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Projector math problem

  1. Oct 12, 2009 #1
    Consider the matrix A=uv* where u and v lie in C^n (C:complex numbers). Under what condition on u and v is A a projector?

    I know that a projector is a square matrix P for which P^2= P.
    Now according to this definition, would that make uv*=Identity. and then u be the inverse of v*?

    Also what is the inverse of a vector that belongs to C^n? I mean what is u^-1? Is u^-1 a vector such that uu^-1 = the (nxn) identity matrix?
    In what set would such a vector (u^-1) lie in?
     
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    Re: Projector

    do you mean an outer product of the column vectors u,v given by
    A = uv*
    where v* is the complex conjugate transposed?

    I haven't done a whole heap with projectors, but here's some ideas based on general linear algebra

    does P2= P, imply P = I?
    if the inverse exists
    P-1.(P.P) = I.P = P = P-1(P) = I
    what if the inverse does not exist?
    I would try writing out the case for C2, and you should be able to see a relation for u & v in terms of their inner product that is easily extendable to the Cn,
     
  4. Oct 13, 2009 #3

    HallsofIvy

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    Re: Projector

    No, the fact that (uv*)^2= uv* does NOT make uv* the identity operator. For example, the operator P((x,y,z))= (0, y, z) is a projection because P^2((x, y, z))= P((0, y, z)= (0, y, z)= P but P does not map (x,y,z) to itself.

    And, in any case, there is no such thing as an "inverse" vector. "Inverse" only applies to functions and operators, not vectors.

     
  5. Oct 13, 2009 #4

    lanedance

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    Re: Projector

    After reading Halls's comment on vectors, consider the case when P is a square matrix defined by the outer product, ie if u,v are column vectors of length n, then P is the nxn matrix given by:

    [tex] P = \textbf{u} \otimes \textbf{v} = \textbf{u} \overline{\textbf{v}}^T [/tex]

    in the case you give Halls (sorry for the poor matrix display)
    P =
    |100|
    |010|
    |000|
    which is clearly not invertible

    From the definition of a projector P^2=P. In the matrix format it is clear, that if P is invertible then P=I.

    So perhaps in general projectors are not invertible except in the trivial case above. This seems to make sense with the example of projecting Rn onto Rm, with m<n (and the 3-space onto a plane example given by Halls).

    In any case, if you perform the outer product defined above for an arbitrary u,v. Then look at what is required to satisfy P2=P, it leads to a simple requirement for u,v in terms of their innner product
     
  6. Oct 14, 2009 #5

    HallsofIvy

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    Re: Projector

    Yes, in general a projector "projects" a vector space onto a subspace of smaller dimension. The entire orthogonal subspace is the kernel of the projector and so the projector is not invertible.

    But my remarks were really adressed, not to a projector, but to math8's talking about the inverse of vectors- which is not defined.
     
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