Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy expansion of determinant of a bordered matrix

  1. Jul 10, 2013 #1
    The Cauchy expansion says that

    [itex] \text{det} \begin{bmatrix}
    A & x \\[0.3em]
    y^T & a
    \end{bmatrix}
    = a \text{det}(A) - y^T \text{adj}(A) x [/itex],

    where A is an n-1 by n-1 matrix, y and x are vectors with n-1 elements, and a is a scalar.
    There is a proof in Matrix Analysis by Horn and Johnson that seems to be based on that A is a principal submatrix. My question is whether some similar result holds if A is not a principal submatrix? Say that we look for

    det[itex] \begin{bmatrix}
    y^T & a \\[0.3em]
    A & x
    \end{bmatrix}
    [/itex].

    Would a similar expression hold?

    Thanks.
     
  2. jcsd
  3. Jul 10, 2013 #2
    Indeed. In fact, it would just be ##\vec{y}^T \operatorname{adj}\textbf{A} \vec{x} - a\operatorname{det}\textbf{A}##. Can you see why? :tongue:
     
  4. Jul 11, 2013 #3
    Hi!

    It just dawned on me that any such matrices (I suppose there are only 4 places A could go ^^, ) are related by simple permutations. Since any permutation matrix has determinant + or - 1 then what you say must be true.

    Thank you for the enlightenment! =)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cauchy expansion of determinant of a bordered matrix
  1. Determinant expansions (Replies: 1)

  2. Determinant of matrix (Replies: 4)

Loading...