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In each case find a basis of the subspace \(U\) a. \(U=\operatorname{span}\\{(1,-1,0,3),(2,1,5,1),(4,-2,5,7)\\}\) b. \(U=\operatorname{span}\\{(1,-1,2,5,1),(3,1,4,2,7), (1,1,0,0,0),(5,1,6,7,8)\\}\) c. \(U=\operatorname{span}\left\\{\left[\begin{array}{l}1 \\ 1 \\ 0 \\\ 0\end{array}\right],\left[\begin{array}{l}0 \\ 0 \\ 1 \\\ 1\end{array}\right],\left[\begin{array}{l}1 \\ 0 \\ 1 \\\ 0\end{array}\right],\left[\begin{array}{l}0 \\ 1 \\ 0 \\\ 1\end{array}\right]\right\\}\) d \(U=\operatorname{span}\left\\{\left[\begin{array}{r}1 \\ 5 \\\ -6\end{array}\right],\left[\begin{array}{r}2 \\ 6 \\\ -8\end{array}\right],\left[\begin{array}{r}3 \\ 7 \\\ -10\end{array}\right],\left[\begin{array}{r}4 \\ 8 \\\ 12\end{array}\right]\right\\}\)

Short Answer

Expert verified
a) \{(1,-1,0,3), (2,1,5,1)\}; b) \{(1,-1,2,5,1), (3,1,4,2,7)\}; c) \{[1,0,1,0], [0,1,0,1]\}; d) \{[1,5,-6], [0,-4,4]\}

Step by step solution

01

Write Vectors as a Matrix (Problem a)

For part (a), arrange the vectors as rows of a matrix:\[A = \begin{bmatrix}1 & -1 & 0 & 3 \2 & 1 & 5 & 1 \4 & -2 & 5 & 7\end{bmatrix}\]
02

Row Reduce the Matrix (Problem a)

Perform row operations to transform the matrix into its row-echelon form:\[\begin{bmatrix}1 & -1 & 0 & 3 \0 & 3 & 5 & -5 \0 & 0 & 0 & -1\end{bmatrix}\]Then continue to reduced row-echelon form:\[\begin{bmatrix}1 & 0 & \frac{5}{3} & \frac{1}{3} \0 & 1 & \frac{5}{3} & -\frac{5}{3} \0 & 0 & 0 & 1\end{bmatrix}\]
03

Identify Basis Vectors (Problem a)

The first two rows (after reduction) are non-zero, so they form a basis. Thus, a basis for \( U \) is:\[ \{ (1, -1, 0, 3), (2, 1, 5, 1) \} \]
04

Write Vectors as a Matrix (Problem b)

Arrange the vectors for part (b):\[B = \begin{bmatrix}1 & -1 & 2 & 5 & 1 \3 & 1 & 4 & 2 & 7 \1 & 1 & 0 & 0 & 0 \5 & 1 & 6 & 7 & 8\end{bmatrix}\]
05

Row Reduce the Matrix (Problem b)

Perform row operations to row-reduce this matrix:\[\begin{bmatrix}1 & -1 & 2 & 5 & 1 \0 & 4 & -2 & -13 & 4 \0 & 0 & 0 & 0 & 0 \0 & 0 & 0 & 0 & 0\end{bmatrix}\]
06

Identify Basis Vectors (Problem b)

The first two non-zero rows form a basis for \( U \):\[ \{ (1, -1, 2, 5, 1), (3, 1, 4, 2, 7) \} \]
07

Write Vectors as a Matrix (Problem c)

For part (c), organize the vectors as a matrix:\[C = \begin{bmatrix}1 & 1 & 0 & 0 \0 & 0 & 1 & 1 \1 & 0 & 1 & 0 \0 & 1 & 0 & 1\end{bmatrix}\]
08

Row Reduce the Matrix (Problem c)

Transform the matrix into its row-echelon form:\[\begin{bmatrix}1 & 0 & 1 & 0 \0 & 1 & 0 & 1 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0\end{bmatrix}\]
09

Identify Basis Vectors (Problem c)

The first two rows after reduction are independent and form a basis:\[ \left\{ \begin{bmatrix} 1 \ 0 \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix} \right\} \]
10

Write Vectors as a Matrix (Problem d)

For part (d), arrange into a matrix:\[D = \begin{bmatrix}1 & 5 & -6 \2 & 6 & -8 \3 & 7 & -10 \4 & 8 & 12\end{bmatrix}\]
11

Row Reduce the Matrix (Problem d)

Row-reduce the matrix:\[\begin{bmatrix}1 & 5 & -6 \0 & -4 & 4 \0 & 0 & 0 \0 & 0 & 0\end{bmatrix}\]
12

Identify Basis Vectors (Problem d)

The first two non-zero rows form a basis:\[ \{ \begin{bmatrix} 1 \ 5 \ -6 \end{bmatrix}, \begin{bmatrix} 0 \ -4 \ 4 \end{bmatrix} \} \]

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Key Concepts

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

Vector Spaces
In linear algebra, a vector space is a collection of vectors that can be added together and multiplied by scalars. Vectors are objects that have both a direction and a magnitude, and scalars are simply real numbers. Vector spaces, sometimes called linear spaces, play a vital role in linear algebra because they provide the settings in which vectors are studied.

