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Chapter 3: Problem 16

Suppose \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k} \in \mathbb{R}^{n}\) form a linearly dependent set. Prove that for some \(j\) between 1 and \(k\) we have \(\mathbf{v}_{j} \in \operatorname{Span}\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{j-1}, \mathbf{v}_{j+1}, \ldots, \mathbf{v}_{k}\right)\). That is, one of the vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\) can be written as a linear combination of the remaining vectors.

Short Answer

Expert verified
In this exercise, we showed that for a linearly dependent set of vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k} \in \mathbb{R}^{n}\), there exists at least one vector \(\mathbf{v}_{j}\) that can be written as a linear combination of the remaining vectors. We used the definition of linearly dependent vectors, which means there exists a nontrivial set of coefficients such that the linear combination of the vectors equals the zero vector. We then showed that if at least one of the coefficients is nonzero, we can express \(\mathbf{v}_{j}\) (in this case, \(\mathbf{v}_{1}\)) as a linear combination of the remaining vectors, completing the proof.

Step by step solution

01

Linearly Dependent Definition

A set of vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k} \in \mathbb{R}^{n}\) is linearly dependent if there exists a nontrivial (not all zero) set of coefficients \(c_{1}, \ldots, c_{k}\) such that \[ c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + \cdots + c_{k}\mathbf{v}_{k} = \mathbf{0}. \]
02

Show that \(\mathbf{v}_{j}\) exists

Since \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\) are linearly dependent, we have a nontrivial linear combination equaling the zero vector. If all coefficients \(c_{1}, \ldots, c_{k}\) were zero, then the set would be linearly independent, which contradicts the problem statement. Thus, at least one of the coefficients must be nonzero. Without loss of generality, we can assume \(c_{1} \neq 0\). Then, we can express \(\mathbf{v}_{1}\) as a linear combination of the remaining vectors: \[ \mathbf{v}_{1} = -\frac{c_{2}}{c_{1}}\mathbf{v}_{2} -\frac{c_{3}}{c_{1}}\mathbf{v}_{3} - \cdots - \frac{c_{k}}{c_{1}}\mathbf{v}_{k}. \] Thus, we have shown that at least one vector \(\mathbf{v}_{j}\) (in this case, \(\mathbf{v}_{1}\)) can be written as a linear combination of the remaining vectors, completing the proof.

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Key Concepts

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

Linear Combinations
A **linear combination** is a quintessential concept in linear algebra. It involves combining several vectors with specific coefficients to obtain another vector. This can be mathematically expressed as:
\[ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_k\mathbf{v}_k \]where \(c_1, c_2, \ldots, c_k\) are coefficients, and \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\) are vectors. In this equation, each vector \(\mathbf{v}_i\) is scaled by the coefficient \(c_i\) and summed to produce a new vector.
In linear dependence, at least one vector in a set is a linear combination of others. This is because there are non-zero coefficients that when applied to the vectors and summed, result in the zero vector. This indicates that the vectors are not giving new directions but rather rely on each other to cover the space.
Vector Spaces
**Vector spaces** are a fundamental structure in mathematics, defining a collection of vectors that can be scaled and added. These vectors form environments where linear combinations can exist.

A vector space follows specific rules, such as:
  • Vector addition: Adding any two vectors in the space results in another vector in the same space.
  • Scalar multiplication: Scaling a vector by a real number yields another vector within the same space.
  • Existence of zero vector: The zero vector must be present in the vector space.
  • Inverses: Each vector in the space must have an inverse such that when added together, they result in the zero vector.
Understanding vector spaces helps contextualize where vectors and their combinations live, supporting the idea that linear dependence notes redundancy within the space.
Span
The **span** of a set of vectors refers to all possible vectors you can reach through linear combinations of those vectors. If you think of vectors as arrows pointing from the origin, the span is like painting all the points you can hit with those arrows in different combinations.
For instance, if we take the vectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\), their span is:
\[ \text{Span}(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k) = \{ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_k\mathbf{v}_k \mid c_1, c_2, \ldots, c_k \text{ are real numbers} \} \]
This concept is crucial to linear dependence. A vector being in the span of others means it can be constructed from them. In linearly dependent sets, at least one vector can be expressed in terms of others in the set, demonstrating its redundancy in spanning the vector space.

