Determinant relations¶

Determinant of a 2 by 2 block matrix¶

The determinant of a \(2\times2\) block matrix can be expressed in terms of its sub-blocks. This result is particularly useful when one of the blocks is invertible.

Consider the matrix

\[\begin{equation} \begin{aligned} Q= \begin{pmatrix} Q_{11} & Q_{12}\\[2mm] Q_{21} & Q_{22} \end{pmatrix} \end{aligned} \end{equation}\]

Under suitable conditions (in particular, assuming the required inverses exist), its determinant can be written as

\[\begin{equation} \label{appendix:det_of_blockmatrix} \det Q = \det(Q_{11}-Q_{12}Q_{22}^{-1}Q_{21})\det Q_{11} \end{equation}\]

This expression allows the determinant of a block matrix to be reduced to determinants of smaller matrices (see [1]).

Determinant identity for scaled matrices¶

A useful identity for determinants of \(2\times2\) matrices is derived in the following.

Consider the expression

\[\begin{equation} \det(\mathbb{1}-\lambda^2 X), \end{equation}\]

where \(X\) is an arbitrary \(2\times2\) matrix. The goal is to express this determinant in terms of invariants of \(X\).

Let

\[\begin{equation} \begin{aligned} X= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \quad (X\in\mathbb{C}^{2\times2}). \end{aligned} \end{equation}\]

Then the matrix \(\mathbb{1}-\lambda^2 X\) takes the form

\[\begin{equation} \begin{aligned} \mathbb{1}-\lambda^2X = \begin{pmatrix} 1-\lambda^2 a & -\lambda^2 b\\ -\lambda^2 c & 1-\lambda^2 d \end{pmatrix}. \end{aligned} \end{equation}\]

The determinant of a general \(2\times2\) matrix is given by

\[\begin{equation} \begin{aligned} \det \begin{pmatrix} p&q\\ r&s \end{pmatrix} =ps-qr, \end{aligned} \end{equation}\]

which can now be applied directly:

\[\begin{equation} \begin{aligned} \det(\mathbb{1}-\lambda^2X) &= (1-\lambda^2 a)(1-\lambda^2 d)-(-\lambda^2 b)(-\lambda^2 c)\\ &= (1-\lambda^2 a)(1-\lambda^2 d)-\lambda^4bc\\ &= 1-\lambda^2(a+d)+\lambda^4(ad-bc). \end{aligned} \end{equation}\]

The result can be expressed more compactly by recalling the definitions

\[\begin{equation} \begin{aligned} \mathrm{Tr}(X)=a+d, \qquad \det(X)=ad-bc. \end{aligned} \end{equation}\]

Substituting these into the previous expression yields

\[\begin{equation} \label{appendix:det_1minusl2X} \begin{aligned} \det(\mathbb{1}-\lambda^2X) = 1-\lambda^2\mathrm{Tr}(X)+\lambda^4\det(X) \end{aligned} \end{equation}\]

This shows that, for any \(2\times2\) matrix \(X\), the determinant of \(\mathbb{1}-\lambda^2X\) depends only on the trace and determinant of \(X\).

References¶

[1] P. D. Powell, Calculating Determinants of Block Matrices, 2011. Available: https://arxiv.org/abs/1112.4379