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In Exercises 5.2 .1-5.2 .6 we write vectors \(\mathbb{R}^{n}\) as rows. Which of the following subsets are independent? Support your answer. $$ \begin{aligned} &\begin{array}{l} \text { a. }\\{(1,-1,0),(3,2,-1),(3,5,-2)\\} \text { in } \mathbb{R}^{3} \\ \text { b. }\\{(1,1,1),(1,-1,1),(0,0,1)\\} \text { in } \mathbb{R}^{3} \end{array}\\\ &\text { c. } \begin{array}{l} \\{(1,-1,1,-1),(2,0,1,0),(0,-2,1,-2)\\} \text { in } \\ \mathbb{R}^{4} \end{array}\\\ &\text { d. }\\{(1,1,0,0),(1,0,1,0), \quad(0,0,1,1),\\\ &(0,1,0,1)\\} \text { in } \mathbb{R}^{4} \end{aligned} $$

Short Answer

Expert verified
Sets b is independent; sets a, c, and d are dependent.

Step by step solution

01

Understand Independence Criterion

To determine if a set of vectors is independent, we must check if the only solution to the equation formed by their linear combination equaling the zero vector is the trivial solution, where all coefficients are zero. In other terms, the vectors are linearly independent if none of them can be expressed as a combination of the others.
02

Analyze Set a

For set a: \( \{(1,-1,0), (3,2,-1), (3,5,-2)\} \). Create a matrix with these vectors as rows and reduce it to row-echelon form:\[\begin{bmatrix}1 & -1 & 0 \3 & 2 & -1 \3 & 5 & -2\end{bmatrix}\rightarrow\left(\text{Row reduction}\right)\rightarrow\begin{bmatrix}1 & -1 & 0 \0 & 5 & -1 \0 & 0 & 0\end{bmatrix}\]There is a row of zeros, indicating dependence. So, set a is linearly dependent.
03

Analyze Set b

For set b: \( \{(1,1,1), (1,-1,1), (0,0,1)\} \). Form a matrix and perform row reduction:\[\begin{bmatrix}1 & 1 & 1 \1 & -1 & 1 \0 & 0 & 1\end{bmatrix}\rightarrow\left(\text{Row reduction}\right)\rightarrow\begin{bmatrix}1 & 1 & 1 \0 & -2 & 0 \0 & 0 & 1\end{bmatrix}\]With no zero rows, this indicates the vectors are independent. Set b is linearly independent.
04

Analyze Set c

For set c: \( \{(1,-1,1,-1), (2,0,1,0), (0,-2,1,-2)\} \). Form a matrix and row reduce:\[\begin{bmatrix}1 & -1 & 1 & -1 \2 & 0 & 1 & 0 \0 & -2 & 1 & -2\end{bmatrix}\rightarrow\left(\text{Row reduction}\right)\rightarrow\begin{bmatrix}1 & -1 & 1 & -1 \0 & 2 & -1 & 2 \0 & 0 & 0 & 0\end{bmatrix}\]A row of zeros indicates dependence, so set c is linearly dependent.
05

Analyze Set d

For set d: \( \{(1,1,0,0), (1,0,1,0), (0,0,1,1), (0,1,0,1)\} \). Form a matrix and check if there is linear independence:\[\begin{bmatrix}1 & 1 & 0 & 0 \1 & 0 & 1 & 0 \0 & 0 & 1 & 1 \0 & 1 & 0 & 1\end{bmatrix}\rightarrow\left(\text{Row reduction}\right)\rightarrow\begin{bmatrix}1 & 0 & 1 & 0 \0 & 1 & 0 & 1 \0 & 0 & 1 & 1 \0 & 0 & 0 & 0\end{bmatrix}\]With a zero row, it indicates the vectors are dependent. Set d is linearly dependent.

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Key Concepts

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

Linear Dependence
Linear dependence is a fundamental concept in vector spaces. Imagine you have a collection of vectors. They are termed as linearly dependent if you can find at least one vector in the set that can be made from adding or scaling the others.
This means that the information contained within that dependent vector is already available in the other vectors.
For instance, consider vectors in a row represented in a matrix. If after you apply row operations, you can find at least one row of zeros in the matrix, it tells you there is linear dependence.
Every dependent vector can be unraveled into simpler pieces using the independent vectors around it.
Row Reduction
Row reduction is a handy tool for uncovering the secrets within a group of vectors. By laying them out in a matrix and performing elementary row operations like switching rows, multiplying rows by a scalar, or adding a multiple of one row to another, you can simplify this matrix into an easier form called row-echelon form.
Through row reduction, we can expose zero rows, which are key in identifying linear dependence among vectors. Each zero row reveals a dependent vector within your set.
  • Step one involves transforming your matrix into this reduced form.
  • Step two is observing the form to check for zero rows.
This method is not only reliable but also provides a visual way to understand the structure of vector arrangements.
Vector Spaces
Vector spaces are like playgrounds where vectors operate within well-established rules. They are collections where you can scale or add vectors together and still remain inside the space.
Each vector space has its own unique characteristics determined by its basis – a set of linearly independent vectors. These basis vectors provide every possible construction within the space without redundancy.
  • The number of vectors in the basis is termed the "dimension" of the space.
  • Vector spaces help in illustrating complex multidimensional ideas in fields like physics and engineering.
Knowing if vectors in your space are independent or dependent is vital for describing how the space behaves. It informs how this space can be utilized or transformed in practical problem-solving.

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Most popular questions from this chapter

Let \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) be a linearly independent set in \(\mathbb{R}^{4}\). Show that \(\left\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{e}_{k}\right\\}\) is a basis of \(\mathbb{R}^{4}\) for some \(\mathbf{e}_{k}\) in the standard basis \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}, \mathbf{e}_{4}\right\\}\).

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(U \neq\\{\mathbf{0}\\}\) is a subspace of \(\mathbb{R}\), show that \(U=\mathbb{R}\).

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Find the least squares approximating function of the form \(r_{0}+r_{1} x^{2}+r_{2} \sin \frac{\pi x}{2}\) for each of the following sets of data pairs. a. (0,3),(1,0),(1,-1),(-1,2) b. \(\left(-1, \frac{1}{2}\right),(0,1),(2,5),(3,9)\)

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