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Is there a way to find the determinant of $X$ in terms of its sub-matrices $A,B,C_0$ and $R_0$?

$$X = \begin{bmatrix} AC_0 & -I_n\\ 0_{(n-1)} & R_0B \end{bmatrix} \in \mathbb{R}^{(2n-1) \times (2n-1)} $$ where $A,B \in \mathbb{R}^{n \times n}$, $C_0 = \begin {bmatrix}0_{1\times (n-1)}\\ I_{(n-1)}\end{bmatrix}$ and $R_0 = \begin {bmatrix}0_{(n-1)\times 1}& I_{(n-1)}\end{bmatrix}$.

The sources I found use Laplace determinant expansion for block matrices for upper traingular matrices like the one above but the diagonal matrices are square and the determinant is the product of the diagonal block determinant.

MATLAB suggests $\det(X) = -\det(R_0BAC_0)$ which is intuitive, but I am not able to find a way to actually prove it.

The more generlized question will be to find the determinant of

$$\begin{vmatrix} A_{m \times n} & B_{m \times m}\\ 0_{n \times n} & C_{n \times m} \end{vmatrix}$$ in terms of $A, B $ and $C$.

Any help will be much appreciated.

Cheers.

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  • $\begingroup$ Do you get my answer? $\endgroup$ – user21820 Aug 1 '16 at 1:39
  • $\begingroup$ Hi, yes I did.. thanks a lot for the help.. I upvoted but apparently it won't be recorded since I am a new user.. will keep the thread updated if I am able to find the answer for the generalized question. $\endgroup$ – user2987008 Aug 2 '16 at 2:41
  • $\begingroup$ Yes you need some reputation before you can upvote/downvote. You can either accept my answer or wait for someone to give you the generalized formula. I doubt it exists because there are too many possibilities. Note that the determinant is easy to compute from $A,C$ alone if $m \le n$, but when $m > n$ the lower-left quarter has nonzero entries and the determinant gets entangled with $B$. $\endgroup$ – user21820 Aug 2 '16 at 2:58
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The simplest way is to use the (Leibniz) definition of determinant, where we need to consider all products of entries that have exactly one in each row and one in each column. Note that any such product that includes an entry in the lower-left part will be zero. But it has exactly $n$ entries in the upper part and exactly $n$ entries in the right part, and hence has exactly $1$ entry in the upper-right part. Hence every product that contributes to $\def\matrix#1{\left[\begin{array}{c}#1\end{array}\right]}$$\det(\matrix{A&-I_n\\0&C})$ (where $A$ is an $n \times (n-1)$ matrix and $C$ is an $(n-1) \times n$ matrix) is the product of a pair of products that contribute to $\det(A)$ and $\det(C)$ respectively multiplied by an entry in $-I_n$. Of course, the only non-zero entries in $-I_n$ are all $-1$ on the diagonal. Note that if the $-1$ is the $k$-th in $-I_n$, then the product comes from $A_k$ and $C_k$ where $A_k$ is $A$ without the $k$-th row and $C_k$ is $C$ without the $k$-th column. Note that the sign of the overall permutation relative to the product of the signs of the corresponding permutations for $A_k,C_k$ is the same regardless of $k$, and when $k=1$ we can easily compute it to be $(-1)^{n-1}$. Therefore we get:

$\det(\matrix{A&-I_n\\0&C}) = \sum_{k=1}^n (-1)^n \det(A_k)\det(C_k) = (-1)^n \det(CA)$ [by Cauchy-Binet].

I don't know what we can say for general $B$ in place of $-I_n$.

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  • $\begingroup$ This formula does not work for $$\begin {bmatrix}1&-1&0\\1&0&-1\\0&1&1\end {bmatrix}, $$ which has determinant $2$ (and not $-2$). $\endgroup$ – Martin Argerami Jul 30 '16 at 10:17
  • $\begingroup$ @MartinArgerami: I've fixed the problem. =) $\endgroup$ – user21820 Jul 30 '16 at 10:31

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