--- title: Determinant (BR Chapter 10) subtitle: Biostat 216 author: Dr. Hua Zhou @ UCLA date: today format: html: theme: cosmo embed-resources: true number-sections: true toc: true toc-depth: 4 toc-location: left code-fold: false jupyter: jupytext: formats: 'ipynb,qmd' text_representation: extension: .qmd format_name: quarto format_version: '1.0' jupytext_version: 1.15.2 kernelspec: display_name: Julia 1.9.3 language: julia name: julia-1.9 --- ```{julia} using Pkg Pkg.activate(pwd()) Pkg.instantiate() Pkg.status() ``` ```{julia} using LinearAlgebra, Plots, Symbolics ``` We review some basic facts about matrix determinant. ## Definition of determinant - The **determinant** of a square matrix $\mathbf{A} \in \mathbb{R}^{n \times n}$ is $$ \det (\mathbf{A}) = \sum (-1)^{\phi(j_1,\ldots,j_n)} \prod_{i=1}^n a_{ij_i}, $$ where the summation is over all permutation $(j_1, \ldots, j_n)$ of the set of integers $(1,\ldots,n)$ and $\phi(j_1,\ldots,j_n)$ is the number of transpositions to change $(1,\ldots,n)$ to $(j_1,\ldots,j_n)$. $(-1)^{\phi(j_1,\ldots,j_n)}$ is also called the **sign of permutation**. - Examples: $n = 2$ and 3. $$ \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = (-1)^{\phi(1,2)} a_{11} a_{22} + (-1)^{\phi(2,1)} a_{12} a_{21} = a_{11} a_{22} - a_{12} a_{21}. $$ ```{julia} # n = 2 @variables A[1:2, 1:2] ``` ```{julia} det(A) |> Symbolics.scalarize ``` ```{julia} # n = 3 @variables A[1:3, 1:3] ``` ```{julia} det(A) |> Symbolics.scalarize |> expand ``` ```{julia} # n = 4 @variables A[1:4, 1:4] ``` ```{julia} det(A) |> Symbolics.scalarize |> expand ``` ## Some interpretations of determinant - Interpretation of the (absolute value of) determinant as the **volume of the parallelotope** defined by the columns of the matrix. For example, if $\mathbf{X} \in \mathbb{R}^2$ has two columns $\mathbf{x}_1$ and $\mathbf{x}_2$, then \begin{eqnarray*} \text{area} &=& bh = \|\mathbf{x}_1\|\|\mathbf{x}_2\| \sin(\theta) \\ &=& \|\mathbf{x}_1\| \|\mathbf{x}_2\| \sqrt{1 - \left( \frac{\langle \mathbf{x}_1, \mathbf{x}_2 \rangle}{\|\mathbf{x}_1\| \|\mathbf{x}_2\|} \right)^2} \\ &=& \sqrt{\|\mathbf{x}_1\|^2 \|\mathbf{x}\|^2 - (\langle \mathbf{x}_1, \mathbf{x}_2\rangle)^2} \\ &=& \sqrt{(x_{11}^2 + x_{12}^2)(x_{21}^2+x_{22}^2) - (x_{11}x_{21} + x_{12}x_{22})^2} \\ &=& |x_{11} x_{22} - x_{12} x_{21}| \\ &=& |\det(\mathbf{X})|. \end{eqnarray*} - Another interpretation of the determinant is the volume changing factor when operating on a set in $\mathbb{R}^n$. $\text{vol}(f(S)) = |\det(\mathbf{A})| \text{vol}(S)$ where $f: \mathbb{R}^n \mapsto \mathbb{R}^n$ is the linear mapping defined by $\mathbf{A}$. - Recall that for differentiable function $f: \mathbb{R}^n \mapsto \mathbb{R}^n$, the **Jacobian matrix** $\operatorname{D} f(\mathbf{x}) \in \mathbb{R}^{n \times n}$ is $$ \operatorname{D} f(\mathbf{x}) = \begin{pmatrix} \frac{\partial f_1}{\partial x_1} (\mathbf{x}) & \frac{\partial f_1}{\partial x_2} (\mathbf{x}) & \cdots & \frac{\partial f_1}{\partial x_n} (\mathbf{x}) \\ \frac{\partial f_2}{\partial x_1} (\mathbf{x}) & \frac{\partial f_2}{\partial x_2} (\mathbf{x}) & \cdots & \frac{\partial f_2}{\partial x_n} (\mathbf{x}) \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_n}{\partial x_1} (\mathbf{x}) & \frac{\partial f_n}{\partial x_2} (\mathbf{x}) & \cdots & \frac{\partial f_n}{\partial x_n} (\mathbf{x}) \end{pmatrix} = \begin{pmatrix} \nabla f_1(\mathbf{x})' \\ \nabla f_2(\mathbf{x})' \\ \vdots \\ \nabla