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Most popular questions from this chapter

Determine the intersection of the subspaces \(\mathcal{P}_{1}\) and \(\mathscr{P}_{2}\) in each case: \({ }^{*}\) a. \(\mathcal{P}_{1}=\operatorname{Span}((1,0,1),(2,1,2)), \mathcal{P}_{2}=\operatorname{Span}((1,-1,0),(1,3,2))\) b. \(\mathscr{P}_{1}=\operatorname{Span}((1,2,2),(0,1,1)), \mathfrak{P}_{2}=\operatorname{Span}((2,1,1),(1,0,0))\) c. \(\mathfrak{P}_{1}=\operatorname{Span}((1,0,-1),(1,2,3)), \mathfrak{P}_{2}=\left\\{\mathbf{x}: x_{1}-x_{2}+x_{3}=0\right\\}\) *d. \(P_{1}=\operatorname{Span}((1,1,0,1),(0,1,1,0)), \mathscr{P}_{2}=\operatorname{Span}((0,0,1,1),(1,1,0,0))\) e. \(\mathcal{P}_{1}=\operatorname{Span}((1,0,1,2),(0,1,0,-1)), \mathfrak{P}_{2}=\operatorname{Span}((1,1,2,1),(1,1,0,1))\)

a. Give a basis for the orthogonal complement of the subspace \(V \subset \mathbb{R}^{4}\) given by $$ V=\left\\{\mathbf{x} \in \mathbb{R}^{4}: x_{1}+x_{2}-2 x_{4}=0, x_{1}-x_{2}-x_{3}+6 x_{4}=0, x_{2}+x_{3}-4 x_{4}=0\right\\} . $$ b. Give a basis for the orthogonal complement of the subspace \(W \subset \mathbb{R}^{4}\) spanned by \((1,1,0,-2),(1,-1,-1,6)\), and \((0,1,1,-4)\). c. Give a matrix \(B\) so that the subspace \(W\) defined in part \(b\) can be written in the form \(W=\mathbf{N}(B)\).

Let \(A\) be an \(m \times n\) matrix and \(B\) be an \(n \times p\) matrix. Prove that a. \(\mathbf{N}(B) \subset \mathbf{N}(A B)\). b. \(\mathbf{C}(A B) \subset \mathbf{C}(A)\). (Hint: Use Proposition 2.1.) c. \(\mathbf{N}(B)=\mathbf{N}(A B)\) when \(A\) is \(n \times n\) and nonsingular. (Hint: See the box on p. 12.) d. \(\mathbf{C}(A B)=\mathbf{C}(A)\) when \(B\) is \(n \times n\) and nonsingular.

Ohm's Law says that \(V=I R\); that is, voltage (in volts) \(=\) current (in amps) \(\times\) resistance (in ohms). Given an electric circuit with \(m\) wires, let \(R_{i}\) denote the resistance in the \(i^{\text {th }}\) wire, and let \(y_{i}\) and \(z_{i}\) denote, respectively, the voltage drop across and current in the \(i^{\text {th }}\) wire, as in the text. Let \(\mathcal{E}=\left(E_{1}, \ldots, E_{m}\right)\), where \(E_{i}\) is the external voltage source in the \(i^{\text {th }}\) wire, and let \(C\) be the diagonal \(m \times m\) matrix whose \(i i\)-entry is \(R_{i}\), \(i=1, \ldots, m\); we assume that all \(R_{i}>0\). Then we have \(\mathbf{y}+\mathcal{E}=C \mathbf{z}\). Let \(A\) denote the incidence matrix for this circuit. a. Prove that for every \(\mathbf{v} \in \mathbf{N}\left(A^{\mathrm{T}}\right)\), we have \(\mathbf{v} \cdot C \mathbf{z}=\mathbf{v} \cdot \varepsilon\), and compare this with the statement of Kirchhoff's second law in Section \(6.3\) of Chapter \(1 .\) b. Assume the network is connected, so that \(\operatorname{rank}(A)=n-1\); delete a column of \(A\) (say the last) and call the resulting matrix \(\bar{A}\). This amounts to grounding the last node. Generalize the result of Exercise 3.4.24 to prove that \(\bar{A}^{\top} C^{-1} \bar{A}\) is nonsingular. (Hint: Write \(C=D^{2}=D D^{\mathrm{T}}\), where \(D\) is the diagonal matrix with entries \(\sqrt{R_{i}}\) ) c. Deduce that for any external voltage sources \(\varepsilon\), there is a unique solution of the equation \(\left(\bar{A}^{\mathrm{T}} C^{-1} \bar{A}\right) \mathbf{x}=\bar{A}^{\mathrm{T}} C^{-1} \mathcal{E}\). d. Deduce that for any external voltage sources \(\mathcal{E}\), the currents in the network are uniquely determined.

Suppose \(U\) and \(V\) are subspaces of \(\mathbb{R}^{n}\). Prove that \((U \cap V)^{\perp}=U^{\perp}+V^{\perp}\). (Hint: Use Exercise 3.1.18 and Proposition 3.6.)
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