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Chapter 10: Problem 24

Suppose \(W\) is a subspace of a vector space \(V .\) Show that the operations in Theorem 10.15 are well defined; namely, show that if \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W,\) then (a) \(\quad(u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W\) and (b) \(\quad k u+W=k u^{\prime}+W \quad\) for any \(k \in K\) (a) Because \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W,\) both \(u-u^{\prime}\) and \(v-v^{\prime}\) belong to \(W .\) But then \((u+v)-\left(u^{\prime}+v^{\prime}\right)=\left(u-u^{\prime}\right)+\left(v-v^{\prime}\right) \in W .\) Hence, \((u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W\) (b) Also, because \(u-u^{\prime} \in W\) implies \(k\left(u-u^{\prime}\right) \in W,\) then \(k u-k u^{\prime}=k\left(u-u^{\prime}\right) \in W ;\) accordingly, \(k u+W=k u^{\prime}+W\).

Short Answer

Expert verified
In summary, given that $u+W=u^{\prime}+W$ and $v+W=v^{\prime}+W$, we have proved that: (a) $(u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W$, and (b) $ku+W=ku^{\prime}+W$ for any $k \in K$. These results show that the operations of addition and scalar multiplication in Theorem 10.15 are well-defined.

Step by step solution

01

Proof for part (a)

Given that both \(u-u^{\prime}\) and \(v-v^{\prime}\) belong to \(W\), we can represent them as: \[u-u^{\prime}=w_1 \quad \text{and} \quad v-v^{\prime}=w_2, \text{ where } w_1, w_2 \in W.\] Now, let's add these two equations: \[(u-u^{\prime})+(v-v^{\prime})=w_1+w_2.\] Rearrange this to obtain: \[(u+v)-(u^{\prime}+v^{\prime})=w_1+w_2.\] Since \(w_1+w_2 \in W\), we have: \[(u+v)-(u^{\prime}+v^{\prime}) \in W.\] Thus, it follows that: \[(u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W.\]
02

Proof for part (b)

We are given that \(u-u^{\prime} \in W\). Thus, multiplying both sides of this equation by a scalar \(k \in K\), we have: \[k(u-u^{\prime})=kw_1, \text{ where } w_1 \in W.\] Now, let's rearrange this to obtain: \[ku-ku^{\prime}=kw_1.\] Since \(kw_1 \in W\), we can conclude that: \[ku-ku^{\prime} \in W.\] Finally, we can write this as: \[ku+W=ku^{\prime}+W.\] This concludes the proof for both parts (a) and (b), establishing that the operations of addition and scalar multiplication in Theorem 10.15 are well-defined.

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Key Concepts

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

Subspace
Understanding the concept of a subspace is crucial in linear algebra. It refers to a set of vectors within a vector space that itself acts like a small vector space. Every subspace must obey three main rules: it must contain the zero vector, it must be closed under vector addition, and it must be closed under scalar multiplication.

When we say a subspace is 'closed' under an operation, we mean that performing that operation on elements within the subspace results in an element that is still within the subspace. For instance, if you take any two vectors from a subspace and add them together, the resulting vector is also in the subspace. Similarly, if you multiply any vector from the subspace by any scalar, the result is still in the subspace.

In the exercise provided, we were asked to show that certain operations are well-defined within a subspace. This involves proving that the subspace is consistent with the vector space it is a part of when performing vector addition and scalar multiplication.
Scalar Multiplication
Scalar multiplication is a fundamental operation in linear algebra that involves multiplying a vector by a scalar (a number). In simpler words, it scales the vector by making it longer or shorter while keeping it's direction unchanged, or flipping it's direction if the scalar is negative.

When we multiply a vector by a scalar, each component of the vector is multiplied by that scalar. Mathematically, for a vector \(v = [v_1, v_2, ..., v_n]\) and a scalar \(k\), scalar multiplication is defined as \(kv = [kv_1, kv_2, ..., kv_n]\). It's a straightforward process, but it's important to ensure that the result of scalar multiplication stays within the vector space or subspace to maintain the structure's integrity.

In the textbook solution, the proof for part (b) revolves around showing that scalar multiplication of a vector by a scalar results in another vector in the subspace \(W\). This demonstrates that \(W\) is indeed a valid subspace.
Cosets
Cosets are not as commonly discussed in introductory linear algebra, but they are an important concept, especially when delving into more abstract areas. A coset is what you get when you take a vector space (or subspace) and 'shift' it by a particular vector that may or may not be inside the space.

