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Chapter 10: Problem 21

Let \(W\) be a subspace of a vector space \(V\). Show that the following are equivalent: (i) \(\quad u \in v+W\) (ii) \(\quad u-v \in W\) (iii) \(\quad v \in u+W\)

Short Answer

Expert verified
We have shown that the conditions (i), (ii), and (iii) are equivalent by proving the following implications: (i) implies (ii), (ii) implies (iii), and (iii) implies (i). These implications are based on the properties of vector addition and subtraction, as well as the properties of a subspace.

Step by step solution

01

Show that (i) implies (ii)

Suppose \(u \in v+W\). This means that there exists some \(w_1 \in W\) such that \(u=v+w_1\). Now we need to show that \(u - v \in W\). So let's subtract \(v\) from both sides of the equation: \[u - v = (v + w_1) - v\] Now use the properties of vector addition and subtraction: \[u - v = w_1\] Since \(w_1 \in W\), we have shown that \(u - v \in W\). Thus, (i) implies (ii).
02

Show that (ii) implies (iii)

Suppose \(u - v \in W\). This means that there exists some \(w_2 \in W\) such that \(u - v = w_2\). Now we need to show that \(v \in u + W\). So let's add \(v\) to both sides of the equation: \[u - v + v = w_2 + v\] Now use the properties of vector addition and subtraction: \[u = w_2 + v\] This equation tells us there exists some \(w_2 \in W\) such that \(u = w_2 + v\), which is equivalent to saying \(v \in u + W\). Thus, (ii) implies (iii).
03

Show that (iii) implies (i)

Suppose \(v \in u + W\). This means that there exists some \(w_3 \in W\) such that \(v = u + w_3\). Now we need to show that \(u \in v + W\). To do this, let's rearrange the equation as follows: \[u = v - w_3\] Since \(w_3 \in W\), and \(W\) is a subspace, its additive inverse \(-w_3 \in W\). Therefore, we have shown that \(u \in v + W\). Thus, (iii) implies (i). Since we have shown that (i) implies (ii), (ii) implies (iii) and (iii) implies (i), we have proved that the conditions (i), (ii), and (iii) are equivalent.

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Key Concepts

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

Vector Addition and Subtraction
Vector addition and subtraction are foundational operations in the study of vector spaces. When we add or subtract vectors, we combine or differentiate their magnitudes and directions. Mathematically, if we have two vectors, say \(v\) and \(w\), we write their sum as \(v + w\), representing a third vector that starts from the tail of \(v\) and ends at the head of \(w\), drawn in a tip-to-tail fashion.

To subtract a vector, such as \(w\) from \(v\), we simply reverse the direction of \(w\) to get its additive inverse, which we denote by \( -w \), and then add it to \(v\), which is \(v + (-w)\), or more simply, \(v - w\). Visualization can greatly assist in understanding these operations: vector addition forms a parallelogram, while subtraction creates a straight line between the terminal points of the vectors involved.

When working within a vector space, or its subspace, these operations adhere to specific properties that ensure the structure of the space is conserved. Understanding how to manipulate vectors through addition and subtraction is crucial to grasping more complex concepts such as linearity, bases, and dimensionality within vector spaces.
Equivalent Conditions in Vector Spaces

Grasping Equivalence

Within vector spaces, certain conditions can be equivalent, meaning they convey the same information or lead to the same result. The equivalence of conditions is a powerful concept as it shows how different-looking statements can fundamentally represent the same thing. In our exercise context, we investigated equivalencies between three conditions involving vector addition and subspace membership.

Each condition conveyed a relationship between vectors and a subspace. However, despite their apparent differences, they were proven to be interchangeable. Through a series of logical steps, we showed that if one condition holds true, so do the others. This reveals an underlying symmetry in the structure of vector spaces and subspaces, demonstrating that the operations and relationships within them are highly interconnected.

This interconnectedness is indicative of the consistent rules governing vector spaces, such as closure under addition and scalar multiplication, which highlights the importance of fully understanding the properties of these mathematical structures to work with vectors effectively.
Properties of Subspaces

Defining a Subspace

A subspace is essentially a 'space within a space' 鈥� a subset of a vector space that is, in itself, a vector space under the same operations of addition and scalar multiplication. However, it's not enough for a subset just be a part of a larger space to be called a subspace. It must satisfy three fundamental properties:
  • The zero vector of the larger space is also in the subspace.
  • It is closed under vector addition, meaning the sum of any two vectors in the subspace is also in the subspace.
  • It is closed under scalar multiplication, meaning the product of any scalar and any vector in the subspace is also in the subspace.
These properties ensure that the subspace maintains the structure and rules necessary to be considered a vector space on its own. During our exercise, we implicitly used these properties; for instance, when we showed that the additive inverse of a vector in a subspace also belongs to the subspace, it was an application of closure under scalar multiplication (since \( -1 \times w_3 \in W\)). Such properties are not just theoretical niceties 鈥� they're practical tools that allow us to predict and confirm the behavior of vectors within subspaces, thereby simplifying and solving complex problems within this mathematical framework.

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Most popular questions from this chapter

Prove Theorem 10.10: Let \(T: V \rightarrow V\) be a nilpotent operator of index \(k .\) Then \(T\) has a unique block diagonal matrix representation consisting of Jordan nilpotent blocks \(N\). There is at least one \(N\) of order \(k,\) and all other \(N\) are of orders \(\leq k\). The total number of \(N\) of all orders is equal to the nullity of \(T\).

Let \(T: V\) be linear. Suppose, for \(v \in V, T^{k}(v)=0\) but \(T^{k-1}(v) \neq 0\).

Let \(W\) be invariant under \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\). Prove \(W\) is also invariant under \(T_{1}+T_{2}\) and \(T_{1} T_{2}\).

Prove Theorem 10.8: In Theorem \(10.7\) (Problem 10.9), if \(f(t)\) is the minimal polynomial of \(T\) (and \(g(t)\) and \(h(t)\) are monic), then \(g(t)\) is the minimal polynomial of the restriction \(T_{1}\) of \(T\) to \(U\) and \(h(t)\) is the minimal polynomial of the restriction \(T_{2}\) of \(T\) to \(W\). Let \(m_{1}(t)\) and \(m_{2}(t)\) be the minimal polynomials of \(T_{1}\) and \(T_{2}\), respectively. Note that \(g\left(T_{1}\right)=0\) and \(h\left(T_{2}\right)=0\) because \(U=\) Ker \(g(T)\) and \(W=\) Ker \(h(T)\). Thus, $$ m_{1}(t) \text { divides } g(t) \quad \text { and } \quad m_{2}(t) \text { divides } h(t) $$ By Problem 10.9, \(f(t)\) is the least common multiple of \(m_{1}(t)\) and \(m_{2}(t) .\) But \(m_{1}(t)\) and \(m_{2}(t)\) are relatively prime because \(g(t)\) and \(h(t)\) are relatively prime. Accordingly, \(f(t)=m_{1}(t) m_{2}(t)\). We also have that \(f(t)=g(t) h(t)\). These two equations together with (1) and the fact that all the polynomials are monic imply that \(g(t)=m_{1}(t)\) and \(h(t)=m_{2}(t)\), as required.

Let \(\left.T: V \text { be linear. Let } X=\operatorname{Ker} T^{i-2}, Y=\operatorname{Ker} T^{i-1}, Z=\operatorname{Ker} T^{i} . \text { Therefore (Problem } 10.14\right)\) \(X \subseteq Y \subseteq Z .\) 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.
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