[Linear Algebra] Singular Value Decomposition and Low Rank Approximation
Recommendations linear algebra
Published : 2021-11-02   Lastmod : 2021-11-29

## Singular value decomposition and low-rank approximation

I am going to summarize my knowledge of linear algebra, which is mainly needed to understand recommendation systems. The user-item matrices (preference matrices) used in recommendation systems are often discussed under the assumption that low-rank approximations hold.

It is based on the implicit assumption that users can often be classified into a finite number of clusters. For example, even if there are one million users of all book e-commerce sites, it is possible to categorize them to some extent, such as users who often buy programming books, users who buy math books, users who buy medical books, and users who buy weekly magazines.

The singular value decomposition is necessary when using the low-rank approximation. It extracts singular value vectors belonging to $k$ singular values with large values, and predicts the user’s preference for the observed item while compressing the data.

### github

• The file in jupyter notebook format is here .

• To run it in google colaboratory here

## singular value decomposition

Singular value decomposition is defined for a general $m \times n$ matrix $A$ that is not a square matrix as follows.

$$A=U\Sigma V^{\mathrm{T}}$$

$$\mathbf{A}=\mathbf{U} \cdot \operatorname{diag}\left(\sigma_{1}, \ldots, \sigma_{r}, \mathbf{0}\right) \cdot \mathbf{V}^{*}$$

The $\sigma_1, \sigma_2 \cdots$ are singular values and are usually defined by numbering them in order from largest to smallest

