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Chapter 3: Problem 3

Show that \(\\{(1,0,0,0,0,0),(1,2,0,0,0,0),(0,1,2,3,0,0),(0,0,1,2,3,4)\\}\) is a linearly independent subset of \(R^{6}\).

Short Answer

Expert verified
The vectors \(\{(1,0,0,0,0,0),(1,2,0,0,0,0),(0,1,2,3,0,0),(0,0,1,2,3,4)\}\) are a linearly independent subset of \(R^{6}\)

Step by step solution

01

Create the matrix

Firstly, form a matrix A where each vector is a column in A. So you will get the following matrix: \[ A = \begin{bmatrix} 1 & 1 & 0 & 0 \ 0 & 2 & 1 & 0 \ 0 & 0 & 2 & 1 \ 0 & 0 & 3 & 2 \ 0 & 0 & 0 & 3 \ 0 & 0 & 0 & 4 \end{bmatrix}\].
02

Apply Gaussian elimination

To solve the system, apply Gaussian elimination to put it into row-echelon form. Firstly, you can multiply the second column by 1/2 to have a 1 in the second position. Secondly, subtract 2 times the second row from the third one and 3 times the third row from the fourth one. This results in the following matrix: \[ R = \begin{bmatrix} 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix} \].
03

Identify the leading 1s

In row-echelon form, the first non-zero element of each row (from left to right) is known as a 'leading 1'. In this case, every row (excluding those with only zeros) has a leading 1, suggesting that the vectors are linearly independent.

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Key Concepts

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

Gaussian Elimination
Gaussian elimination is a powerful algorithm used to solve systems of linear equations. It transforms a given matrix into a simpler form. This simpler form is known as row-echelon form, which makes it easier to analyze the solutions of the system. The process involves using elementary row operations.
These operations include:
  • Swapping two rows
  • Multiplying a row by a non-zero constant
  • Adding or subtracting a multiple of one row to another
By applying these operations, we systematically eliminate variables. This is achieved by creating zeros below each pivot. Each pivot is the first non-zero number in a row when viewed from the top of the matrix down. The converted matrix reveals information about the independence and dependence of vectors in the system.
Row-Echelon Form
Row-echelon form is a specific arrangement of a matrix. In this form, each leading entry of a row is 1, and each leading 1 has only zeros below it. This streamlined form is helpful for determining the rank of a matrix and thereby understanding solutions to linear systems.
Here are characteristics of row-echelon form:
  • All non-zero rows are above any rows containing only zeros.
  • The leading entry of each non-zero row after the first occurs to the right of the leading entry of the previous row.
  • The leading entry in any non-zero row is 1.
In our exercise, transforming the matrix into row-echelon form helps to check if vectors are linearly independent. If there are pivots in every column that represent a vector in our system, it implies no vector can be written as a combination of others鈥攈ence they are independent.
Linear Algebra
Linear algebra is a branch of mathematics that studies vectors, vector spaces, and linear equations. This discipline is foundational to various mathematical and practical applications. Vector spaces such as \(R^6\), which appears in our exercise, consist of all possible vectors of a certain dimension鈥攊n this case, six dimensions.
Fundamental concepts in linear algebra include:
  • Vectors and their operations: addition and scalar multiplication
  • Matrices and determinants
  • Systems of linear equations
  • Linear transformations
Understanding these concepts aids in solving real-world problems involving multidimensional data, such as physics simulations, computer graphics, and more. In our case, deciphering linear independence through Gaussian elimination is a classic application.
R6
R6 refers to \(\mathbb{R}^6\), a six-dimensional Euclidean space. This space is made of all ordered six-tuples of real numbers. Its usage is common in linear algebra when dealing with high-dimensional vectors.
Characteristics of \(\mathbb{R}^n\) spaces include:
  • Dimensionality indicating the number of components in each vector (six in \(\mathbb{R}^6\))
  • The combination of base vectors spans the space, meaning any vector in \(\mathbb{R}^6\) is expressible as a vector sum
  • Applications ranging from realistic simulations in physics to detailed color descriptions in computer graphics
In our exercise, analyzing a subset of \(\mathbb{R}^6\) involves checking for linear independence. It shows whether vectors in this space can be created from others within the subset, using techniques from linear algebra to confirm their independence.

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Most popular questions from this chapter

in the standard form of a plane through the origin. Show that the set $$ \left\\{\left[\begin{array}{rrr} 2 & 1 & 0 \\ -1 & 1 & 0 \end{array}\right] \cdot\left[\begin{array}{lll} 1 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right] \cdot\left[\begin{array}{rrr} -1 & 2 & 1 \\ 0 & 0 & 1 \end{array}\right] \cdot\left[\begin{array}{lll} 0 & -2 & 1 \\ 1 & -1 & 2 \end{array}\right]\right\\} $$ does not span \(\mathbb{M}(2,3)\).

Use the proof of the Expansion Theorem as a recipe for finding a basis for \(\mathbb{P}_{3}\) that contains \(\left\\{x+3, x^{2}+x\right\\}\).

Suppose \(S\) and \(T\) are finite-dimensional subspaces of a vector space \(V\). Suppose \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{k}, \mathbf{v}_{k+1}, \ldots, \mathbf{v}_{m}\right\\}\) is a basis for \(S\) and \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{k}, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_{n}\right\\}\) is a basis for \(T\) chosen as in the previous exercise so that \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{k}\right\\}\) is a basis for \(S \cap T\). a. Prove that \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{k}, \mathbf{v}_{k+1}, \ldots, \mathbf{v}_{m}, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_{n}\right\\}\) is linearly independent. (Suggestion: Start off by assuming a linear combination of the vectors is equal to zero. Use this equation to find a vector that is both in \(S\) and in \(T\). Use Theorem \(3.17\) to derive the conclusion that all the coefficients must be zero.) b. Prove that \(\operatorname{dim}(\operatorname{span}(S \cup T))=\operatorname{dim} S+\operatorname{dim} T-\operatorname{dim}(S \cap T)\).

Show that the functions defined by \(f(x)=e^{x}\) and \(g(x)=e^{-x}\) form a linearly independent subset of \(\mathbb{F}(\mathbb{R})\).

Show that \(\\{\sin , \cos , \exp \\}\) is a linearly independent subset of \(F(\mathbb{R})\).
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