Chapter 4: Problem 14
Prove that if \(\inf _{\lambda \in \mathbb{R}}\|I-\lambda A\|<1\), then \(A\) is invertible.
Short Answer
Expert verified
Given the condition, \(A\) is invertible due to the convergence of the Neumann series for \(\|I-\lambda_0 A\|<1\).
Step by step solution
01
Understanding the Problem
We need to prove that matrix \(A\) is invertible given the condition \(\inf _{\lambda \in \mathbb{R}}\|I-\lambda A\|<1\). The expression involves an infimum and the norm of a matrix expression, hinting at spectral properties.
02
Condition Definition
The condition \(\inf _{\lambda \in \mathbb{R}}\|I-\lambda A\|<1\) implies that there exists some real number \(\lambda_0\) such that \(\|I-\lambda_0 A\|<1\). This condition is crucial because it leads to matrix properties related to invertibility.
03
Apply the Series Expansion
If \(\|I-\lambda_0 A\|<1\), then the series \((I-\lambda_0 A)^{-1}= I + \lambda_0 A + (\lambda_0 A)^2 + \cdots \) converges. This is a Neumann series, and convergence implies that \(\lambda_0 A\) is invertible.
04
Invertibility Deduction
Since \((I - \lambda_0 A)^{-1}\) exists, matrix \(I - \lambda_0 A\) is invertible. Using the identity \(\lambda_0 A = I - (I - \lambda_0 A)\), we conclude that \(A\) itself is invertible because \(A\) can be expressed as an inverse operation of a sum of invertible matrices, confirming the invertibility of \(A\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Norm
The concept of a matrix norm is essential in understanding how the size or length of a matrix is measured. Unlike vectors, which use simple norms like the Euclidean norm, matrix norms evaluate matrices using slightly more complex rules. A matrix norm is generally denoted as \( \|A\| \), where \( A \) is a given matrix. It must satisfy specific properties:
Understanding and calculating matrix norms help us deduce whether certain matrix expressions, like \( I - \lambda A \), can be expanded into a convergent series, which is central to proving invertibility.
- Non-negativity: \( \|A\| \geq 0 \), and \( \|A\| = 0 \) if and only if \( A \) is a zero matrix.
- Scalar multiplication: \( \|cA\| = |c| \cdot \|A\| \) for any scalar \( c \).
- Triangle Inequality: \( \|A + B\| \leq \|A\| + \|B\| \) for any matrices \( A \) and \( B \).
- Submultiplicativity: \( \|AB\| \leq \|A\| \cdot \|B\| \) for any matrices \( A \) and \( B \).
Understanding and calculating matrix norms help us deduce whether certain matrix expressions, like \( I - \lambda A \), can be expanded into a convergent series, which is central to proving invertibility.
Spectral Properties
Spectral properties of matrices refer to characteristics that arise from eigenvalues and eigenvectors. The eigenvalues of a matrix \( A \) can provide us with critical insights into its behavior and are central to understanding conditions for invertibility.
An eigenvalue \( \lambda \) of a matrix \( A \) satisfies the equation \( Av = \lambda v \), where \( v \) is a non-zero eigenvector. The collection of all eigenvalues forms the spectrum of the matrix. Matrix invertibility is tied to spectral properties because a matrix is invertible if zero is not an eigenvalue.
An eigenvalue \( \lambda \) of a matrix \( A \) satisfies the equation \( Av = \lambda v \), where \( v \) is a non-zero eigenvector. The collection of all eigenvalues forms the spectrum of the matrix. Matrix invertibility is tied to spectral properties because a matrix is invertible if zero is not an eigenvalue.
- If \( \lambda_0 \) exists such that \( \|I - \lambda_0 A\| < 1 \), then the eigenvalues of \( \lambda_0 A \) are constrained within a unit circle centered at 1.
- This information implies that the matrix \( I - \lambda_0 A \) cannot have eigenvalue zero, supporting its invertibility.
Neumann Series
A Neumann series is a powerful tool in linear algebra used to find the inverse of a matrix expression under specific conditions. The series is particularly useful when dealing with matrices that resemble the identity matrix. The Neumann series for a matrix \( B \) is given by: \[(I - B)^{-1} = I + B + B^2 + B^3 + \cdots \]This series converges if \( \|B\| < 1 \), implying that each subsequent power of \( B \) becomes increasingly negligible.
This concept is critical when proving the invertibility of \( A \) in the exercise:
This concept is critical when proving the invertibility of \( A \) in the exercise:
- Given \( \|I - \lambda_0 A\| < 1 \), we can safely apply the Neumann series \( (I - \lambda_0 A)^{-1} \).
- This convergent series allows us to express the inverse of \( I - \lambda_0 A \) directly.
- Therefore, \( A \) is invertible, since \( I - \lambda_0 A \) can be inverted using the series.