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Chapter 1: Problem 25

a. If \(A\) is invertible, is \(A+A^{T}\) always invertible? b. If \(A\) is invertible, is \(A+A\) always invertible?

Short Answer

Expert verified
(a) No, (b) Yes

Step by step solution

01

Understanding the Problem

To determine the invertibility of a matrix, remember that a matrix is invertible if it has a non-zero determinant. For part (a), if matrix \(A\) is invertible, we need to check if the matrix \(A+A^T\) is also invertible. For part (b), we need to check if \(A+A\), which simplifies to \(2A\), is invertible when \(A\) is invertible.
02

Consideration of Symmetric Matrices for Part (a)

Since \(A^T\) is the transpose of \(A\), \(A + A^T\) is always a symmetric matrix. A symmetric matrix is not guaranteed to be invertible even if \(A\) itself is invertible. Consider a counterexample: let \(A\) be a matrix where \(A + A^T = 0 \) (null or zero matrix), which is clearly not invertible.
03

Counterexample to Part (a)

Let \(A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}\). \(A^T = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}\). Then \(A + A^T = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}\). The determinant of \(A + A^T\) is \(3\), which confirms invertibility. This matrix is invertible. However, to disprove the general case, use another matrix: \(A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}\), then \(A^T = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}\), making \(A + A^T = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}\), which is not invertible. Thus, \(A + A^T\) may not always be invertible.
04

Simplifying Part (b) to Scalar Multiplication

For part (b), \(A + A\) simplifies to \(2A\). If \(A\) is invertible, multiplying it by a nonzero scalar (here, 2) results in \(2A\), which will still be invertible. The determinant of \(2A\) is \(2^n \cdot \text{det}(A)\), which is non-zero since \(\text{det}(A)eq 0\).
05

Conclusion for Both Parts

Part (a): \(A + A^T\) is not always invertible even if \(A\) is invertible, due to existence of counterexamples.Part (b): \(A + A\) or \(2A\) is always invertible if \(A\) is invertible, since multiplying by 2 does not affect invertibility.

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Key Concepts

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

Matrix Transpose
The matrix transpose is a fundamental concept in linear algebra. When you transpose a matrix, you essentially "flip" it over its diagonal. This means that
  • The element located at row \(i\) and column \(j\) in the original matrix \(A\) becomes the element at row \(j\) and column \(i\) in the transposed matrix \(A^T\).
  • For example, if you have a matrix \(A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}\), the transpose, \(A^T\), is \(\begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix}\).
Matrix transposition is a simple yet powerful tool that often reveals insights into a matrix's structure. It plays a crucial role in symmetric matrices, where a matrix is equal to its transpose.
Symmetric Matrices
Symmetric matrices are of particular interest because they hold unique algebraic properties. A symmetric matrix is one where the matrix is equal to its transpose, or mathematically, \(A = A^T\). This property ensures that the elements across the main diagonal are mirrored.
  • For example, \(\begin{bmatrix} 2 & -1 \ -1 & 3 \end{bmatrix}\) is symmetric since it equals its transpose.
  • Symmetric matrices are often involved in real-valued functions and optimization problems due to their stability and predictability.
Symmetric matrices are not guaranteed to be invertible, despite having pleasing mathematical properties. The invertibility of a matrix primarily depends on the determinant being non-zero, which isn't influenced by symmetry alone.
Matrix Determinant
The matrix determinant is a scalar value with profound importance. It helps determine whether a square matrix is invertible. The determinant can be thought of as a measure of the matrix's volume scaling effect during linear transformation.
  • If the determinant of a matrix is zero, the matrix is not invertible. This situation means the transformation results in volume collapse, losing dimensionality.
  • For a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is \(ad - bc\).
  • An invertible matrix must have a non-zero determinant.
Understanding and calculating determinants are fundamental in solving linear equations and performing eigenvalue analyses, among other applications.
Scalar Multiplication
Scalar multiplication involves multiplying every entry of a matrix by a scalar value. This operation directly affects the matrix's determinant and invertibility.
  • If matrix \(A\) is invertible, and you multiply it by a scalar \(k\), resulting matrix \(kA\) is also invertible as long as \(k eq 0\).
  • The determinant of \(kA\) is \(k^n \times \text{det}(A)\), where \(n\) is the size of the square matrix.
This scalar multiplication property is crucial in numerous areas, particularly in scaling transformations, without affecting matrix invertibility, provided the scalar is non-zero. Therefore, operations like \(A + A\) simplify to scalar multiplication \(2A\), easily maintaining invertibility provided \(A\) itself is invertible.

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