/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Find a basis and dimension of th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 29

Find a basis and dimension of the subspace \(W\) of \(\mathbf{R}^{3}\) where (a) \(\quad W=\\{(a, b, c): a+b+c=0\\}\) (b) \(\quad W=\\{(a, b, c):(a=b=c)\\}\)

Short Answer

Expert verified
For subspace (a), the basis is B = { \((1, -1, 0), (1, 0, -1)\) } and the dimension is 2. For subspace (b), the basis is B = { \((1, 1, 1)\) } and the dimension is 1.

Step by step solution

01

Write the given equation as a system of linear equations

We are given the equation \(a + b + c = 0\). We can rewrite this equation as a matrix equation: \[ \begin{bmatrix} 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix} \]
02

Find a basis for the null space of the matrix

Now, we can find the basis for the null space of the matrix: \[ \begin{bmatrix} 1 & 1 & 1 \\ \end{bmatrix} \] This is a single-dimensional subspace of 鈩澛�, hence the dimension of its null space is 2. Let's find two linearly independent vectors that belong to the null space: \[a + b + c = 0\] We can set \(a = 1, b = -1\), and \(c = 0\), which gives us the vector: \(v_1 = (1, -1, 0)\) We can also set \(a = 1, b = 0\), and \(c = -1\), which gives us the vector: \(v_2 = (1, 0, -1)\)
03

State the basis and dimension of the subspace

The basis for the subspace W is the set of vectors: B = { \(v_1, v_2\) } = { \((1, -1, 0), (1, 0, -1)\) } The dimension of the subspace is 2. For subspace (b):
04

Write the given equation as a system of linear equations

We are given the equation \(a = b = c\). We can rewrite this equation as the following system of linear equations: \[a - b = 0\] \[a - c = 0\] Using matrices, we can rewrite this system as: \[ \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]
05

Find a basis for the null space of the matrix

Now, we can find the basis for the null space of the matrix: \[ \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ \end{bmatrix} \] This is a two-dimensional subspace of 鈩澛�, hence the dimension of its null space is 1. Let's find a vector that belongs to the null space: \[a - b = 0\] \[a - c = 0\] We can set \(a = 1\), which implies \(b = 1\) and \(c = 1\), giving us the vector: \(v_3 = (1, 1, 1)\)
06

State the basis and dimension of the subspace

The basis for the subspace W is the set of vectors: B = { \(v_3\) } = { \((1, 1, 1)\) } The dimension of the subspace is 1.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that \(\operatorname{span}(S)=\operatorname{span}(S \cup\\{0\\}) .\) That is, by joining or deleting the zero vector from a set, we do not change the space spanned by the set.

Suppose \(U\) and \(W\) are subspaces of \(V\) for which \(U \cup W\) is a subspace. Show that \(U \subseteq W\) or \(W \subseteq U\).

Suppose \(\left(a_{11}, \ldots, a_{1 n}\right),\left(a_{21}, \ldots, a_{2 n}\right), \ldots,\left(a_{m 1}, \ldots, a_{m n}\right)\) are linearly independent vectors in \(K^{n},\) and suppose \(v_{1}, v_{2}, \ldots, v_{n}\) are linearly independent vectors in a vector space \(V\) over \(K\). Show that the following vectors are also linearly independent: \\[w_{1}=a_{11} v_{1}+\cdots+a_{1 n} v_{n}, \quad w_{2}=a_{21} v_{1}+\cdots+a_{2 n} v_{n}, \quad \ldots, \quad w_{m}=a_{m 1} v_{1}+\cdots+a_{m n} v_{n}\\]

Find the coordinate vector of \(A=\left[\begin{array}{ll}2 & 3 \\ 4 & -7\end{array}\right]\) in the real vector space \(\mathbf{M}=\mathbf{M}_{2,2}\) relative to (a) the basis \(S=\left\\{\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right],\left[\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right],\left[\begin{array}{rr}1 & -1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\right\\}\), (b) the usual basis \(E=\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\}\)

Let \(v_{1}, \ldots, v_{n}\) belong to a vector space \(V\) over \(K,\) and let \(P=\left[a_{i j}\right]\) be an \(n\) -square matrix over \(K\). Let \\[w_{1}=a_{11} v_{1}+a_{12} v_{2}+\cdots+a_{1 n} v_{n}, \quad \ldots, \quad w_{n}=a_{n 1} v_{1}+a_{n 2} v_{2}+\cdots+a_{n n} v_{n}\\] (a) Suppose \(P\) is invertible. Show that \(\left\\{w_{i}\right\\}\) and \(\left\\{v_{i}\right\\}\) span the same space; hence, \(\left\\{w_{i}\right\\}\) is independent if and only if \(\left\\{v_{i}\right\\}\) is independent. (b) Suppose \(P\) is not invertible. Show that \(\left\\{w_{i}\right\\}\) is dependent (c) Suppose \(\left\\{w_{i}\right\\}\) is independent. Show that \(P\) is invertible.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.