0.1 Basic Content

$$ \begin{align} (\boldsymbol{AB})^{-1} &= \boldsymbol{B}^{-1}\boldsymbol{A}^{-1} \cr (\boldsymbol{ABC\cdots})^{-1} &= \cdots\boldsymbol{C}^{-1}\boldsymbol{B}^{-1}\boldsymbol{A}^{-1} \cr (\boldsymbol{A}^\top)^{-1} &= (A^{-1})^\top \cr (\boldsymbol{A} + \boldsymbol{B})^\top &= \boldsymbol{A}^\top + \boldsymbol{B}^\top \cr (\boldsymbol{AB})^\top &= \boldsymbol{B}^\top\boldsymbol{A}^\top \cr (\boldsymbol{ABC\cdots})^\top &= \cdots\boldsymbol{C}^\top\boldsymbol{B}^\top\boldsymbol{A}^\top \cr (\boldsymbol{A}^{H})^{-1} &= (\boldsymbol{A}^{-1})^{H} \cr (\boldsymbol{A} + \boldsymbol{B})^H &= \boldsymbol{A}^H + \boldsymbol{B}^H \cr (\boldsymbol{AB})^H &= \boldsymbol{B}^H\boldsymbol{A}^H \cr (\boldsymbol{ABC\cdots})^H &= \cdots\boldsymbol{C}^H\boldsymbol{B}^H\boldsymbol{A}^H \end{align} $$

0.2 Trace

$$ \begin{align} \text{TR}(\boldsymbol{A}) &= \sum_{i}A_{ii} \cr \text{TR}(\boldsymbol{A}) &= \sum_{i}\lambda_{i}, \quad \lambda_{i}=\text{eig}(\boldsymbol{A})_{i} \cr \text{TR}(\boldsymbol{A}) &= \text{TR}(\boldsymbol{A}^\top) \cr \text{TR}(\boldsymbol{AB}) &= \text{TR}(\boldsymbol{BA}) \cr \text{TR}(\boldsymbol{A+B}) &= \text{TR}(\boldsymbol{A}) + \text{TR}(\boldsymbol{B}) \cr \text{TR}(\boldsymbol{ABC}) &= \text{TR}(\boldsymbol{BCA}) = \text{TR}(\boldsymbol{CAB}) \cr \boldsymbol{a}^\top\boldsymbol{a} &= \text{Tr}(\boldsymbol{aa}^\top) \end{align} $$

0.3 Determinant

Let $\boldsymbol{A}$ is a $n\times n$ matrix. $$ \begin{align} \det(\boldsymbol{A}) &= \prod_i \lambda_i \quad \lambda_i=\text{eig}(\boldsymbol{A})_i \cr \det(c\boldsymbol{A}) &= c^n \det(\boldsymbol{A})\quad \text{if }\boldsymbol{A}\in\mathbb{R}^{n\times n} \cr \det(\boldsymbol{A}^\top) &= \det(\boldsymbol{A}) \cr \det(\boldsymbol{AB}) &= \det(\boldsymbol{A})\det(\boldsymbol{B}) \cr \det(\boldsymbol{A}^{-1}) &= 1/\det(\boldsymbol{A}) \cr \det(\boldsymbol{A}^n) &= \det(\boldsymbol{A})^n \cr \det(\boldsymbol{I}+\boldsymbol{uv}^\top) &= 1 + \boldsymbol{u}^\top\boldsymbol{v} \end{align} $$

For $n=2$: $$ \begin{equation} \det(\boldsymbol{I}+\boldsymbol{A}) = 1 + \det(\boldsymbol{A}) + \text{Tr}(\boldsymbol{A}) \end{equation} $$

For $n=3$: $$ \begin{equation} \det(\boldsymbol{I}+\boldsymbol{A}) = 1 + \det(\boldsymbol{A}) + \text{Tr}(\boldsymbol{A}) + \frac{1}{2}\text{Tr}(\boldsymbol{A})^2 - \frac{1}{2}\text{Tr}(\boldsymbol{A}^2) \end{equation} $$

For $n=4$: $$ \begin{align} \det(\boldsymbol{I}+\boldsymbol{A}) = &1 + \det(\boldsymbol{A}) + \text{Tr}(\boldsymbol{A}) + \frac{1}{2} \notag\cr &+ \text{Tr}(\boldsymbol{A})^2 - \text{Tr}(\boldsymbol{A}^2) \notag\cr &+ \frac{1}{6}\text{Tr}(\boldsymbol{A})^3 - \frac{1}{2}\text{Tr}(\boldsymbol{A})\text{Tr}(\boldsymbol{A}^2) + \frac{1}{3}\text{Tr}(\boldsymbol{A}^3) \end{align} $$

For small $\epsilon$, the following approximation holds: $$ \begin{equation} \det(\boldsymbol{I}+\boldsymbol{A}) = 1 + \det(\boldsymbol{A}) + \epsilon\text{Tr}(\boldsymbol{A}) + \frac{1}{2}\epsilon^2\text{Tr}(\boldsymbol{A})^2-\frac{1}{2}\epsilon^2\text{Tr}(\boldsymbol{A}^2) \end{equation} $$

0.4 The Special Case $2\times 2$

Consider the matrix $\boldsymbol{A}$: $$ A = \begin{bmatrix} A_{11} & A_{12} \cr A_{21} & A_{22} \end{bmatrix} $$

Determinant and trace: $$ \begin{align} \det(\boldsymbol{A}) &= A_{11}A_{22} - A_{12}A_{21} \cr \text{Tr}(\boldsymbol{A}) &= A_{11} + A_{22} \end{align} $$

Eigenvalues: $$ \lambda^2 - \lambda\cdot\text{Tr}(\boldsymbol{A})+\det(\boldsymbol{A})=0 $$ $$ \lambda_1=\frac{\text{Tr}(\boldsymbol{A}) + \sqrt{\text{Tr}(\boldsymbol{A})^2 - 4\det(\boldsymbol{A})}}{2},\quad \lambda_1=\frac{\text{Tr}(\boldsymbol{A}) - \sqrt{\text{Tr}(\boldsymbol{A})^2 - 4\det(\boldsymbol{A})}}{2} $$ $$ \lambda_1+\lambda_2=\text{Tr}(\boldsymbol{A}) \quad \lambda_1\lambda_2=\det(\boldsymbol{A}) $$

Eigenvectors: $$ \boldsymbol{v}_1\propto [A_{12},\lambda_1-A_{11}]^\top \qquad \boldsymbol{v}_1\propto [A_{12},\lambda_2-A_{11}]^\top $$

Inverse: $$ \begin{equation} \boldsymbol{A}^{-1} = \frac{1}{\det(\boldsymbol{A})}\begin{bmatrix} A_{22} & -A_{12} \cr -A_{21} & A_{11} \end{bmatrix} \end{equation} $$