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Chapter 6: Problem 14

Prove that if \(A\) and \(B\) are unitarily equivalent matrices, then \(A\) is positive definite [semidefinite] if and only if \(B\) is positive definite [semidefinite]. (See the definitions in the exercises in Section 6.4.)

Short Answer

Expert verified
Given that A and B are unitarily equivalent matrices, meaning there exists a unitary matrix U such that \( B = U^* A U \). A is positive definite (or semidefinite), i.e., for any non-zero vector x, \( x^* A x > 0 \) (or \( x^* A x \geq 0 \) for semidefinite). To prove that B is also positive definite (or semidefinite), we analyze \( x^* B x \) by substituting B: \( x^* B x = x^* (U^* A U) x \), then rearrange the expression: \( x^* (U^* A U) x = (x^* U^*) A (Ux) \). Let \( y = Ux \), then the expression becomes: \( y^* A y \). Since A is positive definite (or semidefinite), this implies that \( x^* B x > 0 \) (or \( x^* B x \geq 0 \) for semidefinite) for any non-zero vector x. Thus, if A is positive definite (or semidefinite) and unitarily equivalent to B, then B is also positive definite (or semidefinite).

Step by step solution

01

Given Data

The given data is: 1. A and B are unitarily equivalent matrices. This means there exists a unitary matrix U such that \( B = U^* A U \). 2. A is a positive definite (or semidefinite) matrix, i.e., for any non-zero vector x, \( x^* A x > 0 \) (or \( x^* A x \geq 0 \) for semidefinite).
02

Analyzing B

To prove that B is also positive definite (or semidefinite), we need to show that for any non-zero vector x, \( x^* B x > 0 \) (or \( x^* B x \geq 0 \) for semidefinite).
03

Substituting B

To do this, we substitute the expression for B from the given data, i.e., \( B = U^* A U \). We will analyze \( x^* B x \) by substituting B: \( x^* B x = x^* (U^* A U) x \)
04

Rearranging the expression

Next, we rearrange the expression by using the property \( (AB)^* = B^* A^* \): \( x^* (U^* A U) x = (x^* U^*) A (Ux) \) Now, let's define a new vector y such that \( y = Ux \). Then, \( x^* U^* = y^* \) and the expression becomes: \( y^* A y \)
05

Relating A and B

Since A is positive definite (or semidefinite), the expression \( y^* A y > 0 \) (or \( y^* A y \geq 0 \) for semidefinite) for any non-zero vector y. This implies that \( x^* B x > 0 \) (or \( x^* B x \geq 0 \) for semidefinite) for any non-zero vector x. Thus, we have proved that if A is positive definite (or semidefinite) and unitarily equivalent to B, then B is also positive definite (or semidefinite).

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Key Concepts

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

Positive Definite Matrices
In mathematics, especially in linear algebra, a positive definite matrix is a matrix that generalizes the concept of positive numbers to higher dimensions. For a matrix to be positive definite, it must satisfy certain conditions when paired with a non-zero vector.
  • The quadratic form for any non-zero vector \( x \), denoted as \( x^* A x \), must be greater than zero: \( x^* A x > 0 \).
  • This is true in the context of complex numbers, where \( x^* \) denotes the conjugate transpose of \( x \).
Positive definite matrices have several important properties:
  • All eigenvalues of a positive definite matrix are positive. This assures certain stability and invertibility of matrices.
  • Such matrices are invertible, and their inverses are also positive definite.
  • They play a fundamental role in optimization problems, where minimizing quadratic forms is often involved.
Recognizing whether a matrix is positive definite involves checking these properties, often through determining if all principal minors (determinants of upper-left submatrices) are positive. This concept lays the foundation for many areas in numerical analysis and systematic methods in engineering.
Positive Semidefinite Matrices
Positive semidefinite matrices are a generalization of positive definite matrices. They allow zero outcomes when considering their quadratic form involving non-zero vectors.
  • For a matrix \( A \) to be positive semidefinite, the quadratic form \( x^* A x \) is non-negative for every non-zero vector \( x \): \( x^* A x \geq 0 \).
  • This means the matrix is allowed to have zero eigenvalues but cannot have negative eigenvalues.
Important aspects of positive semidefinite matrices include:
  • They arise naturally in statistics, particularly in covariance matrices where non-negative correlation structures are necessary.
  • If a matrix is positive semidefinite, then so is any of its principal minors.
  • They may not be invertible, especially if zero eigenvalues are present, which requires special handling in computations.
Understanding positive semidefinite matrices is crucial for fields dealing with variances and deviations, such as machine learning and financial risk management. They ensure reliability in probabilistic and statistical models.
Unitary Transformation
A unitary transformation is a type of linear transformation, represented by a unitary matrix, which preserves the lengths of vectors. It is akin to a rotation or reflection that does not alter the magnitude of any vector in the space.
  • A matrix \( U \) is unitary if its conjugate transpose is its inverse, \( U^* U = U U^* = I \), where \( I \) is the identity matrix.
  • Such transformations preserve the inner product, meaning for vectors \( x \) and \( y \), the inner product \( x^* y \) remains unchanged: \( (Ux)^* (Uy) = x^* y \).
The significance of unitary transformations lies in:
  • The preservation of angles and distances, making them ideal in quantum mechanics and signal processing.
  • Facilitating diagonalization of normal matrices, simplifying complex computations.
  • Being instrumental in converting problems into easier equivalent forms without losing essential characteristics of the space.
Unitary equivalence plays an essential role in demonstrating the properties of matrices, such as proving the correspondence of spectral properties which is vital in the provided solution example on matrix equivalence and definiteness.

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Most popular questions from this chapter

Prove that if \(\mathrm{T}\) is a unitary operator on a finite-dimensional inner product space \(\mathrm{V}\), then \(\mathrm{T}\) has a unitary square root; that is, there exists a unitary operator \(U\) such that \(T=U^{2}\). Visit goo.gl/jADTaS for a solution.

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