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Chapter 1: Problem 9

Prove: a) Every square matrix is orthogonally congruent to itself. b) If \(B\) is orthogonally congruent to \(C\), then \(C\) is orthogonally congruent to \(B\). c) If \(B\) is orthogonally congruent to \(C\) and \(C\) is orthogonally congruent to \(D\). then \(B\) is orthogonally congruent to \(D\).

Short Answer

Expert verified
Question: Prove the following properties of orthogonally congruent matrices: a) Every square matrix \(A\) is orthogonally congruent to itself. b) If \(B\) is orthogonally congruent to \(C\), then \(C\) is orthogonally congruent to \(B\). c) If \(B\) is orthogonally congruent to \(C\) and \(C\) is orthogonally congruent to \(D\), then \(B\) is orthogonally congruent to \(D\).

Step by step solution

01

Find an orthogonal matrix Q

To prove that every square matrix \(A\) is orthogonally congruent to itself, we need to find an orthogonal matrix \(Q\) such that \(A = Q^T A Q\).
02

Use the identity matrix

The identity matrix \(I\) is an orthogonal matrix, since \(I^T I = I\). In this case, let's choose \(Q = I\).
03

Prove that A is orthogonally congruent to itself

Since we have chosen \(Q = I\), we need to show that \(A = I^T A I\). We have: \[ A = I^T A I = A \] Therefore, every square matrix \(A\) is orthogonally congruent to itself. Proof of property (b):
04

Assume B is orthogonally congruent to C

Given that \(B\) is orthogonally congruent to \(C\), we know that there exists an orthogonal matrix \(Q_1\) such that \(C = Q_1^T B Q_1\).
05

Find an orthogonal matrix for the reverse congruence

In order to prove that \(C\) is orthogonally congruent to \(B\), we need to find another orthogonal matrix \(Q_2\) such that \(B = Q_2^T C Q_2\).
06

Use the transpose of the original orthogonal matrix

Since \(Q_1\) is orthogonal, its transpose \(Q_1^T\) is also orthogonal. Let's choose \(Q_2 = Q_1^T\).
07

Prove that C is orthogonally congruent to B

Using our choice of \(Q_2 = Q_1^T\), we get: \[ B = Q_2^T C Q_2 = (Q_1^T)^T C Q_1^T = Q_1 C Q_1^T \] Now, we have \(C = Q_1^T B Q_1\) and \(B = Q_1 C Q_1^T\). Substituting the expression for \(C\) into the expression for \(B\), we get: \[ B = Q_1 (Q_1^T B Q_1) Q_1^T \] Thus, if \(B\) is orthogonally congruent to \(C\), then \(C\) is orthogonally congruent to \(B\). Proof of property (c):
08

Assume B is orthogonally congruent to C and C to D

Given that \(B\) is orthogonally congruent to \(C\) and \(C\) is orthogonally congruent to \(D\), we know that there exist orthogonal matrices \(Q_1\) and \(Q_2\) such that \[ C = Q_1^T B Q_1 \quad \text{and} \quad D = Q_2^T C Q_2 \]
09

Find an orthogonal matrix for B to D congruence

To prove that \(B\) is orthogonally congruent to \(D\), we need to find an orthogonal matrix \(Q_3\) such that \(D = Q_3^T B Q_3\).
10

Combine the two orthogonal matrices

We can choose \(Q_3 = Q_2 Q_1\). Since both \(Q_1\) and \(Q_2\) are orthogonal, their product \(Q_3\) is also orthogonal.
11

Prove that B is orthogonally congruent to D

Using our choice of \(Q_3 = Q_2 Q_1\), we get: \[ D = Q_3^T B Q_3 = (Q_2 Q_1)^T B (Q_2 Q_1) = Q_1^T Q_2^T B Q_2 Q_1 \] Now, substitute the expression for \(C\) into the expression for \(D\): \[ D = Q_1^T (Q_1^T B Q_1) Q_1 = (Q_1^T Q_1)^T B (Q_1^T Q_1) = I^T B I = B \] Thus, if \(B\) is orthogonally congruent to \(C\) and \(C\) is orthogonally congruent to \(D\), then \(B\) is orthogonally congruent to \(D\).

