/*! 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 18 Show that if \(U\) and \(V\) are... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 18

Show that if \(U\) and \(V\) are subspaces of \(\mathbb{R}^{n}\) and \(U \cap V=\\{0\\},\) then \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V \\]

Short Answer

Expert verified
Taking the basis of U as \(B_{U}=\{u_{1}, u_{2}, \ldots, u_{k}\}\) and V as \(B_{V}=\{v_{1}, v_{2}, \ldots, v_{l}\}\), we can show that the sum of U and V, U+V, has a basis \(B_{U\oplus V} = B_{U} \cup B_{V}\), formed from linearly independent vectors. Since the intersection is only the zero vector, it follows that the dimension of U+V equals the sum of their individual dimensions, i.e., \(\operatorname{dim}(U+V)=\operatorname{dim}(U)+\operatorname{dim}(V)\).

Step by step solution

01

Write definitions of the sum and dimension of the subspaces U and V

The sum of two subspaces U and V, denoted by U+V, is the set of all vectors in \(\mathbb{R}^{n}\) that can be written as the sum of a vector in U and a vector in V. The dimension of a subspace W, denoted by \(\operatorname{dim}(W)\), is the number of vectors in a basis for W.
02

Relate the dimension of the sum of the subspaces U and V

To find the dimension of U+V, we'll first write down the basis for U+V as a union of the bases of U and V. Let \(B_{U}=\{u_{1}, u_{2}, \ldots, u_{k}\}\) be a basis for U and \(B_{V}=\{v_{1}, v_{2}, \ldots, v_{l}\}\) be a basis for V. Since the intersection of U and V is only the zero vector, the vectors in basis \(B_{U}\) and \(B_{V}\) are linearly independent. Therefore, the set \(B_{U\oplus V} = B_{U} \cup B_{V}\) forms a basis for U+V. Notice that \(\operatorname{dim}(U+V) = k+l = \operatorname{dim}(U) + \operatorname{dim}(V)\).
03

Relate the dimension of U+V to the dimensions of U and V

Now that we have expressed the dimension of U+V in terms of the dimensions of U and V, we can use this to prove the given statement. Since we have shown that the set \(B_{U\oplus V} = B_{U} \cup B_{V}\) forms a basis for U+V, and we know that the union of two linearly independent sets is also linearly independent, it follows that the dimension of the sum of the subspaces U and V is equal to the sum of the dimensions of the individual subspaces. Thus, we have proven that if U and V are subspaces of \(\mathbb{R}^{n}\) with their intersection being the zero vector, then \(\operatorname{dim}(U+V)=\operatorname{dim}(U)+\operatorname{dim}(V)\).

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

Let \(A\) be an \(m \times n\) matrix whose rank is equal to \(n\) If \(A \mathbf{c}=A \mathbf{d},\) does this imply that \(\mathbf{c}\) must be equal to \(\mathbf{d} ?\) What if the rank of \(A\) is less than \(n\) ? Explain your answers.

Let \(\mathbb{R}^{+}\) denote the set of positive real numbers. Define the operation of scalar multiplication, denoted ?, by $$\alpha \circ x=x^{\alpha}$$ for each \(x \in \mathbb{R}^{+}\) and for any real number \(\alpha\). Define the operation of addition, denoted \(\oplus,\) by $$x \oplus y=x \cdot y \quad \text { for all } \quad x, y \in \mathbb{R}^{+}$$ Thus, for this system, the scalar product of -3 \(\operatorname{times} \frac{1}{2}\) is given by $$-3 \circ \frac{1}{2}=\left(\frac{1}{2}\right)^{-3}=8$$ and the sum of 2 and 5 is given by $$2 \oplus 5=2 \cdot 5=10$$ Is \(\mathbb{R}^{+}\) a vector space with these operations? Prove your answer.

Let \(A\) and \(B\) be row-equivalent matrices. (a) Show that the dimension of the column space of \(A\) equals the dimension of the column space of \(B\) (b) Are the column spaces of the two matrices necessarily the same? Justify your answer.

Let \(S\) be the vector space of infinite sequences defined in Exercise 15 of Section \(1 .\) Let \(S_{0}\) be the set of \(\left\\{a_{n}\right\\}\) with the property that \(a_{n} \rightarrow 0\) as \(n \rightarrow \infty\) Show that \(S_{0}\) is a subspace of \(S\).

Determine whether the following vectors are linearly independent in \(\mathbb{R}^{2 \times 2}\) : (a) \(\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right)\) (b) \(\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right),\left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right)\) (c) \(\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right),\left(\begin{array}{ll}2 & 3 \\ 0 & 2\end{array}\right)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.