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Determinant of outer product matrices

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Given [itex]u, v \in \mathbb{R}^{n}[/itex], and [itex]A \in \mathbb{R}^{n \times n}[/itex], [itex]\mathrm{det}\left(A\right) \neq 0[/itex], find [itex]\mathrm{det}\left( A + uv^{T} \right)[/itex]


    2. Relevant equations
    Generic determinant and eigenvalue equations, I suppose.


    3. The attempt at a solution
    Hoping to gain some insight, I found the determinant for the special case [itex] A = I [/itex]. In that case we can see that the eigenvalues of [itex]I + uv^T[/itex] satisfy
    [tex] \mathrm{det}\left(I + uv^T + \lambda I\right) = \mathrm{det}\left( uv^T + \left( \lambda + 1\right) I\right) = 0 [/tex]
    so we see that the eigenvalues of [itex] uv^T + I [/itex] are those of [itex] uv^T [/itex] plus one.

    Outer product matrices have rank 1, so [itex] uv^T [/itex] has only one nonzero eigenvalue (which is pretty easy to see is equal to [itex] \mathrm{tr}\left( uv^T \right) = u^{T}v[/itex]).

    The determinant of [itex] uv^T + I [/itex] is equal to the product of its eigenvalues. Since it only has one eigenvalue not equal to 1, the determinant will equal that eigenvalue, or [itex] \mathrm{tr}\left(uv^T\right) + 1 = 1 + u^Tv [/itex]

    I tried similar methods replacing [itex]I[/itex] with the more generic [itex]A[/itex] without any luck. I also tried starting from the equation
    [tex] A\vec{\eta} = \lambda\vec{\eta} [/tex]
    again without any luck.

    I got to the equation
    [tex] \mathrm{det}\left(A + uv^T\right) = \mathrm{det}(A) \mathrm{det}\left( I + A^{-1} u v^T \right) [/tex]
    but I don't think it's very useful.
     
  2. jcsd
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