To explain this with an example, if \(W\) is our subspace and \(v\) is any vector, then the set \(v + W\) represents a coset of \(W\). It consists of all vectors that can be described as \(v + w\) for each \(w\) in \(W\). Cosets play a pivotal role in understanding the structure and classification of subspaces within a vector space. In the exercise, we looked at when two cosets, \(u + W\) and \(u' + W\), are equal, which helps solidify our understanding of vector addition within the framework of subspaces and cosets.
Linear Algebra
Linear algebra is the branch of mathematics that deals with vectors, vector spaces, linear transformations, and systems of linear equations. It's the backbone of many areas of mathematics and is crucial for fields such as physics, engineering, computer science, and more.

The core operations of linear algebra include vector addition, scalar multiplication, and matrix multiplication. These operations are used to solve linear systems, perform transformations, and analyze data. Linear algebra gives us a language and a set of tools for describing and solving a wide array of problems.

In both parts (a) and (b) of the exercise, the principles of linear algebra are applied to explore and verify the properties of vector space operations within subspaces. It shows how these fundamental concepts of linear algebra work together to form a coherent and powerful mathematical structure.

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Most popular questions from this chapter

Suppose dim \(V=n .\) Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

Let \(W\) be the solution space of the homogeneous equation \(2 x+3 y+4 z=0 .\) Describe the cosets of \(W\) in \(\mathbf{R}^{3}\) \(W\) is a plane through the origin \(O=(0,0,0),\) and the cosets of \(W\) are the planes parallel to \(W\) Equivalently, the cosets of \(W\) are the solution sets of the family of equations \\[ 2 x+3 y+4 z=k, \quad k \in \mathbf{R} \\] In fact, the coset \(v+W\), where \(v=(a, b, c),\) is the solution set of the linear equation \\[ 2 x+3 y+4 z=2 a+3 b+4 c \quad \text { or } \quad 2(x-a)+3(y-b)+4(z-c)=0 \\]

Show that any Jordan nilpotent block matrix \(N\) is similar to its transpose \(N^{T}\) (the matrix with 1 's below the diagonal and \(0^{\prime}\) s elsewhere)

Let \(\left.T: V \text { be linear. Let } X=\operatorname{Ker} T^{i-2}, Y=\operatorname{Ker} T^{i-1}, Z=\text { Ker } T^{i} . \text { Therefore (Problem } 10.14\right)\) \(X \subseteq Y \subseteq Z .\) Suppose \\[ \left\\{u_{1}, \ldots, u_{r}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{t}\right\\} \\] are bases of \(X, Y, Z,\) respectively. Show that \\[ S=\left\\{u_{1}, \ldots, u_{r}, T\left(w_{1}\right), \ldots, T\left(w_{t}\right)\right\\} \\] is contained in \(Y\) and is linearly independent. By Problem \(10.14, T(Z) \subseteq Y\), and hence \(S \subseteq Y\). Now suppose \(S\) is linearly dependent. Then there exists a relation \\[ a_{1} u_{1}+\cdots+a_{r} u_{r}+b_{1} T\left(w_{1}\right)+\cdots+b_{t} T\left(w_{t}\right)=0 \\] where at least one coefficient is not zero. Furthermore, because \(\left\\{u_{i}\right\\}\) is independent, at least one of the \(b_{k}\) must be nonzero. Transposing, we find Hence, \\[ \begin{array}{c} b_{1} T\left(w_{1}\right)+\cdots+b_{t} T\left(w_{t}\right)=-a_{1} u_{1}-\cdots-a_{r} u_{r} \in X=\operatorname{Ker} T^{i-2} \\ T^{i-2}\left(b_{1} T\left(w_{1}\right)+\cdots+b_{t} T\left(w_{t}\right)\right)=0 \end{array} \\] Thus, \\[ T^{i-1}\left(b_{1} w_{1}+\cdots+b_{t} w_{t}\right)=0, \quad \text { and so } \quad b_{1} w_{1}+\cdots+b_{t} w_{t} \in Y=\text { Ker } T^{i-1} \\] Because \(\left\\{u_{i}, v_{j}\right\\}\) generates \(Y,\) we obtain a relation among the \(u_{i}, v_{j}, w_{k}\) where one of the coefficients (i.e., one of the \(b_{k}\) ) is not zero. This contradicts the fact that \(\left\\{u_{i}, v_{j}, w_{k}\right\\}\) is independent. Hence, \(S\) must also be independent.

Let \(W\) be a subspace of \(V\). Suppose the set of cosets \(\left\\{v_{1}+W, \quad v_{2}+W, \ldots, v_{n}+W\right\\}\) in \(V / W\) is linearly independent. Show that the set of vectors \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) in \(V\) is also linearly independent.
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