$$\sigma_{1}\geqq \cdots \geqq \sigma_{r}>0$$

$$r is the rank of the matrix A$$ r=\operatorname{rank}(\mathbf{A}) $$U and V are unitary matrices of m \times m and n \times n$$ \mathbf{U}^{-1}=\mathbf{U}^{*}  \mathbf{V}^{-1}=\mathbf{V}^{*} $$If A is a symmetric matrix, the singular value and eigenvalue of A will be the same. ### Image of the matrix The visual representation of the matrix after decomposition is as follows.$$ A=U\Sigma V^{T}=U\left(\begin{array}{ccc|c} \sigma_{1} & & 0 & \\ & \ddots & & 0 \\ 0 & & \sigma_{r} & \\ \hline & 0 & & 0 \end{array}\right) V^{\mathrm{T}} \\ = \left( \mathbf{u}_{1}\cdots \mathbf{u}_{r}\right) \left(\begin{array}{llll} \sigma_{1} & & & \\ & & \ddots & \\ & & \sigma_{r} \end{array}\right) \left(\begin{array}{c} \mathbf{v}{1}^{T} \\ \vdots \\ \mathbf{v}{r}^{T} \end{array}\right) $$## Eigenvalue decomposition and singular value decomposition Since U and V in A=U\Sigma V^{T} are unitary matrices, we can use the vectors u and v and multiply them as follows.$$ \begin{aligned} \mathbf{A} \mathbf{v} &=\sigma \mathbf{u} \\ \mathbf{A}^{T} \mathbf{u} &=\sigma \mathbf{v} \end{aligned} $$Further transformations.$$ \begin{aligned} &\mathbf{A}^{T} \mathbf{A} \mathbf{v}=\sigma \mathbf{A}^{T} \mathbf{u}=\sigma^{2} \mathbf{v} \\ &\mathbf{A} \mathbf{A}^{T} \mathbf{u}=\sigma \mathbf{A} \mathbf{v}=\sigma^{2} \mathbf{u} \end{aligned} $$\mathbf{u} and \mathbf{v} are eigenvectors of \mathbf{A}^{T} \mathbf{A} and \mathbf{A}^{T} , respectively. \mathbf{u} and \mathbf{v} are called left and right singular vectors, respectively, and \mathbf{u} and \mathbf{v} are eigenvectors of \mathbf{AA^{T}}, \mathbf{A^{T}A}. Also, the square of the singular value of \mathbf{A} becomes the eigenvalue of \mathbf{AA^T} and \mathbf{A^TA} From this property, we can link the singular value decomposition to the eigenvalue decomposition. First, perform the eigenvalue decomposition of \mathbf{A^TA} to find the eigenvectors. The eigenvector is \mathbf{v_i} and the eigenvalue is \lambda_i Since \lambda_i = \sigma^2$$ \mathbf{u}_i = \frac{1}{\sqrt{\lambda_i}}\mathbf{A} \mathbf{v_i} $$and we can find \mathbf{u}_i. It is possible to obtain \mathbf{v}_i in the same way, and the singular value decomposition can be calculated using the eigenvalue decomposition algorithm. ## Operator norm (operator norm) The norm defined for a matrix A, the$$ ||\mathbf{A}||=\max _{x \in \mathbb{C}^{n},|x|=1}||\mathbf{A} \mathbf{x}|| $$is called the operator norm. Using singular value decomposition, we can easily obtain the operator norm.$$ \begin{aligned} \||A x||^{2} &=||U \Sigma V^{} x||^{2} \\ &=x^{*}V \Sigma^{*} U^{*}U \Sigma V^{*} x=||\Sigma y||^{2} \\ &=\sigma_{1}^{2}\left|y\{1}\right|^{2}+\cdots+\sigma_{r}^{2}\left|y\{r}\right|^{2} \quad (y = x^{} V) \end{aligned} $$Therefore.$$ \begin{aligned} ||\mathbf{A}||&=\max_{x \in \mathbb{C}^{n},||x||=1}||A x|| \\ &=\max_{||y||=1} \sqrt{\sigma_{1}^{2}\left|y_{1}\right|^{2}+\sigma_{2}^{2}\left|y_{2}\right|^{2}+\cdots+\sigma_{r}^{2}\left|y_{r}\right|^{2}} \\ &=\sigma_{1} \quad\left(\sigma_{1} \geq \sigma_{2} \geq \cdots \geq \sigma_{r} \right) \end{aligned} $$and the operator norm is the largest singular value. ### Operator norm of unitary matrix The operator norm of a unitary matrix is 1. This is obvious from the definition: if the unitary matrix is U (U^{T}U=I), then$$ ||U x||^{2}=x^{T} U^{T} U x=|||x||^{2} = 1 $$It becomes. ### Trigonometric inequality$$ ||A+B|| \leq ||A||+||B|| $$### Inequalities about products$$ ||A B|| \leq||A|| \cdot||B|| $$## Eckart-Young’s theorem. ### Low rank approximation Approximating A by taking the matrix A from k values with large singular values is called low-rank approximation.$$ A_{k}=U \Sigma_{k} V^{\mathrm{T}} \equiv U \operatorname{diag}\left(\sigma_{1}, \ldots, \sigma_{k}, 0\right) V^{\mathrm{T}}=\sum_{i=1}^{k} \sigma_{i} u_{i} v_{i}^{\mathrm{T}}(0 \leq k<r) $$In other words, we say that \sigma_{k+1}, \cdots, \sigma_{r} is 0. ### Details of the theorem I would like to note that Eckert-Young’s theorem is often cited in recommendation system papers when the discussion is based on low-rank approximations. You can find the details in reference .