Determinant identities

In mathematics the determinant is an operator which has certain useful identities.

Identities

$\det(I_{n})=1$ where I_n is the n × n identity matrix.

$\det(A^{\rm {T}})=\det(A).$

$\det(A^{-1})={\frac {1}{\det(A)}}=\det(A)^{-1}.$

For square matrices A and B of equal size,

$\det(AB)=\det(A)\det(B).$

$\det(cA)=c^{n}\det(A)$ for an n × n matrix.

If A is a triangular matrix, i.e. a_i,j = 0 whenever i > j or, alternatively, whenever i < j, then its determinant equals the product of the diagonal entries: $\det(A)=a_{1,1}a_{2,2}\cdots a_{n,n}=\prod _{i=1}^{n}a_{i,i}.$

Schur complement

The following identity holds for a Schur complement of a square matrix:

The Schur complement arises as the result of performing a block Gaussian elimination by multiplying the matrix M from the right with the "block lower triangular" matrix

$L=\left[{\begin{matrix}I_{p}&0\\-D^{{-1}}C&I_{q}\end{matrix}}\right].$

Here I_p denotes a p×p identity matrix. After multiplication with the matrix L the Schur complement appears in the upper p×p block. The product matrix is

${\begin{aligned}ML&=\left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]\left[{\begin{matrix}I_{p}&0\\-D^{{-1}}C&I_{q}\end{matrix}}\right]=\left[{\begin{matrix}A-BD^{{-1}}C&B\\0&D\end{matrix}}\right]\\&=\left[{\begin{matrix}I_{p}&BD^{{-1}}\\0&I_{q}\end{matrix}}\right]\left[{\begin{matrix}A-BD^{{-1}}C&0\\0&D\end{matrix}}\right].\end{aligned}}$

That is, we have shown that

${\begin{aligned}\left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]&=\left[{\begin{matrix}I_{p}&BD^{{-1}}\\0&I_{q}\end{matrix}}\right]\left[{\begin{matrix}A-BD^{{-1}}C&0\\0&D\end{matrix}}\right]\left[{\begin{matrix}I_{p}&0\\D^{{-1}}C&I_{q}\end{matrix}}\right],\end{aligned}}$