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Angle between vector and its transpose

  1. Nov 21, 2012 #1
    Hi

    What is the angle between a vector (e.g. a row vector A) and it's transpose (a column vector) ? I know what transpose means mathematically but what is the intuition?

    Thanks guys
     
  2. jcsd
  3. Nov 21, 2012 #2
    If you're talking about simple real vectors (e.g. arrows in euclidian space) than the transposed vector is the same as the original. It's merely a matter of notation,whether you write the components in row or column.
    In general case there is a distinction between row vectors and column vectors however in such cases it's hard to give concise meaning to the word "angle".
     
  4. Nov 21, 2012 #3
    I don't understand the question. Usually in vector spaces like this (well Euclidean Vector Spaces), we define "angle" to be the cosine of the inner product of two vectors. However,
    a vector and its transpose are not in the same space. (In the sense that one is a 1xn matrix and one is a nx1 matrix.)
     
  5. Nov 21, 2012 #4

    arildno

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    I agree with Robert.
    The question is meaningless.
     
  6. Nov 21, 2012 #5

    Mark44

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    Consider
    ##u = (1, 2, 3)##
    and
    $$u^T = \begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}$$

    Clearly the angle between the two vectors is 90°:tongue:
     
  7. Nov 21, 2012 #6
    I don't think it's meaningless. In introductory classes (where vectors aren't even defined properly) rows and columns are often just different notations for components of vectors.
     
  8. Nov 21, 2012 #7

    arildno

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    :rofl:
     
  9. Nov 21, 2012 #8

    arildno

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    In such classes, the word "transpose" ought to be banned. THAT word should be introduced AFTER proper definition of a vector.
     
  10. Nov 21, 2012 #9
    Assuming that question is meaningless, what is the intuition behind taking transpose?
     
  11. Nov 21, 2012 #10

    arildno

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    You transpose, in order to make the transpose.
    It is a well-defined matrix operation.
     
  12. Nov 21, 2012 #11
    Are you familiar with the dot product? Well, this can be written like [itex]v^\top w[/itex]. If you have some vector [itex]v[/itex] then the matrix that projects onto that vector is given by [itex]vv^\top[/itex] (if you don't understand what projection means, that's OK, but this example probably doesn't make sense.)

    Additionally, there are important classes of matricies that are defined using transpose (or complex conjugate transpose which is where you do the transpose and take the complex conjugates of the entries of your matrix, if your matrix has real entries then obviously the complex conjugate transpose is just the transpose.)

    For example, if [itex]Q^\top=Q^{-1}[/itex] the matrix is called orthogonal. These matrices preserve angles and lengths of vectors. These are good for numerical applications for that reason, but also it is MUCH easier to compute the transpose than it is to compute the inverse (in a sense, you don't need to "compute" the transpose, your code just needs to "iterate backward" - if you don't understand that part, its OK.)

    Another special class is Symmetric Matrices where [itex]M[/itex] is symmetric if [itex]M=M^\top[/itex]. These are really nice for several reasons. First, they are diagonalisable by an orthogonal matrix. Since they are diagonalisable, there is an eigenbasis and so you can do a "spectral resolution." Also, the eigenvalues are all real, even if the entries are complex.

    This is just *very light* scratching the surface, but there are MANY important topics that involve transposes of matrices and vectors.
     
  13. Dec 5, 2012 #12
    Can you please explain what you mean by above projection example or did you switch [itex]v[/itex] and [itex]w[/itex]? I don't see any use of [itex]w[/itex] in your logic.

    Thanks
     
  14. Dec 5, 2012 #13

    MarneMath

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    It's two different examples. The first is an alternative definition for the dot product (that I believe only applies to column vectors, someone correct me if I'm wrong), the second is an orthogonal projection onto a line. To verify this, take two unit vectors and perform the operation described and draw them!
     
  15. Dec 6, 2012 #14
    The projection of [itex]w[/itex] onto [itex]v[/itex] is given by [itex](v\circ w)v[/itex] which is equal to [itex](v^\top w)v[/itex] which is equal to [itex](vv^\top)w[/itex]. That is, [itex]vv^\top[/itex] is the projection matrix. Note that I have assumed that [itex]v[/itex] has unit length.
     
  16. Dec 6, 2012 #15

    HallsofIvy

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    What kind of "introductory class" are you talking about? Every introductory linear algebra class I heard of defines vectors quite clearly. And if you mean physics classes where they just give vectors as "rows and columns", it still is meaningless to talk about "angles" between them. More rigorously the "transpose" of a vector, v, is its "dual" which is not even in the same vector space as v.
     
  17. Dec 6, 2012 #16
    Meaningless is a strong word that probably does not help the OP much. To most non-mathematicians who use vectors, the the dual space is implicitly identified with the original space via the dot product. "Vectors" are written either as rows or columns depending on convenience. Why do so many people want to "ban" people from using reasonable notation just because they aren't versed with abstract mathematics? It's like being against differentials, or implicit functions, or Leibniz notation.
     
  18. Dec 6, 2012 #17

    HallsofIvy

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    Using that interpretation, you are saying that a "vector" and its "transpose" are just different ways of writing the same vector so the "angle between them" is necessarily 0.

    (Of course "identifying the dual space with the original space via the dot product" depends on the choice of basis so still doesn't have any meaning as a property of the vector space itself.

    Oh, and "differentials", "implicit functions", and "Leibniz notation" all have very specific, rigorous, mathematical definitions.)
     
  19. Dec 6, 2012 #18

    micromass

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    If you are solving a problem and if you eventually have to find the angle between a column and a row vector, then this is a strong indication that you have done something wrong. Using rigorous notation is useful because it prevents people from making mistakes.

    Also, notation that is "reasonable" for one person might not be understandable for another person. We should teach standard notation that can be understood by everybody and not just by the person who invented it.
     
  20. Dec 6, 2012 #19
    I just think that Dead Boss was right when he said that vectors are commonly represented using either rows or columns, even within the same problem. This can be fully justified via the dot product. And if you are working with matrices, you have the "standard basis" to work with, so in context there is rarely any ambiguity. To me this seems like sufficiently common usage to be considered "reasonable notation". I mean, if you want to correct your students in class, that's one thing. But to decry it as meaningless without seeing it in context is different.

    My point about Leibniz notation, differentials, etc. was not that they are ill-defined (though thanks HoI for clearing that up), but that they get unfairly picked on because some profs think their students won't understand their potential ambiguities. I think we can all invent circumstances in which out of context Leibniz notation could be misinterpreted. Regarding differentials, how often is it said that you can't multiply both sides of [itex] dy/dt = y ' [/itex] by dt? And yet, dy = y'(t) dt is perfectly valid despite the fact that it does not, strictly speaking, come from multiplication.

    I'm not saying it is wrong to point out the basic problem with the OPs original question. Rather, that we could at least acknowledge that the question makes sense if we accept something that is more or less conventional in many contexts outside the strict confines of geometry/algebra.
     
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