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Exercises \(22-26\) provide a glimpse of some widely used matrix factorizations, some of which are discussed later in the text. (Rank Factorization) Suppose an \(m \times n\) matrix \(A\) admits a factorization \(A=C D\) where \(C\) is \(m \times 4\) and \(D\) is \(4 \times n .\) a. Show that \(A\) is the sum of four outer products. (See Section 2.4.) b. Let \(m=400\) and \(n=100 .\) Explain why a computer programmer might prefer to store the data from \(A\) in the form of two matrices \(C\) and \(D .\)

Short Answer

Expert verified
A is expressed as four outer products: \( A = \mathbf{c_1}\mathbf{d_1}^T + \mathbf{c_2}\mathbf{d_2}^T + \mathbf{c_3}\mathbf{d_3}^T + \mathbf{c_4}\mathbf{d_4}^T \).\nStoring \( C \) and \( D \) requires far less memory than storing \( A \) directly.

Step by step solution

01

Conceptual Understanding of Outer Product

The outer product of two vectors results in a matrix. For two vectors, say \( \mathbf{u} \) of size \( m \times 1 \) and \( \mathbf{v} \) of size \( 1 \times n \), their outer product \( \mathbf{u} \mathbf{v}^T \) is a matrix of size \( m \times n \), where each element is the product of the corresponding elements of the vectors.
02

Expressing A as Sum of Outer Products

Given \( A = CD \), where \( C \) is \( m \times 4 \) and \( D \) is \( 4 \times n \), we can express each entry of \( A \) as a sum of products of a column from \( C \) and a row from \( D \). Thus, \( A = \begin{bmatrix} \mathbf{c_1} & \mathbf{c_2} & \mathbf{c_3} & \mathbf{c_4} \end{bmatrix} \begin{bmatrix} \mathbf{d_1}^T \ \mathbf{d_2}^T \ \mathbf{d_3}^T \ \mathbf{d_4}^T \end{bmatrix} = \mathbf{c_1}\mathbf{d_1}^T + \mathbf{c_2}\mathbf{d_2}^T + \mathbf{c_3}\mathbf{d_3}^T + \mathbf{c_4}\mathbf{d_4}^T \), showing \( A \) as a sum of four outer products.
03

Understanding Memory Storage with Matrices

For part b, given \( m = 400 \) and \( n = 100 \), storing \( A \) directly would require \( 400 \times 100 = 40,000 \) elements. Storing \( C \) and \( D \) together requires only storing \( m \times 4 \) elements for \( C \) and \( 4 \times n \) elements for \( D \), totaling \( 400(4) + 4(100) = 1,600 + 400 = 2,000 \) elements, which is much less storage.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Outer Product
Understanding the outer product involves multiplying two vectors to obtain a matrix. Imagine you have two vectors, let's say vector \( \mathbf{u} \) which has a size of \( m \times 1 \) and vector \( \mathbf{v} \) with a size of \( 1 \times n \). When you take the outer product of these vectors, you get a new matrix that is \( m \times n \) in size.
Every element in this resulting matrix comes from multiplying corresponding elements from the two vectors:
\[ \mathbf{u} \mathbf{v}^T = \begin{bmatrix} u_1 \ u_2 \ \vdots \ u_m \end{bmatrix} \begin{bmatrix} v_1 & v_2 & \cdots & v_n \end{bmatrix} \]
This process can translate complex interactions and patterns of data into simpler, structured forms, making it a pivotal part of matrix mathematics.
Rank Factorization
Rank factorization provides a unique way to break down any matrix into simpler parts. In our exercise, imagine a matrix \( A \) that can be broken down into two matrices where \( C \) is \( m \times 4 \) and \( D \) is \( 4 \times n \). Here, 4 represents the 'rank' which suggests the number of outer products used in the sum.
These components, \( C \) and \( D \), when multiplied together, exactly recreate the original matrix \( A \). The wonderful part of rank factorization is how it simplifies high-dimensional data by representing it through fewer and simpler components, with each outer product revealing more about the original matrix's characteristics.
In simpler terms, it's like expressing a complex number or equation by the sum of simpler functions that hold the same value.
Matrix Storage Efficiency
Matrix storage efficiency relates directly to how data is stored in computers. When we're working with a large matrix like \( A \) in our exercise, keeping it in memory can require a tremendous amount of space. For example, directly storing a \( 400 \times 100 \) matrix requires 40,000 elements.
Using rank factorization to store \( A \) in the form of matrices \( C \) and \( D \) minimizes this need. Here, you only need to store 2,000 elements in total since \( C \) has \( 400 \times 4 = 1,600 \) elements and \( D \) has \( 4 \times 100 = 400 \) elements.
Key benefits include:
  • Reduced memory usage, which is critical for large-scale computations.
  • Faster computation times as the operations are done with smaller matrices.
  • Less space required makes working with large datasets more feasible.
This efficiency is a major reason developers and programmers favor using matrix factorization techniques in practical applications.

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Most popular questions from this chapter

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