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Homework Help: Projector matrices

  1. May 4, 2010 #1
    Consider the matrix [tex]A = u v^{\ast }[/tex] where [tex]u, v \in \textbf{C}^{n}[/tex]. Under what condition on u and v is A a projector?


    A is a projector if [tex]A^{2}=A [/tex], so we have [tex]u v^{*} u v^{*}= u v^{\ast }[/tex].

    Does this imply [tex] u v^{\ast } = I[/tex] ? And what exactly are the conditions on u and v that they are asking?

    do we have that [tex]u_{i} v^{\ast }_{i}=1[/tex] and [tex]u_{i} v^{\ast }_{j}=0[/tex] for [tex] i\neq j[/tex] ?
     
  2. jcsd
  3. May 4, 2010 #2

    Mark44

    Staff: Mentor

    Can you clarify what uv* means and how this could be a matrix?
     
  4. May 4, 2010 #3

    jbunniii

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    Note that [itex]v^{*}u[/itex] is a scalar, so you rearrange the terms of the product:

    [tex]u v^* u v^* = u (v^* u) v^* = (v^* u) (u v^*)[/tex]
     
  5. May 4, 2010 #4
    To Mark44, u is an nx1 column vector, v* is the conjugate transpose of v, where v is an nx1 column vector. So uv* is an nxn square matrix.
     
  6. May 4, 2010 #5

    Mark44

    Staff: Mentor

    OK, that makes more sense.

    A2 = A <==> A(A - I) = 0. What are the conditions for the last equation? Clearly, one possibility is that A = I (which is to say that uv* = I). But there are other possibilities (plural).
     
  7. May 4, 2010 #6
    The only other possibility that I see is that uv* is the 0 matrix. In this case, [tex]u_{i}v^{*}_{j} = 0[/tex] for all i and all j.
     
  8. May 4, 2010 #7

    Mark44

    Staff: Mentor

    There's another possibility. AB = 0 does not necessarily imply that either A = 0 or B = 0. For example,
    [tex]\left[\begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array}\right]\left[\begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array}\right] = \left[\begin{array}{c c} 0 & 0 \\ 0 & 0 \end{array}\right][/tex]
     
  9. May 4, 2010 #8
    Oh true! But I am lost here, I am not sure what would be the conditions on u and v then.
     
  10. May 4, 2010 #9

    Mark44

    Staff: Mentor

    If AB = 0 (meaning the 0 matrix), then |AB| = 0 (the number).
     
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