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# 线性代数代考_Linear Algebra代考_MTH309 THE SINGULAR VALUE DECOMPOSITION

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## 线性代数代考_Linear Algebra代考_THE SINGULAR VALUE DECOMPOSITION

The diagonalization theorems in Sections $5.3$ and $7.1$ play a part in many interesting applications. Unfortunately, as we know, not all matrices can be factored as $A=P D P^{-1}$ with $D$ diagonal. However, a factorization $A=Q D P^{-1}$ is possible for any $m \times n$ matrix $A$ ! A special factorization of this type, called the singular value decomposition, is one of the most useful matrix factorizations in applied linear algebra.

The singular value decomposition is based on the following property of the ordinary diagonalization that can be imitated for rectangular matrices: The absolute values of the eigenvalues of a symmetric matrix $A$ measure the amounts that $A$ stretches or shrinks certain vectors (the eigenvectors). If $A \mathbf{x}=\lambda \mathbf{x}$ and $|\mathbf{x}|=1$, then
$$|A \mathbf{x}|=|\lambda \mathbf{x}|=|\lambda||\mathbf{x}|=|\lambda|$$
If $\lambda_1$ is the eigenvalue with the greatest magnitude, then a corresponding unit eigenvector $\mathbf{v}_1$ identifies a direction in which the stretching effect of $A$ is greatest. That is, the length of $A \mathbf{x}$ is maximized when $\mathbf{x}=\mathbf{v}_1$, and $\left|A \mathbf{v}_1\right|=\left|\lambda_1\right|$, by (1). This description of $\mathbf{v}_1$ and $\left|\lambda_1\right|$ has an analogue for rectangular matrices that will lead to the singular value decomposition.

## 线性代数代考_Linear Algebra代考_The Singular Values of an m X n Matrix

Let $A$ be an $m \times n$ matrix. Then $A^T A$ is symmetric and can be orthogonally diagonalized. Let $\left{\mathbf{v}_1, \ldots, \mathbf{v}_n\right}$ be an orthonormal basis for $\mathbb{R}^n$ consisting of eigenvectors of $A^T A$, and let $\lambda_1, \ldots, \lambda_n$ be the associated eigenvalues of $A^T A$. Then, for $1 \leq i \leq n$,
\begin{aligned} \left|A \mathbf{v}_i\right|^2 &=\left(A \mathbf{v}_i\right)^T A \mathbf{v}_i=& \mathbf{v}_i^T A^T A \mathbf{v}_i \ &=\mathbf{v}_i^T\left(\lambda_i \mathbf{v}_i\right) & & \text { Since } \mathbf{v}_i \text { is an eigenvector of } A^T A \ &=\lambda_i & & \text { Since } \mathbf{v}_i \text { is a unit vector } \end{aligned}
So the eigenvalues of $A^T A$ are all nonnegative. By renumbering, if necessary, we may assume that the eigenvalues are arranged so that
$$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n \geq 0$$
The singular values of $A$ are the square roots of the eigenvalues of $A^T A$, denoted by $\sigma_1, \ldots, \sigma_n$, and they are arranged in decreasing order. That is, $\sigma_i=\sqrt{\lambda_i}$ for $1 \leq i \leq n$. By equation (2), the singular values of $A$ are the lengths of the vectors $A \mathbf{v}_1, \ldots, A \mathbf{v}_n$.
EXAMPLE 2 Let $A$ be the matrix in Example 1. Since the eigenvalues of $A^T A$ are 360,90 , and 0 , the singular values of $A$ are
$$\sigma_1=\sqrt{360}=6 \sqrt{10}, \quad \sigma_2=\sqrt{90}=3 \sqrt{10}, \quad \sigma_3=0$$
From Example 1, the first singular value of $A$ is the maximum of $|A \mathbf{x}|$ over all unit vectors, and the maximum is attained at the unit eigenvector $\mathbf{v}_1$. Theorem 7 in Section $7.3$ shows that the second singular value of $A$ is the maximum of $|A \mathbf{x}|$ over all unit vectors that are orthogonal to $\mathbf{v}_1$, and this maximum is attained at the second unit eigenvector, $\mathbf{v}_2$ (Exercise 22). For the $\mathbf{v}_2$ in Example 1 ,
$$A \mathbf{v}_2=\left[\begin{array}{rrr} 4 & 11 & 14 \ 8 & 7 & -2 \end{array}\right]\left[\begin{array}{r} -2 / 3 \ -1 / 3 \ 2 / 3 \end{array}\right]=\left[\begin{array}{r} 3 \ -9 \end{array}\right]$$
This point is on the minor axis of the ellipse in Figure 1, just as $A \mathbf{v}_1$ is on the major axis. (See Figure 2.) The first two singular values of $A$ are the lengths of the major and minor semiaxes of the ellipse.

The fact that $A \mathbf{v}_1$ and $A \mathbf{v}_2$ are orthogonal in Figure 2 is no accident, as the next theorem shows.

## 线性代数代考_Linear Algebra代考_THE SINGULARVALUE DECOMPOSITION

$A=P D P^{-1}$ 和 $D$ 对角线。然而，一个因式分解 $A=Q D P^{-1}$ 对任何一个都是可能的 $m \times n$ 矩阵 $A$ ! 这种类型的特殊分解称为 奇异值分解，是应用线性代数中最有用的矩阵分解之一。

$$|A \mathbf{x}|=|\lambda \mathbf{x}|=|\lambda||\mathbf{x}|=|\lambda|$$

## 线性代数代弌_Linear Algebra代考_The Singular Values of an $m \times n$ Matrix

$\left|A \mathbf{v}_i\right|^2=\left(A \mathbf{v}_i\right)^T A \mathbf{v}_i=\quad \mathbf{v}_i^T A^T A \mathbf{v}_i=\mathbf{v}_i^T\left(\lambda_i \mathbf{v}_i\right) \quad$ Since $\mathbf{v}_i$ is an eigenvector of $A^T A \quad=\lambda_i \quad$ Since $\mathbf{v}_i$ is a unit vector

$$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n \geq 0$$

(2)，奇异值 $A$ 是向量的长度 $A \mathbf{v}_1, \ldots, A \mathbf{v}_n$.

$$\sigma_1=\sqrt{360}=6 \sqrt{10}, \quad \sigma_2=\sqrt{90}=3 \sqrt{10}, \quad \sigma_3=0$$

$$A \mathbf{v}_2=\left[\begin{array}{lllll} 4 & 11 & 148 & 7 & -2 \end{array}\right][-2 / 3-1 / 32 / 3]=\left[\begin{array}{ll} 3 & -9 \end{array}\right]$$

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