Quadratic Forms and Positive Definite Matrices

二次型与正定矩阵 | Quadratic Forms and Positive Definite Matrices

定义 | Definitions

二次型 | Quadratic Form

二次型是指形式为 $Q(\mathbf{x})={\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}$ 的表达式，其中 $\mathbf{x}$ 是 $n$ 维列向量，$\mathbf{A}$ 是 $n\times n$ 的对称矩阵

A quadratic form is an expression of the form $Q(\mathbf{x})={\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}$, where $\mathbf{x}$ is an $n$-dimensional column vector and $\mathbf{A}$ is an $n\times n$ symmetric matrix

正定矩阵 | Positive Definite Matrix

正定矩阵是指对于任何非零向量 $\mathbf{x}$，都有 ${\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}>0$ 的对称矩阵

A positive definite matrix is a symmetric matrix $\mathbf{A}$ such that ${\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}>0$ for any non-zero vector $\mathbf{x}$

性质 | Properties

二次型的性质 | Properties of Quadratic Forms

1. 对称性: $Q(\mathbf{x})=Q(-\mathbf{x})$

2. 如果 $\mathbf{A}$ 是正定矩阵，则 $Q(\mathbf{x})$ 为正定二次型

3. $Q(\mathbf{x})$ 可以通过特征值分解写成 $Q(\mathbf{x})={\sum }_{i=1}^n{\lambda }_i{z}_i^2$，其中 ${\lambda }_i$ 是 $\mathbf{A}$ 的特征值，$z_i$ 是线性变换后的变量

4. Symmetry: $Q(\mathbf{x})=Q(-\mathbf{x})$

5. If $\mathbf{A}$ is a positive definite matrix, then $Q(\mathbf{x})$ is a positive definite quadratic form

6. $Q(\mathbf{x})$ can be expressed using eigenvalue decomposition as $Q(\mathbf{x})={\sum }_{i=1}^n{\lambda }_i{z}_i^2$, where ${\lambda }_i$ are the eigenvalues of $\mathbf{A}$ and $z_i$ are the transformed variables

正定矩阵的性质 | Properties of Positive Definite Matrices

1. 所有特征值均为正

2. 所有主子矩阵的行列式均为正

3. $\mathbf{A}$ 可被分解为 $\mathbf{A}=\mathbf{L}{\mathbf{L}}^{\mathrm{T}}$，其中 $\mathbf{L}$ 为下三角矩阵

4. All eigenvalues are positive

5. Determinants of all leading principal minors are positive

6. $\mathbf{A}$ can be decomposed as $\mathbf{A}=\mathbf{L}{\mathbf{L}}^{\mathrm{T}}$, where $\mathbf{L}$ is a lower triangular matrix

计算技巧 | Calculation Techniques

二次型化简 | Simplifying Quadratic Forms

通过正交变换 $\mathbf{P}$ 将 $\mathbf{A}$ 对角化，可以将二次型化简为 $Q(\mathbf{y})={\sum }_{i=1}^n{\lambda }_i{y}_i^2$

By orthogonal transformation $\mathbf{P}$ that diagonalizes $\mathbf{A}$, the quadratic form can be simplified to $Q(\mathbf{y})={\sum }_{i=1}^n{\lambda }_i{y}_i^2$

判定正定矩阵的方法 | Methods to Determine Positive Definite Matrices

1. 确认所有特征值是否均为正
2. 使用主子矩阵行列式进行测试
3. 计算 $\mathbf{A}$ 是否可以进行 Cholesky 分解

4. Cholesky Decomposition Cholesky 分解

1. Check if all eigenvalues are positive
2. Use leading principal minors to test
3. Determine if $\mathbf{A}$ can be decomposed using Cholesky decomposition

坐标变换 | Coordinate Transformations

正交变换 | Orthogonal Transformation

正交变换保持二次型的形式不变，即 $Q(\mathbf{y})={\mathbf{y}}^{\mathrm{T}}({\mathbf{P}}^{\mathrm{T}}\mathbf{AP})\mathbf{y}$，其中 $\mathbf{P}$ 是正交矩阵

An orthogonal transformation preserves the form of the quadratic form, i.e., $Q(\mathbf{y})={\mathbf{y}}^{\mathrm{T}}({\mathbf{P}}^{\mathrm{T}}\mathbf{AP})\mathbf{y}$, where $\mathbf{P}$ is an orthogonal matrix

