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Chapter 3: Problem 13

Prove that any finite set of vectors that contains the zero vector must be linearly dependent.

Short Answer

Expert verified
A finite set of vectors S contains the zero vector \(\vec{0}\) and other vectors. By choosing one of the coefficients to be 1 (say, \(c_1 = 1\)) and the rest to be 0, we can express the zero vector as a linear combination of the vectors in the set: \[1\vec{0} + 0\vec{v}_2 + 0\vec{v}_3 + \cdots + 0\vec{v}_n = \vec{0}\] Since at least one of the coefficients (namely, \(c_1=1\)) is nonzero and the equation holds, the set S is linearly dependent, by the definition of linear dependence. Thus, any finite set of vectors that contains the zero vector must be linearly dependent.

Step by step solution

01

State the given information

We are given a finite set of vectors, say S, that contains the zero vector, denoted as \(\vec{0}\).
02

Define linear dependence

A set of vectors is linearly dependent if at least one vector in the set can be written as a linear combination of the other vectors in the set. In other words, there exist scalar coefficients, say \(c_1\), \(c_2\), \(c_3\), ... \(c_n\), such that at least one of the coefficients is nonzero and the following equation holds: \[c_1\vec{v}_1 + c_2\vec{v}_2 + \cdots + c_n\vec{v}_n = \vec{0}\]
03

Prove that the set containing the zero vector is linearly dependent

Now, let's consider the finite set S that contains the zero vector \(\vec{0}\) along with other vectors. By choosing one of the coefficients to be 1 (say, \(c_1 = 1\)) and the rest to be 0, we can express the zero vector as a linear combination of the vectors in the set: \[1\vec{0} + 0\vec{v}_2 + 0\vec{v}_3 + \cdots + 0\vec{v}_n = \vec{0}\] Since at least one of the coefficients (namely, \(c_1=1\)) is nonzero and the equation holds, the set S is linearly dependent, by the definition of linear dependence. Thus, we have proved that any finite set of vectors that contains the zero vector must be linearly dependent.

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Most popular questions from this chapter

We can define a one-to-one correspondence between the elements of \(P_{n}\) and \(\mathbb{R}^{n}\) by $$\begin{array}{l} p(x)=a_{1}+a_{2} x+\cdots+a_{n} x^{n-1} \\ \leftrightarrow\left(a_{1}, \ldots, a_{n}\right)^{T}=\mathbf{a} \end{array}$$ Show that if \(p \leftrightarrow \mathbf{a}\) and \(q \leftrightarrow \mathbf{b},\) then (a) \(\alpha p \leftrightarrow \alpha\) a for any scalar \(\alpha\) (b) \(p+q \leftrightarrow \mathbf{a}+\mathbf{b}\) [In general, two vector spaces are said to be isomorphic if their elements can be put into a one-to-one correspondence that is preserved under scalar multiplication and addition as in (a) and (b).]

Let \(Z\) denote the set of all integers with addition defined in the usual way, and define scalar multiplication, denoted o, by $$\alpha \circ k=\mathbb{I} \alpha \| \cdot k \quad \text { for all } \quad k \in Z$$ where \([[\alpha]]\) denotes the greatest integer less than or equal to \(\alpha .\) For example, $$2.25 \circ 4=[[2.25] \cdot 4=2 \cdot 4=8$$ Show that \(Z\), together with these operations, is not a yect fail to hold?

Let \(A\) and \(B\) be \(n \times n\) matrices. (a) Show that \(A B=O\) if and only if the column space of \(B\) is a subspace of the null space of \(A\) (b) Show that if \(A B=O\), then the sum of the ranks of \(A\) and \(B\) cannot exceed \(n\).

Determine whether the following are subspaces of \(C[-1,1]:\) (a) The set of functions \(f\) in \(C[-1,1]\) such that \(f(-1)=f(1)\) (b) The set of odd functions in \(C[-1,1]\) (c) The set of continuous nondecreasing functions on [-1,1] (d) The set of functions \(f\) in \(C[-1,1]\) such that \(f(-1)=0\) and \(f(1)=0\) (e) The set of functions \(f\) in \(C[-1,1]\) such that \(f(-1)=0\) or \(f(1)=0\)

Let \(U\) and \(V\) be subspaces of a vector space \(W\) Prove that their intersection \(U \cap V\) is also a subspace of \(W\)
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