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Chapter 4: Problem 18

Let \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{n}\right\\}\) be a basis for a vector space \(V .\) Explain why the \(\mathcal{B}\) -coordinate vectors of \(\mathbf{b}_{1}, \ldots, \mathbf{b}_{n}\) are the columns \(\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\) of the \(n \times n\) identity matrix.

Short Answer

Expert verified
The \( \mathcal{B} \)-coordinate vectors of \( \mathbf{b}_1, \ldots, \mathbf{b}_n \) are standard basis vectors \( \mathbf{e}_1, \ldots, \mathbf{e}_n \) because each \( \mathbf{b}_i \) is its own coordinate with all others zero.

Step by step solution

01

Understanding the Basis

A basis for a vector space \( V \) is a set of vectors \( \mathcal{B}=\{\mathbf{b}_1, \mathbf{b}_2, \ldots, \mathbf{b}_n\} \) such that they are linearly independent and span the space. This means any vector in \( V \) can be uniquely expressed as a linear combination of the basis vectors.
02

The B-coordinates of a Basis Vector

The \( \mathcal{B} \)-coordinate vector of a vector \( \mathbf{v} \) with respect to the basis \( \mathcal{B} \) is a vector \( [\mathbf{v}]_\mathcal{B} \) comprising the coefficients of \( \mathbf{v} \) when expressed as a linear combination of the basis vectors.
03

Expressing Basis Vectors in Terms of Themselves

Each basis vector \( \mathbf{b}_i \) in \( \mathcal{B} \) can be expressed as a linear combination of itself and other basis vectors, but specifically, \( \mathbf{b}_i = 1 \cdot \mathbf{b}_i + 0 \cdot \mathbf{b}_1 + \cdots + 0 \cdot \mathbf{b}_{i-1} + 0 \cdot \mathbf{b}_{i+1} + \cdots + 0 \cdot \mathbf{b}_n \).
04

Forming the Coordinate Vector

The coordinate vector of any basis vector \( \mathbf{b}_i \), \( [\mathbf{b}_i]_{\mathcal{B}} \), becomes \( [0, \ldots, 0, 1, 0, \ldots, 0]^T \) with a 1 in the \( i^{th} \) position, which is exactly the vector \( \mathbf{e}_i \) from the identity matrix.
05

Relating to the Identity Matrix

Since each coordinate vector \( [\mathbf{b}_i]_{\mathcal{B}} \) is precisely \( \mathbf{e}_i \), when assembled into a matrix, these vectors form the columns of the \( n \times n \) identity matrix. Therefore, the \( \mathcal{B} \)-coordinate vectors of the basis vectors equal the standard basis vectors \( \mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n \).

