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Chapter 10: Problem 71

Let \(W\) be a subspace of \(V\). Suppose the set of cosets \(\left\\{v_{1}+W, v_{2}+W, \ldots, v_{n}+W\right\\}\) 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.

Short Answer

Expert verified
Assuming that the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) is linearly dependent, we have scalars \(c_1, c_2, \ldots, c_n\), not all of them zero, such that \(c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0\). Adding subspace W to both sides, we get \(c_1(v_1 + W) + c_2(v_2 + W) + \cdots + c_n(v_n + W) = 0 + W\). Since the set of cosets in \(V / W\) is linearly independent, this implies \(c_1 = c_2 = \cdots = c_n = 0\), contradicting our initial assumption. Thus, the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) in V is linearly independent.

Step by step solution

01

Definition of linear independence

Recall that the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) is linearly independent if for any scalars \(c_1, c_2, \ldots, c_n\), \[ c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0 \implies c_1 = c_2 = \cdots = c_n = 0 \] We want to prove that the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) in V is linearly independent.
02

Assuming the set of vectors is linearly dependent

Let's assume that the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) is linearly dependent. Then, there exist scalars \(c_1, c_2, \ldots, c_n\), not all of them zero, such that \[ c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0 \]
03

Adding subspace W to the equation

Now, let's add the subspace W to both sides of the equation: \[ c_1(v_1 + W) + c_2(v_2 + W) + \cdots + c_n(v_n + W) = 0 + W \] Since we are given that the set of cosets \(\{v_1 + W, v_2 + W, \ldots, v_n + W\}\) in \(V / W\) is linearly independent, this means that \[ c_1(v_1 + W) + c_2(v_2 + W) + \cdots + c_n(v_n + W) = 0 + W \implies c_1 = c_2 = \cdots = c_n = 0 \]
04

Contradiction and conclusion

However, this contradicts our assumption that the coefficients \(c_1, c_2, \ldots, c_n\), are not all zero. Hence, our assumption that the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) in V is linearly dependent must be false. Therefore, the set of vectors \(\{v_1, v_2, \ldots, v_n\}\) in V is linearly independent.

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Key Concepts

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

Subspace
In linear algebra, a subspace is a critical concept that delves into the idea of "space within a space." A subspace is essentially a subset of a vector space that is closed under vector addition and scalar multiplication. That means if you take any two vectors in this subset and add them, the result will still be in the subset. Similarly, if you take any vector in the subspace and multiply it by a scalar, it remains within the same subset.

Subspaces are important because they help us understand more complex spaces by breaking them down into more manageable parts. They must include the zero vector (or the origin in geometrical space), supporting the structure's integrity. Understanding subspaces helps us work with vector spaces more efficiently by allowing us to focus on specific areas that maintain the essential properties of the larger space.
Vector Spaces
At its core, a vector space is a collection of vectors. But more than just a collection, it comes equipped with two operations - vector addition and scalar multiplication - that follow specific rules.

A vector space must fulfill a list of properties, such as associativity, commutativity in addition, and the existence of an additive identity (like the zero vector), among others. Vector spaces can be of any dimension, from the familiar 2D and 3D that we often visualize, to higher dimensions used in advanced mathematical theories.

Understanding vector spaces is foundational for exploring topics like linear transformations, eigenvalues, and matrices. They are pivotal in many fields, including physics, computer science, and engineering, since they provide a framework for modeling linearly structured data or systems.
Cosets
Cosets offer a fascinating way to partition a vector space into subsets of equal size. The concept of a coset comes from group theory and is used in vector spaces to understand equivalence classes better.

Given a subspace \(W\) of a vector space \(V\), a coset is formed by taking a vector \(v\) from \(V\) and adding every element of \(W\) to it. This forms the coset represented by \(v + W\). Each coset has the same number of elements as \(W\) and partitions the vector space into non-overlapping pieces.

Cosets allow us to simplify complex problems by working within these equivalence classes. In the context of the exercise, their linear independence in \(V/W\) helps us deduce properties about the original vectors, showing how crucial and insightful understanding cosets can be.
Linear Dependence
A fundamental theme in linear algebra is understanding when sets of vectors are dependent or independent. Linear dependence refers to a scenario where at least one vector in a set can be written as a combination of others. This means there are coefficients, not all zero, that can relate these vectors by addition to yield the zero vector.

Recognizing linear dependence is vital because it tells us about the redundancy in a set of vectors. If vectors are dependent, not all of them are needed to span the space they reside in. Thus, identifying and removing dependencies allows for simplification and revealing the core structure of the vector space.