f_n(\mathbf{x})' \end{pmatrix}. $$ Its determinant, the **Jacobian determinant**, appears in the higher-dimensional version of **integration by substitution** or **change of variable** $$ \int_{f(U)} \phi(\mathbf{v}) \, \operatorname{d} \mathbf{v} = \int_U \phi(f(\mathbf{u})) | \det \operatorname{D} f(\mathbf{u})| \, \operatorname{d} \mathbf{u} $$ for function $\phi: \mathbb{R}^n \mapsto \mathbb{R}$. This result will be used in transformation of random variables in 202A. For an example of $n=1$, an indefinite integral can be transformed to a definite integral over box [-1,1] via change of variable $v = u / (1-u^2)$: $$ \int_{-\infty}^\infty f(v) \, dv = \int_{-1}^1 f\left(\frac{u}{1-u^2}\right) \frac{1+u^2}{(1-u^2)^2} \, du. $$ ## Some properties of determinant (important) - The determinant of a **lower or upper triangular matrix** $\mathbf{A}$ is the product of the diagonal elements $\prod_{i=1}^n a_{ii}$. (Why?) - Any square matrix $\mathbf{A}$ is singular if and only if $\det(\mathbf{A}) = 0$. Proof (optional): see BR p287. - Product rule: $\det(\mathbf{A} \mathbf{B}) = \det(\mathbf{A}) \det(\mathbf{B})$. Proof (optional): see BR p288-289. Product rule is extremely useful. For example, computer calculates the determinant of a square matrix $\mathbf{A}$ by first computing the LU decomposition $\mathbf{A} = \mathbf{L} \mathbf{U}$ and then $\det(\mathbf{A}) = \det(\mathbf{L}) \det(\mathbf{U})$. - Determinant of an orthogonal matrix is 1 (**rotation**) or -1 (**reflection**). This classifies orthogonal matrices into two classes: rotations and reflections. ```{julia} # a rotator θ = π/4 A = [cos(θ) -sin(θ); sin(θ) cos(θ)] ``` ```{julia} det(A) ``` ```{julia} # a reflector B = [cos(θ) sin(θ); sin(θ) -cos(θ)] ``` ```{julia} B'B ``` ```{julia} det(B) ``` ```{julia} # 3 points for a triangle X = [1 1 2 1; 1 3 1 1] # rotation Xrot = A * X # reflection Xref = B * X ``` ```{julia} plt = plot(X[1, :], X[2, :], color = :blue, legend = :none, xlims = (-3, 3), ylims = (-3, 3), xticks = -3:1:3, yticks = -3:1:3, framestyle = :origin, aspect_ratio = :equal) plot!(plt, Xrot[1, :], Xrot[2, :]) plot!(plt, Xref[1, :], Xref[2, :], annotations = [(-1.5, 2, "rotation"), (2, -1.75, "reflection")]) ``` - $\det(\mathbf{A}') = \det(\mathbf{A})$. - $\det(\mathbf{A}^{-1}) = 1/\det(\mathbf{A})$. - $\det(c\mathbf{A}) = c^n \det(\mathbf{A})$. - Determinant of a permutation matrix is the sign of the corresponding permutation. - Determinant of triangular block matrix \begin{eqnarray*} \det \left( \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{O} & \mathbf{D} \end{pmatrix} \right) = \det (\mathbf{A}) \det (\mathbf{D}). \end{eqnarray*} - For $\mathbf{A}$ and $\mathbf{D}$ square and nonsingular, \begin{eqnarray*} \det \left( \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix} \right) = \det (\mathbf{A}) \det (\mathbf{D} - \mathbf{C} \mathbf{A}^{-1} \mathbf{B}) = \det(\mathbf{D}) \det(\mathbf{A} - \mathbf{B} \mathbf{D}^{-1} \mathbf{C}). \end{eqnarray*} Proof: Take determinant on the both sides of the matrix identity \begin{eqnarray*} \begin{pmatrix} \mathbf{A} & \mathbf{0} \\ \mathbf{0} & \mathbf{D} - \mathbf{C} \mathbf{A}^{-1} \mathbf{B} \end{pmatrix} = \begin{pmatrix} \mathbf{I} & \mathbf{0} \\ - \mathbf{C} \mathbf{A}^{-1} & \mathbf{I} \end{pmatrix} \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix} \begin{pmatrix} \mathbf{I} & - \mathbf{A}^{-1} \mathbf{B} \\ \mathbf{0} & \mathbf{I} \end{pmatrix}. \end{eqnarray*}