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Chapter 2: Problem 9

Show that if \(S\) is real and \(S^{T}=-S\), then \(I-S\) is nonsingular and the matrix \((I-S)^{-1}(I+S)\) is orthogonal. This is known as the Cayley transform of \(S\).

Short Answer

Expert verified
\( (I-S)^{-1}(I+S) \) is orthogonal as shown by proving \( A^T A = I \).

Step by step solution

01

Define the problem and given conditions

We are given a real matrix \( S \) such that \( S^T = -S \). This implies \( S \) is a skew-symmetric matrix. We need to show that \( I - S \) is nonsingular and \( (I-S)^{-1}(I+S) \) is an orthogonal matrix.
02

Show that \( I - S \) is nonsingular

To prove \( I - S \) is nonsingular, we must show that its determinant is non-zero. For any skew-symmetric matrix \( S \), all eigenvalues are purely imaginary or zero. Because \( I - S \) shifts the eigenvalues of \( S \) by 1, all eigenvalues of \( I - S \) have a positive real part. Thus, \( I - S \) is invertible, implying it is nonsingular.
03

Use properties of orthogonal matrices

A matrix \( A \) is orthogonal if \( A^T A = I \). We need to check this property for \( A = (I-S)^{-1}(I+S) \). We compute \( A^T A \) and verify whether it equals the identity matrix.
04

Compute transpose \((I-S)^{-1}(I+S)\)

Given \( A = (I-S)^{-1}(I+S) \), its transpose is \( A^T = (I+S)^T((I-S)^{-1})^T = (I+S)^T(I-S)^{-T) \). Since \( (I-S)^T = I+S \), we have \( A^T = (I+S)^T(I+S)^{-1} \).
05

Simplify \( A^T A \) and verify identity

Compute \( A^T A \ = (I+S)^{-1}(I-S)(I-S)^{-1}(I+S)\ = (I+S)^{-1}(I-S^2)(I+S)^{-1}. \) Since \( S^2 = -SS^T = -(SS^T) = I \), we find \( A^T A = I \), confirming that \( (I-S)^{-1}(I+S) \) is orthogonal.

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Key Concepts

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

Skew-Symmetric Matrix
A matrix is called skew-symmetric if its transpose equals its negative. In mathematical terms, a matrix \( S \) is skew-symmetric if \( S^T = -S \).
This property implies that the diagonal elements of a skew-symmetric matrix must be zero because, for these diagonal elements, which are equal to their own negatives, this is only possible if they are zero.
A characteristic feature of skew-symmetric matrices is their eigenvalues.
  • All eigenvalues of a skew-symmetric matrix are purely imaginary or zero. This is important when analyzing matrices like \( S \) given its structural properties.
Because real skew-symmetric matrices force the eigenvalues off the real axis into the imaginary plane, they play a vital role in various transformations, such as rotations in vector spaces.
Cayley Transform
The Cayley transform is a fascinating mathematical operation that relates a skew-symmetric matrix to an orthogonal one.
For a given skew-symmetric matrix \( S \), the Cayley transform is defined as \( (I-S)^{-1}(I+S) \). An important property to note is that the matrix \( I-S \) is nonsingular, meaning it can be inverted, because all its eigenvalues have a positive real part.
The significance of the Cayley transform in matrix theory lies in its ability to represent transformations while maintaining orthogonality.
  • It preserves the essential characteristics necessary for orthogonal transformations, crucial in applications like signal processing and computer graphics.
Orthogonal Matrix
An orthogonal matrix \( A \) is one that satisfies the condition \( A^T A = I \), where \( I \) is the identity matrix. This property ensures that orthogonal matrices represent rigid transformations, such as rotations and reflections, which conserve distances and angles.
Orthogonal matrices have several intriguing properties:
  • The inverse of an orthogonal matrix is equal to its transpose, i.e., \( A^{-1} = A^T \).
  • All columns (and rows) of an orthogonal matrix are orthonormal vectors. This means they are perpendicular to each other and have unit length.
These traits make orthogonal matrices especially valuable in numerical computations, ensuring stability and consistency when transforming datasets. They serve as the cornerstone for algorithms in applications like 3D modeling and machine learning.

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Most popular questions from this chapter

Show that any matrix in \(\mathbf{R}^{m \times n}\) is the limit of a sequence of full rank matrices.

Let \(\|\). \(\|\) be a vector norm on \(\mathbf{R}^{m}\) " and assume \(A \in \mathbf{R}^{m \times n}\). Show that if \(\operatorname{rank}(A)=n\), then \(\|x\|_{A}=\|A x\|\) is a vector norm on \(\mathbf{R}^{4}\).

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