1
$\begingroup$

I have a matrix of the form

$B = \left[\begin{array}{cccc} A_1 & C^T\\ C & A_2\\ \end{array} \right]$ Where $A_1,A_2$ and $C$ are all square, and $A_1,A_2$ are symmetric positive definite. Therefore the whole matrix is symmetric. I'm wonder if there is a formula for the determinant of this kind of square symmatric block matrix in terms of the determinants of the individual matrices?

Note, this is not a duplicate of this because in that case, the matrix is not symmetrix, and $A_1, A_2$ are not necessarily invertible. In our case, $A_1, A_2$ are positive definite, and therefore invertible. However, similar to that question, the matrix C has determinant 0.

Any help is appreciated!

$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

Since $A_1$ is invertible, you get $\det(B) = \det(A_1)\det( A_2-CA_1^{-1}C^T)$. This is in Block matrix

$\endgroup$
5
  • $\begingroup$ Yes, I see this from wikipedia. I was hoping that there could be some further simplification. One thing I should note is that the matrix C not only has determinante 0, but is expected to have very small matrix norm. Can I argue in that case that $det(B) \approx det(A_1)det(A_2)$? $\endgroup$
    – Paul
    Aug 11 at 20:15
  • $\begingroup$ @Paul This completely changes the question. As far as exact formulas go, $\det(B) = \det(A_1) \det(A_2 - CA_1^{-1} C^T)$ can't be simplified further, given what we know about $B$. But if you are only looking for an approximation $\det(B) \approx \det(A_1) \det(A_2)$, then there might be more we can say. In this case, you need to provide more information, e.g., a quantitative bound on the norm of $C$, the form of approximation that you are looking for, additional structure on $B$. If you like, we could discuss these things in a chat room. $\endgroup$
    – nowhere
    Aug 11 at 22:49
  • $\begingroup$ sure, that would be great. how do we do that? $\endgroup$
    – Paul
    Aug 12 at 1:52
  • $\begingroup$ @Paul Can you join this room (link below)? Any time today is fine. If you prefer a different time, just type your availabilities in the chat. chat.stackexchange.com/rooms/13473/linear-abstract-algebra $\endgroup$
    – nowhere
    Aug 12 at 20:16
  • $\begingroup$ @nowhere FYI I posted a more detailed version of the question on math overflow here: mathoverflow.net/questions/452769/… $\endgroup$
    – Paul
    Aug 15 at 14:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .