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Chapter 20: Problem 14

If a vector space \(V\) is spanned by \(n\) vectors, show that any set of \(m\) vectors in \(V\) must be linearly dependent for \(m>n\).

Short Answer

Expert verified
Question: Prove that if a vector space V is spanned by n vectors, then any set of m vectors with m > n in V is linearly dependent. Answer: We proved that when a vector space V is spanned by n vectors and we have a set S with m vectors, where m > n, there exists a nontrivial linear combination of the m vectors in S that results in the zero vector. This shows that the set S is linearly dependent.

Step by step solution

01

Understand the given information

Given a vector space \(V\) that is spanned by \(n\) vectors. It means that any vector in \(V\) can be represented as a linear combination of these \(n\) vectors. When a set of vectors in \(V\) has \(m\) elements, where \(m > n\), we want to prove that this set must be linearly dependent.
02

Define linear dependence

A set of vectors is linearly dependent if there is a nontrivial linear combination of the vectors that results in the zero vector. A nontrivial linear combination means that not all of the coefficients in the linear combination are zero.
03

Consider a set with \(m\) vectors where \(m>n\)

Let's consider a set \(S\) with \(m\) vectors in \(V\), where \(m > n\). We can represent each of these \(m\) vectors as a linear combination of the \(n\) vectors that span \(V\). Let the set \(B = \{b_1, b_2, ..., b_n\}\) be the basis of \(V\) that has \(n\) elements, and set \(S = \{v_1, v_2, ..., v_m\}\) have \(m\) elements. Then, we can write each \(v_i\) as: \(v_i = \sum_{j=1}^n a_{ij} b_j\) for every \(i=1,2,...,m\).
04

Form a linear combination of the \(m\) vectors

We want to find a linear combination of the \(m\) vectors in \(S\) that results in the zero vector. In other words, we need to find coefficients \(c_1, c_2, ..., c_m\) such that: \(\sum_{i=1}^m c_i v_i = 0.\) Substitute each \(v_i\) with its representation using the basis vectors \(b_j\): \(\sum_{i=1}^m c_i (\sum_{j=1}^n a_{ij} b_j) = 0.\)
05

Rearrange the expression and analyze the coefficients

Rearrange the expression, grouping the basis vectors: \(\sum_{j=1}^n (\sum_{i=1}^m c_i a_{ij}) b_j = 0.\) By definition, the basis vectors \(b_j\) are linearly independent. So, the only way for the linear combination to equal zero is if each of the coefficients is zero: \(\sum_{i=1}^m c_i a_{ij} = 0\) for every \(j=1,2,...,n\). Since this system of linear equations contains \(n\) equations with \(m\) unknowns (\(c_1, c_2, ..., c_m\)) and \(m > n\), there are more unknowns than equations. According to linear algebra, such a system has a nontrivial solution, meaning that at least one of the coefficients \(c_i\) is nonzero.
06

Conclude that the set is linearly dependent

We found a nontrivial linear combination of the \(m\) vectors in \(S\) that results in the zero vector. This means the set of \(m\) vectors (\(m>n\)) in \(V\) is linearly dependent.

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Key Concepts

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

Vector space
A vector space is a fundamental concept in linear algebra and is essentially a collection of vectors. These vectors abide by specific mathematical rules for addition and scalar multiplication. In simpler terms, a vector space is a playground where vectors can be added together, and they can be stretched or shrunk by numbers, known as scalars. Let鈥檚 look at some key properties that define a vector space:

  • Closure under addition and scalar multiplication: This means that if you take any two vectors in the vector space and add them, their sum is also in the vector space. Likewise, if a vector is multiplied by a scalar, the product is still within the space.
  • Existence of a zero vector: This zero vector is like adding zero in numbers. Any vector added to the zero vector results in the original vector again.
  • Existence of additive inverses: For any vector in the space, there must be another vector which, when added to it, will give the zero vector.
Understanding a vector space is crucial as it lays the groundwork for exploring linear combinations, bases, and the overall structure of the space.
Basis of a vector space
To navigate and describe a vector space efficiently, we need a basis. A basis is a set of vectors that are both linearly independent and span the entire vector space. Think of it like the building blocks of the space.

  • Linearly independent: This means that none of the vectors in the basis can be written as a linear combination of the other vectors. They are unique in terms of direction.
  • Spanning the vector space: Every vector in the vector space can be expressed as a linear combination of the basis vectors. Essentially, with the basis vectors, you can create or describe any vector within that space.
An ideal way to conceptualize a basis is to consider the natural coordinate system in a 3-dimensional space. The vectors \(\vec{i}, \vec{j}, \vec{k}\) along the x, y, and z axes form a basis. Every point or vector in 3D space can be described using these three vectors.
Linear combination
A linear combination involves combining several vectors using scalar multipliers, called coefficients. This combination can form new vectors in a vector space. Let's break this down:

  • Scalars: These are numbers that are used to multiply each of the vectors in the combination.
  • Combining vectors: To form a linear combination, you multiply each vector with a corresponding scalar and then add them up. For example, in a simple vector space with vectors \(\vec{v}_1\) and \(\vec{v}_2\), a linear combination might look like \((a_1 \vec{v}_1 + a_2 \vec{v}_2)\), where \(a_1\) and \(a_2\) are scalars.
Linear combinations allow us to analyze vector spaces and find relationships between vectors, such as determining linear dependence or independence. Understanding linear combinations is like having the tools to sculpt the landscape of vector spaces.

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Most popular questions from this chapter

Prove that the set \(P_{n}\) of all polynomials of degree less than \(n\) form a subspace of the vector space \(F[x]\). Find a basis for \(P_{n}\) and compute the dimension of \(P_{n}\).

Let \(F\) be a field and denote the set of \(n\) -tuples of \(F\) by \(F^{n}\). Given vectors \(u=\) \(\left(u_{1}, \ldots, u_{n}\right)\) and \(v=\left(v_{1}, \ldots, v_{n}\right)\) in \(F^{n}\) and \(\alpha\) in \(F,\) define vector addition by $$ u+v=\left(u_{1}, \ldots, u_{n}\right)+\left(v_{1}, \ldots, v_{n}\right)=\left(u_{1}+v_{1}, \ldots, u_{n}+v_{n}\right) $$ and scalar multiplication by $$ \alpha u=\alpha\left(u_{1}, \ldots, u_{n}\right)=\left(\alpha u_{1}, \ldots, \alpha u_{n}\right) $$ Prove that \(F^{n}\) is a vector space of dimension \(n\) under these operations.

Prove that any set of vectors containing \(\mathbf{0}\) is linearly dependent.

If \(F\) is a field, show that \(F[x]\) is a vector space over \(F\), where the vectors in \(F[x]\) are polynomials. Vector addition is polynomial addition, and scalar multiplication is defined by \(\alpha p(x)\) for \(\alpha \in F\).

Direct Sums. \(\quad\) Let \(U\) and \(V\) be subspaces of a vector space \(W\). The sum of \(U\) and \(V,\) denoted \(U+V\), is defined to be the set of all vectors of the form \(u+v,\) where \(u \in U\) and \(v \in V\) (a) Prove that \(U+V\) and \(U \cap V\) are subspaces of \(W\). (b) If \(U+V=W\) and \(U \cap V=\mathbf{0},\) then \(W\) is said to be the direct sum. In this case, we write \(W=U \oplus V\). Show that every element \(w \in W\) can be written uniquely as \(w=u+v,\) where \(u \in U\) and \(v \in V\) (c) Let \(U\) be a subspace of dimension \(k\) of a vector space \(W\) of dimension \(n\). Prove that there exists a subspace \(V\) of dimension \(n-k\) such that \(W=U \oplus V\). Is the subspace \(V\) unique? (d) If \(U\) and \(V\) are arbitrary subspaces of a vector space \(W\), show that $$ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V-\operatorname{dim}(U \cap V) $$
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