Cauchy expansion of determinant of a bordered matrix

  • Thread starter ekkilop
  • Start date
  • #1
29
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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.
 

Answers and Replies

  • #2
611
24
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.
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:
 
  • #3
29
0
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! =)
 

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