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Let \(D, I,\) and \(X\) denote finite, nonempty sets of vectors in a vector space \(V\). Assume that \(D\) is dependent and \(I\) is independent. In each case answer yes or no, and defend your answer. a. If \(X \supseteq D,\) must \(X\) be dependent? b. If \(X \subseteq D,\) must \(X\) be dependent? c. If \(X \supseteq I,\) must \(X\) be independent? d. If \(X \subseteq I,\) must \(X\) be independent?

Short Answer

Expert verified
a. Yes. b. No. c. No. d. Yes.

Step by step solution

01

Identify the dependent set

By definition, if a set \(D\) is dependent, it means that there is at least one vector in \(D\) that can be written as a linear combination of the others. Thus, \(D\) is not linearly independent.
02

Identify the independent set

An independent set \(I\) has the property that no vector in \(I\) can be expressed as a linear combination of the others. Therefore, \(I\) is linearly independent.
03

Evaluate part (a)

Given that \(X \supseteq D\), all vectors in \(D\) are in \(X\), and since \(D\) is dependent, \(X\) must also be dependent because it contains all vectors of a dependent set.
04

Evaluate part (b)

For \(X \subseteq D\), since \(X\) is only a subset of the dependent set \(D\), \(X\) does not necessarily have to be dependent. It could be a smaller set that happens to be independent. So, the answer is no.
05

Evaluate part (c)

If \(X \supseteq I\), then \(X\) includes all vectors from the independent set \(I\). However, \(X\) might have additional vectors that create dependencies. Thus, \(X\) is not necessarily independent. The answer is no.
06

Evaluate part (d)

For \(X \subseteq I\), any subset of an independent set is itself independent because if any of the vectors in \(X\) could be written as a combination of others, \(I\) wouldn't be independent. Hence, the answer is yes.

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Key Concepts

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

Dependent Sets in Vector Spaces
In vector spaces, a set of vectors is termed "dependent" if there is at least one vector in the set that can be expressed as a linear combination of the others. This essentially means that one or more vectors are redundant as they do not contribute additional dimensions to the span of the set. Dependent sets have the inherent property that the vectors involve do not hold the capability to maintain the distinction among their respective spatial directions.
Here are some key points about dependent sets:
  • If a set of vectors is dependent, you can remove at least one vector without changing the span of that set.
  • Any set containing the zero vector is dependent, as the zero vector can be zero times any vector plus other scaled vectors, maintaining a zero sum.
  • If you add more vectors to a dependent set, the new set remains dependent since the original dependencies still exist.
Independent Sets and Their Properties
Independent sets in vector spaces have a distinct characteristic where no vector in the set can be constructed as a linear combination of the other vectors. This property ensures that each vector adds a new independent direction to the span, effectively expanding the dimensionality of the space covered by the set.
Consider the following properties of independent sets:
  • In an independent set, removing any vector changes the span of the set, implying each vector contributes uniquely.
  • An independent set containing \( n \) vectors will span an \( n \)-dimensional space if they belong to a vector space of at least dimension \( n \).
  • Every subset of an independent set is also independent, preserving the lack of redundancy among vectors.
These properties make independent sets fundamental in defining basis sets, which are the minimal representations that can span vector spaces without redundancy.
Understanding Linear Combinations
A linear combination in vector spaces refers to an expression where vectors are multiplied by scalars (real numbers), and their results are summed. This concept is crucial in vector spaces as it helps describe how vectors can be combined or formed into new vectors, thereby creating spans, which can potentially cover entire spaces.
Here’s a deeper dive into linear combinations:
  • Given vectors \( v_1, v_2, ..., v_n \) in a space, a linear combination is any vector of the form \( c_1v_1 + c_2v_2 + ... + c_nv_n \) where \( c_1, c_2, ..., c_n \) are scalars.
  • If a vector can be written as a linear combination of others, those vectors actually determine the same directions in the space, hinting at dependencies.
  • To form the span of a vector set, you consider all possible linear combinations of the vectors in the set, which can represent any vector in the space when the set is large enough.
Mastery of linear combinations is pivotal in understanding the structure and potential transformations within vector spaces, making this a cornerstone concept in linear algebra.

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Most popular questions from this chapter

Find the Lagrange interpolation expansion of \(f(x)\) relative to \(a_{0}=1, a_{1}=2,\) and \(a_{2}=3\) if: a. \(f(x)=x^{2}+1\) b. \(f(x)=x^{2}+x+1\)

Describe span \(\\{\mathbf{0}\\}\).

\(\mathbf{} \quad\) Let \(\\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\\}\) be independent. Which of the following are dependent? a. \(\\{\mathbf{u}-\mathbf{v}, \mathbf{v}-\mathbf{w}, \mathbf{w}-\mathbf{u}\\}\) b. \(\\{\mathbf{u}+\mathbf{v}, \mathbf{v}+\mathbf{w}, \mathbf{w}+\mathbf{u}\\}\) c. \(\\{\mathbf{u}-\mathbf{v}, \mathbf{v}-\mathbf{w}, \mathbf{w}-\mathbf{z}, \mathbf{z}-\mathbf{u}\\}\) d. \(\\{\mathbf{u}+\mathbf{v}, \mathbf{v}+\mathbf{w}, \mathbf{w}+\mathbf{z}, \mathbf{z}+\mathbf{u}\\}\)

Show that each of the following sets of vectors is independent. a. \(\left\\{1+x, 1-x, x+x^{2}\right\\}\) in \(\mathbf{P}_{2}\) b. \(\left\\{x^{2}, x+1,1-x-x^{2}\right\\}\) in \(\mathbf{P}_{2}\) \(\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{rr}0 & 0 \\ 1 & -1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right]\right\\}\) \(\left\\{\left[\begin{array}{ll}1 & 1 \\ 1 & 0 \\ \text { in } & \mathbf{M}_{22}\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\right\\}\)

Let \(A\) be an \(m \times n\) matrix. For which columns b in \(\mathbb{R}^{m}\) is \(U=\left\\{\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^{n}, A \mathbf{x}=\mathbf{b}\right\\}\) a subspace of \(\mathbb{R}^{n} ?\) Support your answer.

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