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Linear Algebra; Transformation of cross product

  1. May 19, 2015 #1
    A matrix is a linear transformation if, T(u+v)= T(u) +T(v) and T(cu)=cT(u).

    Theorem 8.4.2 If V is a finnite dimensional vector space, and T: V-> V is a linear operator then the following are equivalent.
    a) T is one to one, b) ker(T)=0, c) nuttility(T)=0 and d) the Range of T is V; R(T)=V

    1- Question
    Let V be a fixed vector in R^3. a)Show that the transformation defined by T( u)= v X u is a linear transformation.
    b) Find the range ot the linear transformation
    c) If v=i , find the matrix for this linear transformation.

    2- Answer
    a) I proved that T(u+v)= T(u) +T(v) and T(cu)=cT(u), through cross product properties, And therefore proved its linear transformation.

    b) This one i am not entirely sure, however with the Theroem 8.4.2 d) I concluded that R(T) = R^3

    c) This one is where I am unsure where to begin. Since T(u) = v X u

    Then T(u) = { Det ( v2, v3 - Det ( v1, v3 Det ( v1, v2
    u2, u3 ) , u1, u3) , u1, u2) }

    I am not sure where to go from there, and unsure what they mean by v=i

    Thank you,
  2. jcsd
  3. May 19, 2015 #2
    To answer part(c), just plug in corresponding unit basis vectors into the linear transformation formula, and you will obtain each column of the matrix.
  4. May 19, 2015 #3
    I am not sure what you mean.
  5. May 19, 2015 #4
    Linear transformation can be written into a matrix form. When you times a matrix with a unit basis vector, if the unit basis vector has a 1 as its first element and 0s as its other elements, then you obtain the first column of the matrix as a result. Keep doing this you can calculate all columns of the corresponding matrix, thus obtain the matrix itself.
  6. May 19, 2015 #5
    ok, so v= i means i=( 1
    0) ?

    But then I don't know what u is . So would I not just have u= u1
    u3 ?

    And then T(u) = v X u = (0, -1u3, 1u2).
  7. May 19, 2015 #6
    No, here i is just an example. I couldn't tell what i is from your information. Is this the whole problem?
  8. May 19, 2015 #7
    yes, that is all the information I am given ?
  9. May 20, 2015 #8


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    I think you meant "function" or "map" here, not "matrix". If T is a matrix, then the map that takes x to Tx is always linear.


    This is wrong. Is there something you can say about how T(u) is related to u?

    You will need to learn how the matrix of a linear transformation is defined. It's described in one of our FAQ posts: https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/ [Broken]
    Last edited by a moderator: May 7, 2017
  10. May 21, 2015 #9
    sorry , for theroem 8.4.2 c) I meant, nullity.

    If I prove that a,b or c is true, then i can state that d is true also, right?
  11. May 21, 2015 #10
    If A is an mXn matrix and TA:Rn-->Rm is multiplication by A then, nullity(TA)= nullity(A) and the rank(TA)=rank(A)
  12. May 21, 2015 #11


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    Correct. But if you think about what the cross product does, you should find it easy to see that none of the statements is true for the specific T you're working with.
  13. May 22, 2015 #12
    Ok, so the range of T is not R^3? And the nullity and kernel is not zero? I had "figured" out that the ker(T)=0, but I must be miss understanding the kernel concept.
  14. May 22, 2015 #13


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    The kernel is the set of vectors that are mapped to the 0 vector, i.e. the set of all x such that Tx=0. You should have no difficulty finding a non-zero vector that's mapped to 0 by this T.
  15. May 22, 2015 #14
    Oh ok, ya makes sense, thanks :).. So how would I determine what the kernel and the range is here then ?
  16. May 22, 2015 #15


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    Are you familiar with the geometric interpretation of the cross product? The right-hand rule and all that? If yes, then you should be able to correctly guess what the kernel and range are. Then you can try to prove that your guesses are correct.
  17. May 23, 2015 #16
    Ok, So i'll do this one at a time.

    I am kinda familiar with the geometric interpration of the cross product, the right hand rule. I also just revisited it.

    In order to find the Kernel I need to find when does vXu=0, correct?

    If that's correct then this is the case when; vector u, equals zero; vector v equals zero; when vector v is in the exact opposite direction of u; and also when vector v is in the exact same direction of u.

    What I understand is that uXv=0 when the angle between them is 0. Is this correct?

    If so how would I write that?
    Last edited: May 23, 2015
  18. May 23, 2015 #17
    So your conjecture is that ##\mathbf{u}\times \mathbf{v} =\mathbf{0}## iff ##\mathbf{u} = t\mathbf{v}## for some ##t\in \mathbb{R}##. This would make the kernel ##\{t\mathbf{v}~\vert~t\in \mathbb{R}\}##. So now try to prove this. This involves two steps:
    1) Each ##t\mathbf{v}## is in the kernel. I.e. for each ##t##, we have ##(t\mathbf{v})\times \mathbf{v} =\mathbf{0}##.
    2) There are no other vectors in the kernel. I.e. if ##\mathbf{u}\times\mathbf{v} = \mathbf{0}##, then ##\mathbf{u} = t\mathbf{v}## for some ##t\in \mathbb{R}##.
  19. May 26, 2015 #18
    Does this make sense?
    What I am understanding the kernel of T is always a subset of V. And it would be either the zero subspace,the entire vector space or a line through the origin. Ane from my understanding that I explained above, I conclude that,
    The Kernel of T will be the line through the origin spanned by V? and so, Ker(T)=Span(V)

    And the Range of T has to be a subspace of the new space. And the new space is an orthogonal projection of V.
    Range would therefore be a plane through the origin orthogonal to v. and so R(T)= Span( VxU) ??
  20. May 26, 2015 #19
    Yes, but what if ##\mathbf{v} = \mathbf{0}##? Is it still true then?

    A plane? Is there more than one such plane?

    That doesn't really make much sense. The vector ##\mathbf{u}## is the variable in the function. So I don't know what ##\text{Span}(\mathbf{v}\times \mathbf{u})## means. Whatever it means, it's not correct since ##\text{Span}(\mathbf{v}\times \mathbf{u})## is 1-dimensional, while you just said that the range is a plane, which would make it two-dimensional.
    Last edited: May 26, 2015
  21. May 26, 2015 #20
    Ok- Can someone help me find what the kernel and the range of T is?
  22. May 26, 2015 #21
    Your attempt in your previous post was very good, don't give up now! Try to think about the remarks I gave you.
  23. May 26, 2015 #22
    ok...I am really confused!
    I thought the line through the origin would include when v=0.
    And for the second part, I really don't know what to do, I understand what you mean by span (vXu) would be a line and not a plane. But I don't know if what I said about the plane was correct and how I would express that.
  24. May 26, 2015 #23
    Does it? It's not unusual to get separate cases for extreme situations.

    What you said about the plane is certainly correct. The plane orthogonal to ##\mathbb{v}## is the range of ##T## (but again, is this correct if ##\mathbf{v}= \mathbf{0}##)?
  25. May 26, 2015 #24
    ok...ummm.. I don't know.. I'm not even sure where to go from there lol
  26. May 26, 2015 #25
    any chance you can you give me little more guidance? :/
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