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Suppose the last column of \(A B\) is entirely zero but \(B\) itself has no column of zeros. What can you say about the columns of \(A ?\)

Short Answer

Expert verified
The columns of \( A \) are linearly dependent.

Step by step solution

01

Understand the Problem Statement

We are given that the last column of the matrix product \( AB \) is entirely zero. However, matrix \( B \) itself has no columns that are completely zero. We need to determine what this implies about the columns of matrix \( A \).
02

Matrix Multiplication Background

Recall that the product \( AB \) is derived from multiplying each column of \( B \) by the corresponding columns of \( A \). Specifically, the column in \( AB \) results from a linear combination of the columns of \( A \), where the coefficients are elements from a column in \( B \).
03

Analyze the Last Column of \( AB \)

The last column of \( AB \), being entirely zero, suggests that the linear combination of the columns of \( A \) that corresponds with the coefficients from the last column of \( B \) equals the zero vector.
04

Interpret \( B \'s \) Contribution

Since \( B \) has no zero columns, the coefficients from the last column of \( B \) are not all zero. This indicates that some non-trivial linear combination of the columns of \( A \) results in the zero vector.
05

Conclusion About Columns of \( A \)

The fact that a non-trivial linear combination of \( A \)'s columns results in the zero vector suggests that the columns of \( A \) are linearly dependent, as there exists non-zero scalars that produce the zero vector when applied.

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra, combining two matrices to form a new matrix. To multiply two matrices, say matrix \( A \) and matrix \( B \), follow these steps:
  • Ensure the number of columns in matrix \( A \) matches the number of rows in matrix \( B \).
  • Take each row from matrix \( A \) and multiply it with each column from matrix \( B \).
  • Add up the products from each row-column multiplication, placing the result in the corresponding position in the product matrix \( AB \).
Matrix multiplication is not commutative, meaning \( AB eq BA \) in general. The process can be visualized as creating a series of linear combinations using rows from \( A \) and columns from \( B \). This is why understanding it as a link between linear combinations and vectors is crucial. Moreover, the resulting matrix \( AB \) is influenced by both matrices' structure.
Linear Combination
A linear combination in mathematics refers to an expression constructed from a set of terms multiplied by constants, called coefficients. In the context of matrices, each column in a product like \( AB \) is generated as a linear combination of the columns from matrix \( A \). The coefficients in this case are elements from the corresponding column in matrix \( B \).

Let's outline the key ideas:
  • The goal is to combine columns from matrix \( A \) such that their weighted sum aligns with the coefficients from matrix \( B \).
  • If these coefficients can form the zero vector without all being zero, it implies a specific relationship about the columns, revealing their dependencies.
Linear combinations underpin many concepts in vector spaces, determining the span, basis, and ultimately the structure of vector sets.
Zero Vector
Within linear algebra, the zero vector is a special vector where every component is zero. It is the additive identity in vector space, meaning any vector added to the zero vector remains unchanged. In the context of matrix multiplication, when a linear combination of columns results in a zero vector, it indicates a crucial signal about dependence relations among those columns.

A zero vector is represented as \( \mathbf{0} = [0, 0, \ldots, 0] \). This situation, especially when obtained through non-zero coefficients, signifies that the columns are not entirely independent. Hence, detecting a zero vector in a matrix product alerts us that at least one set of vectors used is redundant in explaining the span, or range, of the vector space involved.
Non-Trivial Solution
The term "non-trivial solution" refers to a solution of a homogeneous equation or system of equations that is not the all-zero solution, which would be the trivial solution. In matrices, finding a non-trivial solution indicates that a linear combination of vectors, using some non-zero coefficients, results in the zero vector.

For example, consider a matrix \( A \) whose columns produce the zero vector when combined with non-zero weights. This confirms linear dependence because the existence of non-zero solutions (those other than the zero vector) implies there is redundancy among the column vectors. Such solutions contribute dramatically to understanding the vector space's dimension and potential compression of information, denoting relations not immediately visible otherwise.

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Most popular questions from this chapter

Exercises 1–4 refer to an economy that is divided into three sectors—manufacturing, agriculture, and services. For each unit of output, manufacturing requires .10 unit from other companies in that sector, .30 unit from agriculture, and .30 unit from services. For each unit of output, agriculture uses .20 unit of its own output, .60 unit from manufacturing, and .10 unit from services. For each unit of output, the services sector consumes .10 unit from services, .60 unit from manufacturing, but no agricultural products. Determine the production levels needed to satisfy a final demand of 18 units for manufacturing, with no final demand for the other sectors. (Do not compute an inverse matrix.)

Determine which sets in Exercises \(15-20\) are bases for \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\) . Justify each answer. $$ \left[\begin{array}{r}{-4} \\ {6}\end{array}\right],\left[\begin{array}{r}{2} \\\ {-3}\end{array}\right] $$

A \(2 \times 200\) data matrix \(D\) contains the coordinates of 200 points. Compute the number of multiplications required to transform these points using two arbitrary \(2 \times 2\) matrices \(A\) and \(B\) . Consider the two possibilities \(A(B D)\) and \((A B) D .\) Discuss the implications of your results for computer graphics calculations.

In Exercises 1 and \(2,\) find the vector \(\mathbf{x}\) determined by the given coordinate vector \([\mathbf{x}]_{\mathcal{B}}\) and the given basis \(\mathcal{B}\) . Illustrate your answer with a figure, as in the solution of Practice Problem \(2 .\) $$\mathcal{B}=\left\\{\left[\begin{array}{r}{-2} \\\ {1}\end{array}\right],\left[\begin{array}{l}{3} \\\ {1}\end{array}\right]\right\\},[\mathbf{x}]_{\mathcal{B}}=\left[\begin{array}{r}{-1} \\\ {3}\end{array}\right]$$

In Exercises 17 and \(18,\) mark each statement True or False. Justify each answer. Here \(A\) is an \(m \times n\) matrix. a. If \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is a basis for a subspace \(H\) and if \(\mathbf{x}=c_{1} \mathbf{v}_{1}+\cdots+c_{p} \mathbf{v}_{p},\) then \(c_{1}, \ldots, c_{p}\) are the coordinates of \(\mathbf{x}\) relative to the basis \(\mathcal{B}\) . b. Each line in \(\mathbb{R}^{n}\) is a one-dimensional subspace of \(\mathbb{R}^{n}\) . c. The dimension of Col \(A\) is the number of pivot columns of \(A .\) d. The dimensions of Col \(A\) and Nul \(A\) add up to the number of columns of \(A\) . e. If a set of \(p\) vectors spans a \(p\) -dimensional subspace \(H\) of \(\mathbb{R}^{n},\) then these vectors form a basis for \(H .\)

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