$$ \min_{\operatorname{rank}||X|| \leq k}||X-A||=\left||A_{k}-A\right||=\sigma_{k+1}(A)=\min _{\operatorname{rank}(X)=k}||X-A|| $$This theorem states that there is a low-rank approximation matrix A_k on a sphere of radius \sigma_{k+1} centered at A. It holds for all k. I believe that defining the norm in terms of the operator norm shows a very clear and intuitive result. The proof is on , please refer to it.$$ \begin{aligned} \left|A_{k}-A\right| &=\left|U\left(\Sigma_{k}-\right) V^{T}\right|=\left|\left(\Sigma_{k}-\right) \right| \\ &=\left\|\operatorname{diag}\left(0, \sigma_{k+1}, \ldots, \sigma_{r}, 0\right)\right|=\sigma_{k+1} \end{aligned} $$## Frobenius norm The Frobenius norm is a measure of the size of a matrix, defined as the sum of the squares of all its components. It is written as ||A||_{\mathrm{F}}.$$ ||A||_{\mathrm{F}}=\sqrt{\sum_{i, j} a_{i j}^{2}} $$### Relationship to Singular Values The Frobenius norm is equal to the sum of squares of singular values.$$ ||A||_{\mathrm{F}}^{2}=\sum_{i=1}^{r} \sigma_{i}^{2} $$### Relationship with traces The trace of the product with the transpose matrix is the Frobenius norm, as is clear from the actual calculation here.$$ ||A||_{\mathrm{F}}^{2}=\operatorname{tr}\left(A A^{\top}\right)=\operatorname{tr}\left(A^{\top} A\right) $$## Implementation with python Let’s try to implement the singular value decomposition using Python. ### Author’s environment The author’s OS is macOS, and the options are different from those of Linux and Unix commands. !sw_vers  ProductName: Mac OS X ProductVersion: 10.14.6 BuildVersion: 18G103  Python -V  Python 3.8.5  %matplotlib inline %config InlineBackend.figure_format = 'svg' import matplotlib import matplotlib.pyplot as plt import scipy import numpy as np import pandas as pd print('matplotlib version :', matplotlib.__version__) print('scipy version :', scipy.__version__) print('numpy version :', np.__version__)  matplotlib version : 3.3.2 scipy version : 1.5.2 numpy version : 1.19.2  ### Prepare A as desired. Create a 4x5 matrix of integers. A = np.array([]) for i in range(4): for j in range(5): A = np.append(A, np.random.randint(1, 10)) A = A.reshape((4,5)) A  array([[2., 2., 4., 4., 3.], [4., 7., 8., 4., 6.], [6., 7., 5., 6., 4.], [5., 7., 4., 1., 2.]])  Check \text{rank}(A). np.linalg.matrix_rank(A)  4  ### Singular Value Decomposition We will now perform the singular value decomposition. u, S, v_T = np.linalg.svd(A)  Check U, \Sigma, and V. u.round(2)  array([[-0.3 , 0.57, 0.14, 0.75], [-0.62, 0.27, -0.66, -0.33], [-0.59, -0.01, 0.73, -0.36], [-0.43, -0.78, -0.14, 0.44]])  S.round(2)  array([21.37, 4.48, 3.14, 0.55])  v_T.T.round(2)  array([[-0.41, -0.39, 0.41, 0.42, 0.58], [-0.56, -0.56, -0.07, -0.44, -0.41], [-0.51, 0.28, -0.52, 0.59, -0.22], [-0.36, 0.55, 0.68, -0.08, -0.31], [-0.37, 0.38, -0.29, -0.53, 0.59]])  ### Reproduction check Verify that the original matrix can be reproduced by multiplying it by three matrices. (u @ np.append(np.diag(S), [, , , ], axis=1) @ v_T).round(2)  array([[2., 2., 4., 4., 3.], [4., 7., 8., 4., 6.], [6., 7., 5., 6., 4.], [5., 7., 4., 1., 2.]])  Subtract from the original A. A.round(2) - (u @ np.append(np.diag(S), [, , , ], axis=1) @ v_T).round(2)  array([[0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.]])  and the result is the original A. We also check that U and V are unitary matrices. (u @ u.T).round(2)  array([ 1., 0., -0., 0.], [ 0., 1., 0., 0.], [-0., 0., 1., -0.], [ 0., 0., -0., 1.]])  (v_T @ v_T.T).round(2)  array([[ 1., 0., 0., 0., 0.]], [ 0., 1., 0., 0., -0.], [ 0., 0., 1., -0., -0.], [ 0., 0., -0., 1., -0.], [ 0., -0., -0., -0., 1.]])  This is the unit matrix, and we have successfully confirmed it. ### Calculating the operator norm To calculate the operator norm in python, we just need to calculate the largest singular value. Let A be the matrix we defined earlier. S.round(2)  21.37  ### Calculating the Frobenius norm The Frobenius norm can be easily calculated with numpy’s linalg. np.linalg.norm(A, 'fro').round(2)  22.07  It can also be computed using the following property. You can see that the two results are consistent.$$ \|A|_{\mathrm{F}}^{2}=\operatorname{tr}\left(A A^{\top}\right)=\operatorname{tr}\left(A^{\top} A\right) 

np.sqrt(np.trace(A.T @ A)).round(2)

22.07


np.abs(np.max(np.linalg.eig(np.array([[1,2,1], [2,1,3], [3,1,2]]))))).round(2)

5.24