2
$\begingroup$

We have the following determinant property

$$\det \begin{bmatrix} U & O \\ V & W \end{bmatrix} = \det(U) \cdot \det(W)$$

where $U \in R^{n\times n}$, $V \in R^{m\times n}$, $W \in R^{m\times m}$ and $O \in R^{n\times m}$ (the zero matrix).

Now suppose the zero block appears in the top left corner instead. Does there in that case also exist a rule to calculate the determinant of the matrix more easily?

The matrices I am thinking of here are of the form

$$Z = \begin{bmatrix} O & A \\ A^T & B \end{bmatrix}$$

with all matrices conformable. An example would be

$$Z = \begin{bmatrix} 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & -9 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & -1 \\ 1 & -9 & 0 & -1 & 2 & 0 \\ 1 & 0 & 0 & 2 & 1 & 0 \\ 1 & 1 & -1 & 0 & 0 & 1 \end{bmatrix}$$

$\endgroup$
  • $\begingroup$ @daw Doesn't it have a 2x2 zero block? $\endgroup$ – Rodrigo de Azevedo Feb 23 '17 at 9:50
1
$\begingroup$

There is no such rule to calculate determinant easy as in the case the zero block is in the top right or bottom left corner. Here you can see all rules you can apply on block matrices https://en.wikipedia.org/wiki/Determinant. Instead, you can transform your matrix with Gaussian transformations to an upper triangular matrix and just multiply elements on diagonal.

$\endgroup$
  • $\begingroup$ Ah that's a pity. Thanks for the hint. $\endgroup$ – Anna Feb 20 '17 at 15:22
0
$\begingroup$

Let $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm B \in \mathbb R^{n \times n}$. Assuming that $\mathrm B$ is invertible, we use the Schur complement

$$\begin{bmatrix} \mathrm O_m & \mathrm A\\ \mathrm A^{\top} & \mathrm B \end{bmatrix} \begin{bmatrix} \mathrm I_m & \mathrm O_{m \times n}\\ -\mathrm B^{-1} \mathrm A^{\top} & \mathrm I_n\end{bmatrix} = \begin{bmatrix} -\mathrm A \mathrm B^{-1} \mathrm A^{\top} & \mathrm A\\ \mathrm O_{n \times m} & \mathrm B\end{bmatrix}$$

and, taking the determinant, we obtain

$$\det \begin{bmatrix} \mathrm O_m & \mathrm A\\ \mathrm A^{\top} & \mathrm B \end{bmatrix} \cdot \underbrace{\det \begin{bmatrix} \mathrm I_m & \mathrm O_{m \times n}\\ -\mathrm B^{-1} \mathrm A^{\top} & \mathrm I_n\end{bmatrix}}_{=1} = \det \begin{bmatrix} -\mathrm A \mathrm B^{-1} \mathrm A^{\top} & \mathrm A\\ \mathrm O_{n \times m} & \mathrm B\end{bmatrix} = \color{blue}{(-1)^m \det (\mathrm A \mathrm B^{-1} \mathrm A^{\top}) \, \det(\mathrm B)}$$

$\endgroup$
0
$\begingroup$

Let us assume that all matrices are square and $n \times n$. Permuting the columns,

$$\det \begin{bmatrix} \mathrm O_n & \mathrm A\\ \mathrm A^{\top} & \mathrm B \end{bmatrix} = \det \begin{bmatrix} \mathrm A & \mathrm O_n\\ \mathrm B & \mathrm A^{\top}\end{bmatrix} \cdot \underbrace{\det \begin{bmatrix} \mathrm O_n & \mathrm I_n\\ \mathrm I_n & \mathrm O_n\end{bmatrix}}_{= \pm 1} = \pm \left( \det (\mathrm A) \right)^2$$

$\endgroup$
  • $\begingroup$ But you cannot calculate the determinant of the nonsquare matrix A right? $\endgroup$ – Anna Feb 20 '17 at 16:05
  • $\begingroup$ Is there any way to determine the determinant of the matrix $Z$ above more easily? $\endgroup$ – Anna Feb 20 '17 at 16:06
  • $\begingroup$ This idea does not seem to work for $Z$. The determinant of $AA^T$ is 313 or -313 while the determinant of $Z$ should be 24. $\endgroup$ – Anna Feb 20 '17 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.