Matrix Theory
Given, $A$ is a square matrix.
Determinant (in-general)
- Laplace expansion formula: $\text{Det}(A) = \sum_{k=1}^{n} (-1)^{k+i} a_{ik} \text{Det}(A(i,k))$
- Laplace formula: $A_{i_{1}j_{1}}A_{i_{2}j_{2}}\cdots A_{i_{n}j_{n}}\epsilon_{j_{1}\cdots j_{n}} = \text{Det}(A)\epsilon_{i_{1}\cdots i_{n}}$
- $\text{Det}(1 + A) = 1 + \text{Tr}(A)$ given matrix $A_{ij}$ is very small
- $\text{Det}(A) = e^{\text{Tr}(\ln(A))}$
- Generalized Kronecker delta:
$\delta^{ij}_{kl} := \text{Det}\begin{pmatrix}\delta^{i}_{k} & \delta^{i}_{l} \\ \delta^{j}_{k} & \delta^{j}_{l}\end{pmatrix} = \delta^{i}_{k}\delta^{j}_{l} - \delta^{i}_{l}\delta^{j}_{k} \in \{-1, 0, 1\}$
Matrix Derivative
- $\frac{d A^{-1}}{dx} = - A^{-1} \frac{dA}{dx} A^{-1}$
- Jacobi Formula:
$\frac{d \text{Det}(A(t))}{dt} = \text{Tr}(\text{adj}(A(t)) \frac{d A(t)}{dt}) = \text{Det}(A(t)) \text{Tr}(A(t)^{-1} \frac{d A(t)}{dt})$
Matrix exponential
- $e^A = \sum_{k=0}^{\infty} \frac{1}{k!} A^k$
- $e^A = \lim_{m \to \infty} \left(1 + \frac{1}{m} A \right)^m$
- $\text{Det}(e^{A})= e^{\text{Tr}(A)}$
Matrix logarithm
- $A^{x} = e^{x \ln(A)}$ where $x \in \mathbb{R}$
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Published on May 7, 2021
Last revised on Jul 2, 2023