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Chapter 20: Problem 15

(Definiteness) Let A and B be positive definite \(n \times n\) matrices. Are \(-A, A^{\top}, A+B, A-B\) positive definite?

Short Answer

Expert verified
\(-A\) is not positive definite; \(A^{\top}\) and \(A+B\) are positive definite; \(A-B\) is not guaranteed to be positive definite.

Step by step solution

01

Understand Positive Definiteness

A matrix \(A\) is positive definite if for all vectors \(x\) (excluding zero vector), the expression \(x^{\top}Ax > 0\) holds. This implies that the eigenvalues of a positive definite matrix are all positive.
02

Determine if \(-A\) is Positive Definite

If \(A\) is positive definite, then for all \(x\), \(x^{\top}Ax > 0\). Therefore, \(-A\) results in \(x^{\top}(-A)x = -x^{\top}Ax < 0\). Thus, \(-A\) cannot be positive definite because this inequality shows that all eigenvalues are negative.
03

Determine if \(A^{\top}\) is Positive Definite

For a matrix \(A\), if \(A\) is positive definite, then so is \(A^{\top}\). This is because the eigenvalues of \(A\) and \(A^{\top}\) are the same, and therefore, \(A^{\top}\) must also have all positive eigenvalues.
04

Determine if \(A+B\) is Positive Definite

Both \(A\) and \(B\) are positive definite, meaning all their eigenvalues are positive. The sum \(A+B\) also results in a positive definite matrix, since adding two matrices with positive eigenvalues results in a matrix with positive eigenvalues.
05

Determine if \(A-B\) is Positive Definite

To determine the definiteness of \(A-B\), consider that if \(A\) and \(B\) are positive definite, this does not inherently guarantee \(A-B\) is positive definite, since the eigenvalues of \(A\) minus those of \(B\) could yield non-positive values. Without additional information about the relative magnitudes, \(A-B\) is not guaranteed to be positive definite.

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Key Concepts

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

Eigenvalues
Eigenvalues are fundamental in understanding the properties of matrices, particularly in determining whether a matrix is positive definite. In simple terms, eigenvalues are scalars associated with a matrix that give insight into the matrix's intrinsic characteristics. Here's how they work:
  • An eigenvalue corresponds to an eigenvector such that when the matrix is applied to the eigenvector, the output is simply a scaled version of the eigenvector itself.
  • If a matrix is known to be positive definite, all its eigenvalues are strictly greater than zero. This is because the positive definiteness of a matrix implies that it "stretches out" space, giving values greater than zero when applied.
  • The eigenvalues not only help in defining definiteness but also play a crucial role in applications such as stability analysis and optimization problems.
To check whether a given matrix is positive definite, determining if all eigenvalues are positive is a crucial step. If any eigenvalue is non-positive, the matrix cannot be positive definite.
Matrix Definiteness
Matrix definiteness is a concept that describes how matrices transform vectors in terms of size and direction. There are different types of matrix definiteness based on the values they return when interacting with vectors:
  • Positive Definite: A matrix is positive definite if, for any non-zero vector \(x\), the quadratic form \(x^{\top} Ax > 0\). This implies that the matrix essentially transforms all vectors to have a positive magnitude.
  • Negative Definite: Conversely, a matrix is negative definite if \(x^{\top} Ax < 0\) for all non-zero \(x\). All eigenvalues of this matrix are negative.
  • Positive Semi-Definite: Here, \(x^{\top} Ax \geq 0\) for all \(x\). While all eigenvalues must be non-negative, some can be zero.
  • Indefinite: If a matrix returns both positive and negative values, it is termed indefinite. This occurs when there are both positive and negative eigenvalues.
Understanding definiteness helps in many mathematical and engineering applications, such as system stability and stress-strain calculations in materials science.
Matrix Transpose
The transpose of a matrix is an operation that flips the matrix over its diagonal. Denoted as \(A^{\top}\), the transpose of a matrix \(A\) converts its rows into columns and its columns into rows:
  • The element at row \(i\) and column \(j\) of matrix \(A\) becomes the element at row \(j\) and column \(i\) in \(A^{\top}\).
  • For instance, the transpose of \ \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \] is: \ \[ \begin{bmatrix} a & c \ b & d \end{bmatrix} \]
One intriguing property of transpose in positive definite matrices is that if a matrix \(A\) is positive definite, its transpose \(A^{\top}\) will also be positive definite. This is because the process of transposition does not alter the eigenvalues; it only rearranges the matrix, hence maintaining the conditions for positive definiteness. This feature is particularly useful because it often simplifies calculations in systems analysis and provides symmetry.

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Most popular questions from this chapter

Solve the following linear systems by Gauss elimination, with partial pivoting if necessary (but without scaling). Show the intermediate steps. Check the result by substitution. If no solution or more than one solution exists, give a reason. $$\begin{aligned} 4 x_{1}+10 x_{2}-2 x_{3} &=-20 \\ -x_{1}-15 x_{2}+3 x_{2} &=30 \\ 25 x_{2}-5 x_{3} &=-50 \end{aligned}$$

Show the factorization and solve. $$\begin{array}{l} 4 x_{1}+6 x_{2}+8 x_{3}=0 \\ 6 x_{1}+34 x_{2}+52 x_{3}=-80 \\ 8 x_{1}+52 x_{2}+129 x_{3}=-226 \end{array}$$

Solve the following linear systems by Gauss elimination, with partial pivoting if necessary (but without scaling). Show the intermediate steps. Check the result by substitution. If no solution or more than one solution exists, give a reason. $$\begin{aligned} 5 x_{1}+3 x_{2}+x_{3} &=2 \\ -4 x_{2}+8 x_{3} &=-3 \\ 10 x_{1}-6 x_{2}+26 x_{3} &=0 \end{aligned}$$

Show the factorization and solve by Doolitule's method. $$\begin{aligned} 2 x_{1}+x_{2}+2 x_{3} &=0 \\ -2 x_{1}+2 x_{2}+x_{3} &=0 \\ x_{1}+2 x_{2}-2 x_{3} &=36 \end{aligned}$$

Illustrate with a \(2 \times 2\) matrix that an eigenvalue may very well lie on a Gerschgorin circle (so that Gerschgorin disks can generally not be replaced with smaller disks without losing the inclusion property).
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