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Key Concepts

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

Linear Independence
Understanding the concept of linear independence is crucial when discussing vector spaces and their bases. A set of vectors is said to be linearly independent if no vector in the set can be written as a linear combination of the others. This means that each vector adds a unique direction in space that cannot be recreated by the others.
Here is why linear independence is important:
  • It ensures that every vector in the set is necessary to span the vector space.
  • A linearly independent set of vectors that spans a vector space forms a basis for that space.
  • This unique combination allows us to represent any vector in the space as a unique sum of the basis vectors.
In simple terms, linearly independent vectors can be thought of as building blocks that cannot be broken down using other blocks in the same set, making them fundamental components of the vector space.
Identity Matrix
The identity matrix is an essential concept in linear algebra that acts as the identity element for matrix multiplication, similar to how the number 1 acts in ordinary multiplication. The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere.
Let's explore the significance of the identity matrix:
  • Its main property is that when multiplied by any compatible matrix, the original matrix is unchanged: if \( I \) is an identity matrix, then \( AI = A \) and \( IA = A \).
  • It plays an instrumental role in transformations, particularly in maintaining the structure of vector spaces during operations.
  • In the context of vector spaces, the identity matrix helps us understand how basis vectors are connected to coordinate vectors. The columns of the identity matrix represent the "standard" basis vectors.
Thus, in vector spaces where identity matrix columns correspond to basis coordinate vectors, it simplifies understanding and manipulating these spaces.
Coordinate Vectors
Coordinate vectors express how a vector in a vector space can be represented as a combination of basis vectors. When you take a vector and express it in terms of a basis, the coefficients used form its coordinate vector.
Understanding how coordinate vectors work:
  • The coordinate vector \([\mathbf{v}]_\mathcal{B}\) gives the coefficients for the linear combination of basis vectors that recreate \(\mathbf{v}\).
  • For basis vectors themselves, the coordinate vector is very simple: it's an identity vector. This means each basis vector in its own coordinate system looks like \([0, \, ..., \, 1, \, ..., \, 0]^T\).
  • Coordinate vectors make it straightforward to transform vectors between different bases, a process integral to applications in physics, computer graphics, and engineering.
Coordinate vectors offer a powerful tool for simplification and manipulation in multi-dimensional spaces, making complex calculations more intuitive.
Linear Combinations
A linear combination involves creating new vectors by multiplying each vector in a set by a corresponding scalar and then adding the results. Spread across numerous applications in linear algebra, linear combinations represent the essence of expressing vectors in terms of others.
Key points about linear combinations:
  • They allow us to express complex vectors as simple combinations of basis vectors.
  • In the context of bases, linear combinations show how each vector in the space can be uniquely formed by its basis vectors.
  • The coefficients you use in a linear combination are crucial as they define the vector's position in the vector space.
By mastering linear combinations, you create a foundation for understanding vector spaces and their inherent structures, enabling easier navigation of advanced concepts in mathematics and related disciplines.

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Most popular questions from this chapter

In Exercises \(7-12\) , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. $$ (-1)^{k}, 3^{k} ; y_{k+3}+y_{k+2}-9 y_{k+1}-9 y_{k}=0 $$

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Show that if \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent in \(V,\) then the set of images, \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\},\) is linearly dependent in \(W .\) This fact shows that if a linear transformation maps a set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) onto a linearly independent set \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{1}\right)\right\\},\) then the original set is linearly independent, too (because it cannot be linearly dependent).

In Exercises \(15-18,\) find a basis for the space spanned by the given vectors, \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{5}\) $$ \left[\begin{array}{r}{1} \\ {0} \\ {-3} \\\ {2}\end{array}\right],\left[\begin{array}{r}{0} \\ {1} \\ {2} \\\ {-3}\end{array}\right],\left[\begin{array}{r}{-3} \\ {-4} \\ {1} \\\ {6}\end{array}\right],\left[\begin{array}{r}{1} \\ {-3} \\ {-8} \\\ {7}\end{array}\right],\left[\begin{array}{r}{2} \\ {1} \\ {-6} \\\ {9}\end{array}\right] $$

On any given day, a student is either healthy or ill. Of the students who are healthy today, 95\(\%\) will be healthy tomorrow. Of the students who are ill today, 55\(\%\) will still be ill tomorrow. a. What is the stochastic matrix for this situation? b. Suppose 20\(\%\) of the students are ill on Monday. What fraction or percentage of the students are likely to be ill on Tuesday? On Wednesday? c. If a student is well today, what is the probability that he or she will be well two days from now?

In Exercises 17 and 18 , \(A\) is an \(m \times n\) matrix. Mark each statement True or False. Justify each answer. a. If \(B\) is any echelon form of \(A\) , then the pivot columns of \(B\) form a basis for the column space of \(A\) . b. Row operations preserve the linear dependence relations among the rows of \(A\) . c. The dimension of the null space of \(A\) is the number of columns of \(A\) that are not pivot columns. d. The row space of \(A^{T}\) is the same as the column space of \(A .\) e. If \(A\) and \(B\) are row equivalent, then their row spaces are the same.
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