Problem 14 Let \(U\) be any invertible \(n ... [FREE SOLUTION]

Chapter 9: Problem 14

Let \(U\) be any invertible \(n \times n\) matrix, and let \(D=\left\\{\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n}\right\\}\) where \(\mathbf{f}_{j}\) is column \(j\) of \(U\). Show that \(M_{B D}\left(1_{\mathbb{R}^{n}}\right)=U\) when \(B\) is the standard basis of \(\mathbb{R}^{n}\).

Short Answer

Expert verified

\( M_{BD}(1_{\mathbb{R}^n}) = U \), as it maps standard basis to columns of \( U \).

Step by step solution

Understanding the Problem

To solve the problem, we need to express the identity map on \( \mathbb{R}^n \) in terms of the matrix transformation with respect to the bases \( B \) and \( D \), and show that the resulting matrix is \( U \).

Define the Standard Basis \( B \)

The standard basis \( B \) for \( \mathbb{R}^n \) consists of \( n \) vectors: \( \mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n \), where each \( \mathbf{e}_j \) has a 1 in the \( j \)-th position and zeros elsewhere.

Evaluate \( M_{BD}(1_{\mathbb{R}^n}) \)

The matrix representation \( M_{BD}(1_{\mathbb{R}^n}) \) corresponds to the transformation matrix that takes coordinates from basis \( D \) to basis \( B \). Since \( 1_{\mathbb{R}^n} \) is the identity map, this matrix should map each vector \( \mathbf{e}_j \) to \( \mathbf{f}_j \), which is the \( j \)-th column of \( U \).

Construct Matrix \( U \)

Each column \( \mathbf{f}_j \) of \( U \) is expressed in the standard basis \( B \). Consequently, the transformation matrix \( M_{BD}(1_{\mathbb{R}^n}) \) has columns \( \mathbf{f}_1, \mathbf{f}_2, \ldots, \mathbf{f}_n \), which exactly reconstructs the matrix \( U \).

Conclude the Solution

Since the matrix representation of the identity map from basis \( B \) to \( D \) is \( U \) itself (as the transformation does not change the vectors), we have shown that \( M_{BD}(1_{\mathbb{R}^n}) = U \).

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Key Concepts

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

Invertible Matrices

Invertible matrices play a crucial role in linear algebra. An invertible matrix, also known as a nonsingular or nondegenerate matrix, is a square matrix that has a unique inverse. In mathematical terms, if a matrix \( U \) is invertible, there exists another matrix \( U^{-1} \) such that:\[U \cdot U^{-1} = U^{-1} \cdot U = I\]where \( I \) is the identity matrix of the same dimension as \( U \). Remember, not every matrix is invertible. For a matrix to have an inverse, it must be square (same number of rows and columns) and have a non-zero determinant.

Invertibility is essential because invertible matrices preserve vector space structure, making them a key part of solving systems of linear equations.
They help determine whether a linear transformation is bijective, implying a one-to-one mapping between vector spaces.

While working with invertible matrices, it's helpful to understand that they allow transformations to be reversible.

Standard Basis

The standard basis for a vector space like \( \mathbb{R}^n \) is one of the simplest yet most important concepts in linear algebra. In \( \mathbb{R}^n \), the standard basis is composed of vectors \( \mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n \), where the vector \( \mathbf{e}_j \) has all components zero except for a 1 in the \( j \)-th position.

For example, in \( \mathbb{R}^3 \), you have \( \mathbf{e}_1 = (1, 0, 0), \mathbf{e}_2 = (0, 1, 0), \mathbf{e}_3 = (0, 0, 1) \).
This forms a basis because these vectors are linearly independent and span the space \( \mathbb{R}^n \).

One of the key reasons why the standard basis is so fundamental is due to its simplicity. It makes the calculation of coordinates straightforward in vector transformations and simplifies the visualization of geometric objects and operations in vector spaces.

Matrix Representation

Matrix representation in linear transformations is central for performing calculations and solving problems in linear algebra. When you have two bases, like the standard basis \( B \) and another basis \( D \), the matrix representation of a transformation can be found using these bases.

A transformation matrix \( M_{BD} \) maps vectors expressed in basis \( D \) to their new form in basis \( B \).
Understanding how to switch between bases gives insight into different perspectives of vector spaces and functional mappings.

For example, if you consider the identity map \( 1_{\mathbb{R}^n} \), mapping all vectors \( \mathbf{e}_j \) to themselves, the transformation matrix with standard basis \( B \) and basis \( D \), composed of columns of an invertible matrix \( U \), becomes matrix \( U \). This shows how basis transformations impact the way we represent transformations mathematically.

Identity Map

The identity map is a particularly simple yet fundamental concept in linear transformations. It essentially signifies a transformation that does nothing more than return the vectors exactly as they were input.

Mathematically, the identity map \( 1_{\mathbb{R}^n} \) maps every vector \( \mathbf{v} \) in \( \mathbb{R}^n \) to itself.
Its matrix representation with respect to the basis \( B \) and \( D \) is simply the identity matrix \( I \), given that \( B \) and \( D \) are the same basis.

When \( D \) is formed from the columns of an invertible matrix \( U \), the identity map \( 1_{\mathbb{R}^n} \) maps basis \( B \) to \( D \) using the matrix \( U \), as verified in the exercise. The process of transformation, when visualized as an identity map, exemplifies how unchanged mappings are represented when changing basis perspectives.

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