x= 1, y= 0, z= 0 for example) that a vector in the image is <1, 2, -1> and you should be able to see that if you use any other values for x, y, and z, you get multiples of that. The image is spanned by the single vector <1, 2, -1>. {<1, 2, -1>} is a basis for the image which has dimension 1.

    The "dimension theorem" says that if T is a linear transformation from U to V, the dimension of the kernel of T plus the dimension of the image of T is equal to the dimension of U. In your example, that was 3.
     
    Last edited: Feb 23, 2008
  12. Feb 23, 2008 #11
    And how we will use the dimension theorem on the last example?
    [tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^2\ ,f(x,y,z)=(x-2y,x-2y)[/tex]
     
  13. Feb 23, 2008 #12

    Tom Mattson

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    I have to ask, what textbook are you using? That is, what is the title and author?

    The reason I ask is that if I know the level of the book, I might have a link to a free online book that is better.
     
  14. Feb 23, 2008 #13

    HallsofIvy

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    You really don't need the "dimension theorem" except as a check.

    I have already shown that the kernel has basis {<2, 1, 0>, <0, 0, 1>} and so has dimension 2, and that the image has basis {< 1, 1>} and so has dimension 1. We can use that fact that dim of ker (f)+ dim of img(f)= 2+ 1= 3 = dimension of R3 as a check.
     
  15. Feb 23, 2008 #14
    But in previous posts you said me that I should use R2 (dimension 2) instead of R3, because of the linear copying. Why now R3?
    Tom, I use text book in my native language, which is called,
    Linear Algebra and Analytic Geometry for 3rd grade high school. In my country, there are (now), 1-9 grades primary school, and 1-4 grades high school. Here are some screen shots:
    Picture 1
    Picture 2
    Picture 3
     
  16. Feb 23, 2008 #15

    HallsofIvy

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    Where did I say that?

    In post 10 of this thread I said:
    "The "dimension theorem" says that if T is a linear transformation from U to V, the dimension of the kernel of T plus the dimension of the image of T is equal to the dimension of U. In your example, that was 3."
     
  17. Feb 23, 2008 #16

    HallsofIvy

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    Oh, you're from Macedonia! Lovely country. I was there the summer of 2006.

    Spent most of my time in Skopje, but managed a few days in Ohrid.
     
    Last edited: Feb 23, 2008
  18. Feb 23, 2008 #17
    Thanks, I am glad, that you have visited Macedonia. What were you doing in Macedonia?
    Btw- Isn't [tex]\mathbb{R}^3[/tex] linear copying into [tex]\mathbb{R}^2[/tex], so the dimension should be 2?
     
  19. Feb 23, 2008 #18
    Why we get <x,y> for image and <x,y,z> (three coordinates) for kernel?
     
  20. Feb 24, 2008 #19

    HallsofIvy

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    Look at the definitions! If T is a linear transformation from U to V, the kernel of T is the set of all vectors u in U such that Tu= 0. They must be in U so that T can be applied to them. The image of T is the set of all vectors v in V such that v= Tx for some x in U. They must be in V since they are the result of applying T to a vector.

    If T is from R3 to R2, then the kernel is a subspace of R3 and the image is a subspace of R2.

    I was visiting my finacee who worked in the U.S. Embassy there.
     
  21. Feb 24, 2008 #20
    Does [tex]0\in V[/tex], for the kernel is T(u)=0?
    Ok, and for this example:
    [tex]f:\mathbb{R}^3\rightarrow\mathbb{R}^3\ ,f(x,y,z)=(x-2y,x-3y,x+y-z)[/tex]
    For kerF, (x-2y,x-3y,x+y-z)=(0,0,0)
    x=0
    y=0
    z=0
    Why in my text book result, they wrote [itex]\varnothing[/itex] instead of (0,0,0)?
     
    Last edited: Feb 24, 2008
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