Linear independence, on the other hand, implies that no vector in the set can be replicated through others. This independence characterizes a complete set of vectors, vital for bases in vector spaces, ensuring that each dimension is represented uniquely and without overlap.

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Most popular questions from this chapter

Determine all possible Jordan canonical forms for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{3}(t-5)^{2} .\) In each case, find the minimal polynomial \(m(t)\) Because \(t-2\) has exponent 3 in \(\Delta(t), 2\) must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\), (b) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right],[5],[5]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{rr}2 & 1 \\ & 2\end{array}\right],[2],\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\) (d) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\ & 2\end{array}\right],[2],[5],[5]\right)\), (e) \(\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\\ 5\end{array}\right]\right)\), (f) \(\operatorname{diag}([2],[2],[2],[5],[5])\) The exponent in the minimal polynomial \(m(t)\) is equal to the size of the largest block. Thus, (a) \(m(t)=(t-2)^{3}(t-5)^{2}\), (b) \(m(t)=(t-2)^{3}(t-5)\), (c) \(m(t)=(t-2)^{2}(t-5)^{2}\), (d) \(m(t)=(t-2)^{2}(t-5)\), (e) \(m(t)=(t-2)(t-5)^{2}\), (f) \(m(t)=(t-2)(t-5)\)

Prove that \(Z(u, T)=Z(v, T)\) if and only if \(g(T)(u)=v\) where \(g(t)\) is relatively prime to the \(T\)-annihilator of \(u\).

Prove Theorem \(10.11\) on the Jordan canonical form for an operator \(T\). By the primary decomposition theorem, \(T\) is decomposable into operators \(T_{1}, \ldots, T_{r}\); that is, \(T=T_{1} \oplus \cdots \oplus T_{r}\), where \(\left(t-\lambda_{i}\right)^{m_{i}}\) is the minimal polynomial of \(T_{i}\). Thus, in particular, $$ \left(T_{1}-\lambda_{1} I\right)^{m_{1}}=\mathbf{0}, \ldots, \quad\left(T_{r}-\lambda_{r} I\right)^{m_{r}}=\mathbf{0} $$ Set \(N_{i}=T_{i}-\lambda_{i} I .\) Then, for \(i=1, \ldots, r\), $$ T_{i}=N_{i}+\lambda_{i} I, \quad \text { where } \quad N_{i}^{m^{i}}=\mathbf{0} $$ That is, \(T_{i}\) is the sum of the scalar operator \(\lambda_{i} I\) and a nilpotent operator \(N_{i}\), which is of index \(m_{i}\) because \(\left(t-\lambda_{i}\right)_{i}^{m}\) is the minimal polynomial of \(T_{i}\). Now, by Theorem \(10.10\) on nilpotent operators, we can choose a basis so that \(N_{i}\) is in canonical form. In this basis, \(T_{i}=N_{i}+\lambda_{i} I\) is represented by a block diagonal matrix \(M_{i}\) whose diagonal entries are the matrices \(J_{i j}\). The direct sum \(J\) of the matrices \(M_{i}\) is in Jordan canonical form and, by Theorem \(10.5\), is a matrix representation of \(T\). Last, we must show that the blocks \(J_{i j}\) satisfy the required properties. Property (i) follows from the fact that \(N_{i}\) is of index \(m_{i}\). Property (ii) is true because \(T\) and \(J\) have the same characteristic polynomial. Property (iii) is true because the nullity of \(N_{i}=T_{i}-\lambda_{i} I\) is equal to the geometric multiplicity of the eigenvalue \(\lambda_{i}\). Property (iv) follows from the fact that the \(T_{i}\) and hence the \(N_{i}\) are uniquely determined by \(T\).

Prove Lemma 10.13: Let \(T: V \rightarrow V\) be a linear operator whose minimal polynomial is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial. Then \(V\) is the direct sum of \(T\) -cyclic subspaces \(Z_{i}=Z\left(v_{i}, T\right), i=1, \ldots, r,\) with corresponding \(T\) -annihilators \\[ f(t)^{n_{1}}, f(t)^{n_{2}}, \ldots, f(t)^{n_{r}}, \quad n=n_{1} \geq n_{2} \geq \cdots \geq n_{r} \\]

Suppose \(V=W_{1} \oplus \cdots \oplus W_{r} .\) Let \(E_{i}\) denote the projection of \(V\) into \(W_{i} .\) Prove (i) \(E_{i} E_{j}=0, i \neq j\) (ii) \(I=E_{1}+\cdots+E_{r}\)
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