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Chapter 4: Problem 6

(Continuation) Let \(\|\cdot\|\) be a subordinate matrix norm, and let \(S\) be a nonsingular matrix. Define \(\|A\|^{\prime}=\left\|S A S^{-1}\right\|\), and show that \(\|\cdot\|^{\prime}\) is a subordinate matrix norm.

Short Answer

Expert verified
\(\|A\|' = \|SAS^{-1}\|\) is a subordinate matrix norm because it satisfies norm properties and subordination condition.

Step by step solution

01

Understanding Subordinate Matrix Norms

A matrix norm \(\| \cdot \|\) is subordinate to a vector norm \(\|\cdot\|_v\) if \(\|Ax\|_v \leq \|A\|\|x\|_v\) for all vectors \(x\). We need to show that \(\|A\|'\) defined by \(\|A\|' = \|SAS^{-1}\|\) satisfies the properties of a subordinate matrix norm.
02

Property 1: Non-negativity and Definiteness

A matrix norm \(\|A\|'\) should satisfy \(\|A\|' \geq 0\) and \(\|A\|' = 0\) only if \(A = 0\). We know \(\|SAS^{-1}\|\) satisfies this since it's a matrix norm. Thus, \(\|A\|' \geq 0\) and \(\|A\|' = 0\) implies \(SAS^{-1} = 0\), leading to \(A = 0\) because \(SAS^{-1}\) is just a similarity transform.
03

Property 2: Homogeneity

A matrix norm \(\|A\|'\) should satisfy \(\|cA\|' = |c|\|A\|'\) for any scalar \(c\). Since \(\|S(cA)S^{-1}\| = \|cSAS^{-1}\| = |c|\|SAS^{-1}\|\), it follows that \(\|cA\|' = |c|\|A\|'\).
04

Property 3: Triangle Inequality

A matrix norm \(\|A\|'\) should satisfy \(\|A + B\|' \leq \|A\|' + \|B\|'\) for all matrices \(A\) and \(B\). Since \(\|S(A+B)S^{-1}\| = \|SAS^{-1} + SBS^{-1}\|\), and because \(\|\cdot\|\) is a norm, it follows that \(\|SAS^{-1} + SBS^{-1}\| \leq \|SAS^{-1}\| + \|SBS^{-1}\|\). This implies \(\|A + B\|' \leq \|A\|' + \|B\|'\).
05

Property 4: Subordination

To be a subordinate matrix norm with respect to a vector norm \(\|\cdot\|_v\), \(\|Ax\|'_v \leq \|A\|'\|x\|_v\) should hold. By defining \(y = S^{-1}x\), we have \(SAS^{-1}y = Sy\) so \(\|Ax\|' \leq \|A\|'\|x\|_v\) holds as a result of the properties of \(SAS^{-1}\) and the subordinate nature of \(\|\cdot\|\).

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Key Concepts

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

Subordinate Norm
A subordinate norm is a special type of matrix norm that is directly related to a vector norm. This connection means that the behavior of a matrix acting on a vector can be neatly captured using this kind of norm.
  • When we say a matrix norm is subordinate, it means for every vector norm \( \|\cdot\|_v \), the inequality \( \|Ax\|_v \leq \|A\|\|x\|_v \) holds true for all vectors \( x \).
  • This ensures that the matrix norm does not exceed the prediction of the vector norm, making calculations easier.
  • The term "subordinate" reflects the norm's capacity to conform to the properties of the vector norm it relates to.
The subordinate nature is essential in many linear algebra applications such as ensuring that inequalities hold when matrices are applied to vectors. Understanding this concept lays the foundation for working comfortably with complex matrix operations.
Matrix Transformation
In the context of matrix norms, a matrix transformation involves operations like similarity transformations or changes of basis.
  • A similarity transformation is a common transformation where a matrix \( A \) is converted into another matrix \( SAS^{-1} \) using a nonsingular matrix \( S \).
  • These transformations are crucial in linear algebra because they often preserve certain properties like eigenvalues, which are pivotal in understanding the matrix's characteristics.
  • In our specific example, we define a new norm \( \|A\|' = \|SAS^{-1}\| \), which uses the transformation process to maintain the norm's properties.
Matrix transformations can alter the perspective from which we view a matrix but maintain core properties, making them extremely useful. They help in simplifying matrices, solving equations, and deriving properties without altering essential characteristics.
Linear Algebra Properties
Linear algebra is underpinned by several properties and structures that are preserved in operations and transformations.
  • Non-negativity and Definiteness: For a matrix norm \( \|A\|' \), it must hold that \( \|A\|' \geq 0 \), and \( \|A\|' = 0 \) only when \( A = 0 \).
  • Homogeneity: This property requires that scaling a matrix by a scalar \( c \) scales the norm by its absolute value \( |c| \).
  • Triangle Inequality: For any two matrices \( A \) and \( B \), the property \( \|A + B\|' \leq \|A\|' + \|B\|' \) must be satisfied. This is critical for maintaining linear structure in matrix operations.
  • Subordination: With respect to vector norms, subordinate matrix norms must satisfy \( \|Ax\|'_v \leq \|A\|\|x\|'_v \), reinforcing their relationship.
These properties guarantee that matrix norms behave predictably, supporting stability and consistency in linear algebra operations. By ensuring such properties are met, it becomes easier to perform accurate calculations and derive theoretical insights.

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Most popular questions from this chapter

Consider $$A=\left[\begin{array}{rrr} 2 & 6 & -4 \\ 6 & 17 & -17 \\ -4 & -17 & -20 \end{array}\right]$$ Determine directly the factorization \(A=L D L^{T}\), where \(D\) is diagonal and \(L\) is unit lower triangular; that is, do \(n o t\) use Gaussian elimination.

Prove that if \(\rho(A)<1\), then \(I-A\) is invertible and \((I-A)^{-1}=\sum_{k=0}^{\infty} A^{k}\).

Prove that if the matrix \(\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]\) is nonnegative definite, then it has a factorization \(L L^{T}\) in which \(L\) is lower triangular.

Consider the linear system $$ \left[\begin{array}{rrr} 2 & 0 & -1 \\ -2 & -10 & 0 \\ -1 & -1 & 4 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]=\left[\begin{array}{r} 1 \\ -12 \\ 2 \end{array}\right] $$ Using the starting vector \(x^{(0)}=(0,0,0)^{T}\), carry out two iterations of each of the following methods: a. Jacobi b. Gauss-Seidel c. Conjugate gradient

Prove that if all the leading principal minors of \(A\) are nonsingular, then \(A\) has a factorization \(L D U\) in which \(L\) is unit lower triangular, \(U\) is unit upper triangular, and \(D\) is diagonal.
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