# Fun with determinants

Especially Jacobian determinants

April 6, 2011 — October 12, 2021

algebra
functional analysis
Gaussian
Hilbert space
linear algebra

Petersen and Pedersen (2012) note the standard identities:

Let $$\mathbf{A}$$ be an $$n \times n$$ matrix. \begin{aligned} \operatorname{det}(\mathbf{A}) &=\prod_{i} \lambda_{i} \quad \lambda_{i}=\operatorname{eig}(\mathbf{A}) \\ \operatorname{det}(c \mathbf{A}) &=c^{n} \operatorname{det}(\mathbf{A}), \quad \text { if } \mathbf{A} \in \mathbb{R}^{n \times n} \\ \operatorname{det}\left(\mathbf{A}^{T}\right) &=\operatorname{det}(\mathbf{A}) \\ \operatorname{det}(\mathbf{A} \mathbf{B}) &=\operatorname{det}(\mathbf{A}) \operatorname{det}(\mathbf{B}) \\ \operatorname{det}\left(\mathbf{A}^{-1}\right) &=1 / \operatorname{det}(\mathbf{A}) \\ \operatorname{det}\left(\mathbf{A}^{n}\right) &=\operatorname{det}(\mathbf{A})^{n} \\ \operatorname{det}\left(\mathbf{I}+\mathbf{u v}^{T}\right) &=1+\mathbf{u}^{T} \mathbf{v} \end{aligned} For $$n=2$$ : $\operatorname{det}(\mathbf{I}+\mathbf{A})=1+\operatorname{det}(\mathbf{A})+\operatorname{Tr}(\mathbf{A})$ For $$n=3$$ : $\operatorname{det}(\mathbf{I}+\mathbf{A})=1+\operatorname{det}(\mathbf{A})+\operatorname{Tr}(\mathbf{A})+\frac{1}{2} \operatorname{Tr}(\mathbf{A})^{2}-\frac{1}{2} \operatorname{Tr}\left(\mathbf{A}^{2}\right)$ For $$n=4$$ : \begin{aligned} \operatorname{det}(\mathbf{I}+\mathbf{A})=& 1+\operatorname{det}(\mathbf{A})+\operatorname{Tr}(\mathbf{A})+\frac{1}{2} \\ &+\operatorname{Tr}(\mathbf{A})^{2}-\frac{1}{2} \operatorname{Tr}\left(\mathbf{A}^{2}\right) \\ &+\frac{1}{6} \operatorname{Tr}(\mathbf{A})^{3}-\frac{1}{2} \operatorname{Tr}(\mathbf{A}) \operatorname{Tr}\left(\mathbf{A}^{2}\right)+\frac{1}{3} \operatorname{Tr}\left(\mathbf{A}^{3}\right) \end{aligned} For small $$\varepsilon$$, the following approximation holds $\operatorname{det}(\mathbf{I}+\varepsilon \mathbf{A}) \cong 1+\operatorname{det}(\mathbf{A})+\varepsilon \operatorname{Tr}(\mathbf{A})+\frac{1}{2} \varepsilon^{2} \operatorname{Tr}(\mathbf{A})^{2}-\frac{1}{2} \varepsilon^{2} \operatorname{Tr}\left(\mathbf{A}^{2}\right)$

For a block matrix we have For $$n=4$$ : \begin{aligned} \operatorname{det}(\mathbf{I}+\mathbf{A})=& 1+\operatorname{det}(\mathbf{A})+\operatorname{Tr}(\mathbf{A})+\frac{1}{2} \\ &+\operatorname{Tr}(\mathbf{A})^{2}-\frac{1}{2} \operatorname{Tr}\left(\mathbf{A}^{2}\right) \\ &+\frac{1}{6} \operatorname{Tr}(\mathbf{A})^{3}-\frac{1}{2} \operatorname{Tr}(\mathbf{A}) \operatorname{Tr}\left(\mathbf{A}^{2}\right)+\frac{1}{3} \operatorname{Tr}\left(\mathbf{A}^{3}\right) \end{aligned} For small $$\varepsilon$$, the following approximation holds $\operatorname{det}(\mathbf{I}+\varepsilon \mathbf{A}) \cong 1+\operatorname{det}(\mathbf{A})+\varepsilon \operatorname{Tr}(\mathbf{A})+\frac{1}{2} \varepsilon^{2} \operatorname{Tr}(\mathbf{A})^{2}-\frac{1}{2} \varepsilon^{2} \operatorname{Tr}\left(\mathbf{A}^{2}\right)$

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