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Chapter 1: Problem 22

Show that if \(A\) is a symmetric nonsingular matrix then \(A^{-1}\) is also symmetric.

Short Answer

Expert verified
We know that \(A\) is symmetric and nonsingular, so \(A^T = A\). To show that \(A^{-1}\) is also symmetric, we need to prove that \((A^{-1})^T = A^{-1}\). First, we multiply the equation \(A^T = A\) by \((A^T)^{-1}\) on the left, giving \((A^T)^{-1} A^T = (A^T)^{-1} A\), which simplifies to \(A^{-1} A^T = (A^{-1})^T A\). We then multiply both sides by \(A^{-1}\) on the right and use the fact that \(AA^{-1}=I\), where \(I\) is the identity matrix. This gives us \(A^{-1} A^{-1} = (A^{-1})^T A^{-1}\). Thus, we have shown that \((A^{-1})^T = A^{-1}\), and so \(A^{-1}\) is also symmetric.

Step by step solution

01

Recall the definition of a symmetric matrix

A square matrix \(A\) is symmetric if its transpose \(A^T\) is equal to the matrix \(A\) itself, i.e., \(A^T = A\).
02

Recall the properties of matrix transposes and inverses

The transpose and inverse of a matrix have the following properties: 1. If \(A\) and \(B\) are matrices of the same size, then \((AB)^T = B^T A^T\). 2. If \(A\) is a nonsingular matrix, then \((A^{-1})^{-1} = A\). 3. If \(A\) is a nonsingular matrix, then \((A^T)^{-1} = (A^{-1})^T\).
03

Prove that \(A^{-1}\) is symmetric

Since \(A\) is symmetric, we have \(A^T = A\). Now we want to show that \((A^{-1})^T = A^{-1}\). First, multiply both sides of the equation \(A^T = A\) by \((A^T)^{-1}\) on the left: \((A^T)^{-1} A^T = (A^T)^{-1} A\). Using the properties of matrix transposes and inverses, we have: \(A^{-1} A^T = (A^{-1})^T A\). Since \(A\) is nonsingular, we can multiply both sides by \(A^{-1}\) on the right to get: \(A^{-1} A^T A^{-1} = (A^{-1})^T A A^{-1}\). Using the fact that \(AA^{-1}=I\), where \(I\) is the identity matrix, we obtain: \(A^{-1} I A^{-1} = (A^{-1})^T I A^{-1}\). Simplifying using the identity matrix gives us: \(A^{-1} A^{-1} = (A^{-1})^T A^{-1}\). Thus, we have shown that \((A^{-1})^T = A^{-1}\), and so \(A^{-1}\) is also symmetric.

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Key Concepts

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

transpose matrix
In linear algebra, the transpose of a matrix involves flipping it over its diagonal. Essentially, it means taking the rows of the matrix and turning them into columns, and vice versa. For a matrix \( A \), the transpose is denoted as \( A^T \). This operation is crucial because it retains the original matrix's dimensions while rearranging its elements, which is important for various calculations.
  • The transpose of a transpose returns the original matrix: \((A^T)^T = A\).
  • Transposing a product of matrices reverses the order: \((AB)^T = B^T A^T\).
  • A symmetric matrix, like the one in the exercise, satisfies the condition \( A^T = A \).
Understanding transpose allows you to better manipulate and work with matrices, which is vital in solving linear equations and other algebraic operations.
inverse matrix
An inverse matrix is essentially the "opposite" of a given square matrix \( A \). When a square matrix has an inverse, it is referred to as nonsingular or invertible, and its inverse is denoted as \( A^{-1} \). The purpose of finding a matrix's inverse is primarily to solve systems of linear equations, among other applications.
  • The product of a matrix and its inverse yields the identity matrix: \( AA^{-1} = A^{-1}A = I \).
  • The inverse of a product of matrices is the product of the inverses in reverse order: \( (AB)^{-1} = B^{-1}A^{-1} \).
  • An important property to remember is \( (A^T)^{-1} = (A^{-1})^T \).
Understanding how to compute and use inverses is critical for matrix algebra and solving equations with unique solutions.
nonsingular matrix
A nonsingular matrix, also known as an invertible matrix, is a square matrix that has a unique inverse. The existence of an inverse is contingent on the matrix having a non-zero determinant. Nonsingular matrices are vital in linear algebra because they allow for the unique solutions of linear systems.
  • If \( A \) is a nonsingular matrix, then there exists an inverse \( A^{-1} \) such that \( AA^{-1} = A^{-1}A = I \).
  • A matrix is nonsingular if its determinant is non-zero.
  • Nonsingular matrices are often associated with full-rank or linearly independent row and column vectors.
The property of being nonsingular makes a matrix robust and stable for use in mathematical computations, such as those found in the exercise provided.
identity matrix
The identity matrix is one of the most fundamental elements in matrix algebra. It is a square matrix with ones on the diagonal and zeros elsewhere. In mathematical notation, it is often represented by \( I \). The identity matrix plays a crucial role similar to the number one in multiplication.
  • Multiplying any matrix by the identity matrix leaves the original matrix unchanged: \( AI = IA = A \).
  • In the case of matrix inverses, \( AA^{-1} = A^{-1}A = I \) is a cornerstone principle.
  • For a \( n \times n \) matrix, the identity matrix is of the form \( I_n \) with dimension \( n \times n \).
The identity matrix serves as a benchmark in verifying operations like the existence of inverses and acts as the neutral element in matrix multiplication.

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Most popular questions from this chapter

Consider the matrix $$A=\left(\begin{array}{lllll} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \end{array}\right)$$ (a) Draw a graph that has \(A\) as its adjacency matrix. Be sure to label the vertices of the graph. (b) By inspecting the graph, determine the number of walks of length 2 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\) (c) Compute the second row of \(A^{3}\) and use it to determine the number of walks of length 3 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\)

If $$A=\left(\begin{array}{rrr} 3 & 1 & 4 \\ -2 & 0 & 1 \\ 1 & 2 & 2 \end{array}\right) \text { and } B=\left(\begin{array}{rrr} 1 & 0 & 2 \\ -3 & 1 & 1 \\ 2 & -4 & 1 \end{array}\right)$$ compute (a) \(2 A\) (b) \(A+B\) (c) \(2 A-3 B\) (d) \((2 A)^{T}-(3 B)^{T}\) (e) \(A B\) (f) \(B A\) (g) \(A^{T} B^{T}\) (h) \((B A)^{T}\)

Let \(A\) be a \(5 \times 3\) matrix. If $$\mathbf{b}=\mathbf{a}_{1}+\mathbf{a}_{2}=\mathbf{a}_{2}+\mathbf{a}_{3}$$ then what can you conclude about the number of solutions of the linear system \(A \mathbf{x}=\mathbf{b} ?\) Explain.

Let \(D\) be an \(n \times n\) diagonal matrix whose diagonal entries are either 0 or 1 (a) Show that \(D\) is idempotent. (b) Show that if \(X\) is a nonsingular matrix and \(A=X D X^{-1},\) then \(A\) is idempotent.

Let $$A=\left(\begin{array}{rrrr} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{array}\right)$$ Compute \(A^{2}\) and \(A^{3} .\) What will \(A^{2 n}\) and \(A^{2 n+1}\) turn out to be?
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