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Chapter 8: Problem 2

Let \(A\) be an invertible symmetric matrix. Show that \(A^{-1}\) is symmetric.

Short Answer

Expert verified
The inverse of an invertible symmetric matrix A is symmetric. We showed this by starting with the equation \(AA^{-1} = I\), where A is symmetric and invertible, and its transpose \(A^T = A\). We then transposed the equation to get \((A^{-1})^T A^T = I^T\). Using properties of symmetric and identity matrices, we wrote the equation as \((A^{-1})^T A = I\). By comparing this equation with the original one, we concluded that \((A^{-1})^T = A^{-1}\), proving that \(A^{-1}\) is symmetric.

Step by step solution

01

Write down the given information

We are given that A is an invertible symmetric matrix. This means: 1. A is symmetric: \(A^T = A\) 2. A is invertible: There exists an inverse matrix \(A^{-1}\) such that \(AA^{-1} = A^{-1}A = I\), where I is the identity matrix.
02

Transpose the equation \(AA^{-1}\)

We know that \(AA^{-1} = I\). Let's transpose both sides of this equation: \((AA^{-1})^T = (I)^T\) Remember that the transpose of a product of two matrices is the product of the transpose of the matrices in the reverse order. So we have: \((A^{-1})^T A^T = I^T\)
03

Use the properties of symmetric matrices and identity matrices

We know that A is symmetric, so its transpose \(A^T\) is equal to A. Also, since the transpose of the identity matrix I is itself, \(I^T = I\). Using these properties, we can rewrite the equation from step 2 as: \((A^{-1})^T A = I\)
04

Compare the equations

Now, compare the equation from step 3 with the given equation \(AA^{-1} = I\). We have: \((A^{-1})^T A = I\) and \(AA^{-1} = I\) Looking at these two equations, it's clear that they share the same structure. In other words, the inverse of the transpose of A shares the same relationship with A as A^-1 does. Because inverses are unique, we can conclude that \((A^{-1})^T = A^{-1}\).
05

State the conclusion

We have shown that the transpose of the inverse of A is equal to the inverse of A, or \((A^{-1})^T = A^{-1}\). This proves that the inverse of an invertible symmetric matrix A is also symmetric.

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Key Concepts

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

Invertible Matrix
An invertible matrix is a type of square matrix that has a special property. It means that there exists another matrix, known as the inverse matrix, which cancels the effects of the original matrix when multiplied together. Simply put, if you have a matrix A, and it is invertible, there is another matrix, denoted as \(A^{-1}\), such that:
  • \(AA^{-1} = I\)
  • \(A^{-1}A = I\)
Here, \(I\) is the identity matrix. The identity matrix acts like the number 1 in normal multiplication. So, when any matrix multiplies with the identity matrix, it remains unchanged.
A matrix can only be invertible if it's square, meaning it has the same number of rows and columns. Also, not every square matrix is invertible; it must fulfill specific criteria, like having a non-zero determinant. If a matrix's determinant is zero, it means the matrix is singular and thus not invertible.
Matrix Transpose
The transpose of a matrix is a simple yet powerful concept in linear algebra. Transposing essentially involves flipping a matrix over its diagonal. To create the transpose of a matrix, denote it by \(A^T\) if the original matrix is \(A\), you swap its rows and columns.
For example, if matrix \(A\) has its elements arranged as:
  • \(A = \begin{pmatrix}1 & 2 \3 & 4\end{pmatrix}\)
The transpose \(A^T\) will be structured as:
  • \(A^T = \begin{pmatrix}1 & 3 \2 & 4\end{pmatrix}\)
The transpose operation is critical in many mathematical operations and plays a key role when determining the symmetry of matrices. For instance, a matrix \(A\) is deemed symmetric if \(A^T = A\). Therefore, understanding how transposing affects matrix operations is essential for tasks such as solving systems of equations or analyzing matrix properties.
Matrix Inverse
The matrix inverse is like a magic key that unlocks solutions in algebra. An inverse of a matrix, typically denoted as \(A^{-1}\) when talking about matrix \(A\), is a matrix that reverses the effects of \(A\). When you multiply a matrix by its inverse, you get the identity matrix:
  • \(AA^{-1} = I\)
  • \(A^{-1}A = I\)
Calculating the inverse of a matrix involves a specific method, often using cofactors or row operations, and requires the matrix to be square and have a non-zero determinant.
One particularly interesting property arises when dealing with symmetric matrices. When a symmetric matrix is invertible, its inverse also tends to be symmetric. This property simplifies many mathematical and computational tasks, making inverse calculations more straightforward within this context.

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Most popular questions from this chapter

Show that if \(f_{1}\) is the quadratic form of the bilinear form \(g_{1}\), and \(f_{2}\) the quadratic form of the bilinear form \(g_{2}\), then \(f_{1}+f_{2}\) is the quadratic form of the bilinear form \(g_{1}+g_{2}\)

Let \(V\) be a finite dimensional space over \(\mathbf{C}\), with a positive definite hermitian form \(\langle,\),\(rangle . Let A: V \rightarrow V\) be a linear map. Show that the following conditions are equivalent: (i) We have \(A A^{*}=A^{*} A\). (ii) For all \(v \in V,\|A v\|=\left\|A^{*} v\right\|(\) where \(\|v\|=\sqrt{\langle v, v\rangle})\). (iii) We ean write \(A=B+i C\), where \(B, C\) are Hermitian, and \(B C=C B\).

Define a rotation of \(V\) to be a real unitary map \(A\) of \(V\) whose determinant is 1. Show that the matrix of \(A\) relative to an orthogonal basis of \(V\) is of type $$ \left(\begin{array}{rr} a & -b \\ b & a \end{array}\right) $$ for some real numbers \(a, b\) sueh that \(a^{2}+b^{2}=1 .\) Also prove the converse, that any linear map of \(V\) into itself represented by such a matrix on an orthogonal basis is unitary, and has determinant 1. Using ealeulus, one ean then eonelude that there exist a number \(\theta\) such that \(a=\cos \theta\) and \(b=\sin \theta\). 13\. Show that there exists a complex unitary matrix \(U\) such that, if $$ A=\left(\begin{array}{lr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{cc} e^{i \theta} & 0 \\ 0 & e^{-i \theta} \end{array}\right) $$ then \(U^{-1} A U=B\).

(a) A matrix \(A\) is called skew-symmetric if ' \(A=-A\). Show that any matrix M can be expressed as a sum of a symmetric matrix and a skewsymmetric one, and that these latter are uniquely determined. [Hint: Let \(\left.A=\frac{1}{2}\left(M+{ }^{t} M\right) .\right]\) (b) If \(A\) is skew-symmetric then \(A^{2}\) is symmetric. (c) Let \(A\) be skew-symmetric. Show that \(\operatorname{Det}(A)\) is 0 if \(A\) is an \(n \times n\) matrix and \(n\) is odd.

A square matrix \(C\) is said to be skew-symmetric if ' \(C=-C\). A bilinear form \(g\) on \(K^{n}\) is said to be alternating if \(g(X, X)=0\) for all \(X \in K^{n}\). Prove that a matrix \(C\) represents an alternating form if and only if it is skew-symmetric.
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