A vector space must satisfy certain properties for addition and scalar multiplication, such as associativity, commutativity, and distributivity. More formally, if you take two vectors from a vector space and add them or multiply them by a scalar, the result should also be a vector that resides in the same vector space.

Examples of vector spaces include:
  • The set of all real numbers forms a vector space.
  • Two-dimensional and three-dimensional space are also vector spaces.
  • The set of all polynomials with real coefficients forms a vector space.
Bases
A basis of a vector space is a set of vectors in that space that are linearly independent and span the entire space. To be linearly independent, no vector in the base can be written as a combination of other vectors in the set. Moreover, to span the space means any vector in the vector space can be expressed as a combination of the basis vectors.

Finding a basis is crucial because it helps fully describe the vector space in the simplest way possible. The number of vectors in the basis is called the dimension of the vector space. A common approach to identify a basis from a given set of vectors is using the row reduction method, which we'll discuss soon.

For example, in a plane, two non-parallel vectors can form a basis. In three-dimensional space, you need three such non-coplanar vectors to establish a basis.
Row Reduction
Row reduction is an algorithmic process used to simplify matrices, mainly by using row operations to transform a matrix into a preferable form, such as the row-echelon form or reduced row-echelon form. This technique is widely used in linear algebra to solve systems of equations or find the basis of a vector space.

The primary row operations permitted during row reduction include:
  • Swapping two rows.
  • Multiplying a row by a non-zero scalar.
  • Adding a multiple of one row to another row.
Row reduction primarily aims to make it easier to analyze and manipulate matrices by reducing the complexity of the system. By reducing a matrix, you can more easily identify linearly independent rows, which assists in finding a basis for the corresponding vector space.
Matrix Representation
Matrices are powerful tools in linear algebra that allow us to represent complex information simply and consistently. A matrix is a rectangular table of numbers arranged in rows and columns. It serves as a compact way to handle linear transformations and solve linear equations.

In the context of vector spaces, matrices can be used to organize vectors to make operations like row reduction possible. The rows of a matrix can represent vectors, and using transformations, you can alter this matrix to reveal more about the vector space they form, such as its dimensionality and possible bases.

Using matrices is essential for elegantly performing calculations and obtaining reliable solutions to algebraic problems. Representing vectors as rows in a matrix helps simplify big data sets and gives structure to raw data so that they can be analyzed using mathematical, statistical, and computational tools.

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Most popular questions from this chapter

We often write vectors in \(\mathbb{R}^{n}\) as rows. Suppose that \(\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\) are vectors in \(\mathbb{R}^{n} .\) If \(\mathbf{y}=a_{1} \mathbf{x}_{1}+a_{2} \mathbf{x}_{2}+\cdots+a_{k} \mathbf{x}_{k}\) where \(a_{1} \neq 0,\) show that \(\operatorname{span}\left\\{\mathbf{x}_{1} \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}=\operatorname{span}\left\\{\mathbf{y}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\)

In each case, find all \((a, b, c, d)\) in \(\mathbb{R}^{4}\) such that the given set is orthogonal. a. \(\\{(1,2,1,0),(1,-1,1,3),(2,-1,0,-1),\) \((a, b, c, d)\\}\) b. \(\\{(1,0,-1,1),(2,1,1,-1),(1,-3,1,0),\) \((a, b, c, d)\\}\)

Newton's laws of motion imply that an object dropped from rest at a height of 100 metres will be at a height \(s=100-\frac{1}{2} g t^{2}\) metres \(t\) seconds later, where \(g\) is a constant called the acceleration due to gravity. The values of \(s\) and \(t\) given in the table are observed. Write \(x=t^{2},\) find the least squares approximating line \(s=a+b x\) for these data, and use \(b\) to estimate \(g\). Then find the least squares approximating quadratic \(s=a_{0}+a_{1} t+a_{2} t^{2}\) and use the value of \(a_{2}\) to estimate \(g\). $$ \begin{array}{|c|c|c|c|} \hline t & 1 & 2 & 3 \\ \hline s & 95 & 80 & 56 \\ \hline \end{array} $$

Let \(\lambda\) be an eigenvalue of \(A\) with corresponding eigenvector \(\mathbf{x}\). If \(B=P^{-1} A P\) is similar to \(A\), show that \(P^{-1} \mathbf{x}\) is an eigenvector of \(B\) corresponding to \(\lambda\)

Let \(A\) be any \(m \times n\) matrix, and let \(\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}, \ldots, \mathbf{b}_{k}\) be columns in \(\mathbb{R}^{m}\) such that the system \(A \mathbf{x}=\mathbf{b}_{i}\) has a solution \(\mathbf{x}_{i}\) for each \(i .\) If \(\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}, \ldots, \mathbf{b}_{k}\right\\}\) is independent in \(\mathbb{R}^{m},\) show that \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \ldots, \mathbf{x}_{k}\right\\}\) is independent in \(\mathbb{R}^{n}\)

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