仿射变换 | Affine Transformation

仿射变换包括旋转、平移等操作，可以用于化简或标准化二次型

Affine transformations include operations such as rotation and translation, and can be used to simplify or standardize quadratic forms

例子 | Examples

二次型的例子 | Example of a Quadratic Form

设 $\mathbf{x}=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$，$\mathbf{A}=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$，则二次型为

$$Q(\mathbf{x})={\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}=\left(\begin{array}{cc}{x}_1& x_2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=2x_1^2+2x_1{x}_2+3x_2^2$$

Let $\mathbf{x}=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$ and $\mathbf{A}=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$, then the quadratic form is

$$Q(\mathbf{x})={\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}=\left(\begin{array}{cc}{x}_1& x_2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=2x_1^2+2x_1{x}_2+3x_2^2$$

正定矩阵的例子 | Example of a Positive Definite Matrix

考虑矩阵 $\mathbf{A}=\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$，计算特征值

$$\det (\mathbf{A}-\lambda \mathbf{I})=\det \left(\begin{array}{cc}2-\lambda & -1\\ -1& 2-\lambda \end{array}\right)=(2-\lambda {)}^2-1={\lambda }^2-4\lambda +3=0$$

特征值为 ${\lambda }_1=1,{\lambda }_2=3$，均为正值，因此 $\mathbf{A}$ 为正定矩阵

Consider the matrix $\mathbf{A}=\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and compute the eigenvalues

$$\det (\mathbf{A}-\lambda \mathbf{I})=\det \left(\begin{array}{cc}2-\lambda & -1\\ -1& 2-\lambda \end{array}\right)=(2-\lambda {)}^2-1={\lambda }^2-4\lambda +3=0$$

The eigenvalues are ${\lambda }_1=1,{\lambda }_2=3$, both positive, hence $\mathbf{A}$ is a positive definite matrix

坐标变换中的应用 | Applications in Coordinate Transformations

对角化二次型 | Diagonalizing Quadratic Forms

为了将二次型 $Q(\mathbf{x})={\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}$ 对角化，我们寻找正交矩阵 $\mathbf{P}$ 使得 ${\mathbf{P}}^{\mathrm{T}}\mathbf{AP}$ 为对角矩阵 $\mathbf{D}$，于是 $Q(\mathbf{y})={\mathbf{y}}^{\mathrm{T}}\mathbf{Dy}$

To diagonalize the quadratic form $Q(\mathbf{x})={\mathbf{x}}^{\mathrm{T}}\mathbf{Ax}$, we find an orthogonal matrix $\mathbf{P}$ such that ${\mathbf{P}}^{\mathrm{T}}\mathbf{AP}$ is a diagonal matrix $\mathbf{D}$, thus $Q(\mathbf{y})={\mathbf{y}}^{\mathrm{T}}\mathbf{Dy}$

坐标变换举例 | Example of Coordinate Transformation

假设 $\mathbf{A}=\left(\begin{array}{cc}4& 2\\ 2& 3\end{array}\right)$，则特征值为 ${\lambda }_1=5,{\lambda }_2=2$，对应的特征向量为 ${\mathbf{v}}_1=\left(\begin{array}{c}2\\ 1\end{array}\right)$，${\mathbf{v}}_2=\left(\begin{array}{c}-1\\ 2\end{array}\right)$。构造正交矩阵 $\mathbf{P}$ 并进行对角化

$$\mathbf{P}=\left(\begin{array}{cc}\frac{2}{\sqrt{5}}& \frac{-1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{array}\right),{\mathbf{P}}^{\mathrm{T}}\mathbf{AP}=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right)$$

新的二次型为 $Q(\mathbf{y})=5y_1^2+2y_2^2$

Suppose $\mathbf{A}=\left(\begin{array}{cc}4& 2\\ 2& 3\end{array}\right)$, the eigenvalues are ${\lambda }_1=5,{\lambda }_2=2$, with corresponding eigenvectors ${\mathbf{v}}_1=\left(\begin{array}{c}2\\ 1\end{array}\right)$ and ${\mathbf{v}}_2=\left(\begin{array}{c}-1\\ 2\end{array}\right)$. Construct the orthogonal matrix $\mathbf{P}$ and diagonalize

$$\mathbf{P}=\left(\begin{array}{cc}\frac{2}{\sqrt{5}}& \frac{-1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{array}\right),{\mathbf{P}}^{\mathrm{T}}\mathbf{AP}=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right)$$

The new quadratic form is $Q(\mathbf{y})=5y_1^2+2y_2^2$