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Key Concepts

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

Orthogonal matrices
Orthogonal matrices are fascinating structures in linear algebra, known for preserving lengths and angles during transformations. To visualize what this means, imagine a shape on a plane. When an orthogonal matrix is applied to the shape, its orientation might change, but the distances between points within the shape remain the same. This is because orthogonal matrices are essentially rotation matrices that can also include reflections.
Orthogonal matrices have an important property: they are the inverse of their own transpose. Let me explain. If you have an orthogonal matrix \( Q \), then \( Q^T Q = QQ^T = I \), where \( I \) is the identity matrix. This means that multiplying an orthogonal matrix by its transpose results in an identity matrix.
  • The columns of an orthogonal matrix are orthonormal, which means each column vector has a length of one and is orthogonal (perpendicular) to other columns.
  • They are used extensively in applications requiring stability in computations, such as in numerical analysis.
  • Simplification of matrix equations is another practical use for orthogonal matrices.
Understanding these key concepts will help you grasp why orthogonally congruent matrices exhibit such essential properties in math.
Square matrices
Square matrices are ubiquitous in linear algebra and mathematics. They are matrices with the same number of rows and columns, dimensionally represented by \( n \times n \). These matrices play a significant role since they provide a framework for many computations and transformations.
For square matrices:
  • Determinants are one of their critical attributes. A non-zero determinant implies that the matrix is invertible.
  • Eigenvalues and eigenvectors are applicable to square matrices and aid in understanding matrix transformations.
  • The identity matrix is a special type of square matrix where the diagonal elements are one, and other elements are zero.
Square matrices are an essential building block, especially for operations like finding the inverse, determinants, or applying matrix functions.
Matrix transpose
The transpose of a matrix is a simple yet powerful concept in linear algebra. It involves flipping a matrix over its diagonal, effectively swapping the row and column indices of each element. If you begin with a matrix \( A \), its transpose is denoted \( A^T \).
Here are a few important aspects:
  • The transpose operation does not change the eigenvalues of the matrix.
  • It plays a significant role in various matrix equations and computations, especially when dealing with symmetric matrices, where \( A^T = A \).
  • For orthogonal matrices, remember that the transpose is the same as the inverse, which simplifies many calculations.
Transposing matrices can solve specific types of matrix equations and help tackle problems involving symmetry and orthogonality. Understanding matrix transposes will thus enrich your problem-solving toolkit in algebra.

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Most popular questions from this chapter

Let \(A\) be an \(m \times n\) matrix. Let \(C=\left(c_{i j}\right)\) be the \(m \times m\) matrix which differs from the identity only in that \(c_{h h}=\lambda \neq 0\). Let \(B\) be the \(m \times m\) matrix which differs from the identity only in the \(k\) th column, in which \(b_{k k}=1\). a) Show that \(C A\) is obtained from \(A\) by multiplying the \(h\) th row by \(\lambda\). b) Show that \(B A\) is obtained from \(A\) by adding \(b_{j k}\) times the \(k\) th row to the \(j\) th row for all \(j\) except \(k\). c) Show that \(B\) and \(C\) are nonsingular. Chapter 1 Vectors and Matrices Remark These results show that steps I and III of Section \(1.10\) are obtained by multiplying A on the left by a nonsingular matrix. The same assertion applies to step II (Problem 11 following Section 1.9).

In \(\underset{E^{n}}{\vec{P}}\) the line segment \(P_{1} P_{2}\) joining points \(P_{1}, P_{2}\) is formed of all points \(P\) such that \(P_{1} P=t \quad \overline{P_{1}} P_{2}\), where \(0 \leq t \leq 1\). a) In \(E^{4}\), show that \((5,10,-1,8)\) is on the line segment from \((1,2,3,4)\) to \((7,14,-3,10)\). b) In \(E^{4}\), is \((2,8,1,6)\) on the line segment joining \((1,7,2,5)\) to \((9,2,0,7)\) ? c) In \(E^{5}\), find the midpoint of the line segment from \(P_{1}:(2,0,1,3,7)\) to \(P_{2}:(10,4,-3\), 3 , 5); that is, find the point \(P\) on the segment such that \(\left|\overrightarrow{P_{1} P}\right|=\left|\overrightarrow{P_{2} P}\right|\). d) In \(E^{5}\), trisect the line segment joining \((7,1,7,9,6)\) to \((16,-2,10,3,0)\). e) In \(E^{n}\), show that if \(P\) is on the line segment \(P_{1} P_{2}\), then \(\left|\overrightarrow{P_{1} P}\right|+\left|\overrightarrow{P_{2} P}\right|=\left|\overrightarrow{P_{1} P_{2}}\right|\).

Prove the identities: a) \(|\mathbf{u} \times \mathbf{v}|^{2}=\left|\begin{array}{ll}\mathbf{u} \cdot \mathbf{u} & \mathbf{u} \cdot \mathbf{v} \\ \mathbf{v} \cdot \mathbf{u} & \mathbf{v} \cdot \mathbf{v}\end{array}\right|\) b) \((\mathbf{u} \times \mathbf{v}) \cdot(\mathbf{w} \times \mathbf{z})=(\mathbf{u} \cdot \mathbf{w})(\mathbf{v} \cdot \mathbf{z})-(\mathbf{u} \cdot \mathbf{z})(\mathbf{v} \cdot \mathbf{w})\) c) \((\mathbf{u} \times \mathbf{v}) \times(\mathbf{u} \times \mathbf{w})=(\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}) \mathbf{u}\) d) \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})+\mathbf{v} \times(\mathbf{w} \times \mathbf{u})+\mathbf{w} \times(\mathbf{u} \times \mathbf{v})=0\)

Let points \(P_{1}:(1,0,2), P_{2}:(2,1,3), P_{3}:(1,5,4)\) be given in space. a) From a rough graph, verify that the points are vertices of a triangle. b) Find the lengths of the sides of the triangle. c) Find the angles of the triangle. d) Find the area of the triangle. e) Find the length of the altitude on side \(P_{1} P_{2}\). f) Find the midpoint of side \(P_{1} P_{2}\). g) Find the point where the medians meet.

Consider two equations in three unknowns: $$ a_{1} x+b_{1} y+c_{1} z=k_{1}, \quad a_{2} x+b_{2} y+c_{2} z=k_{2} . $$ a) Show that if Gaussian elimination can be carried out to solve for \(x\) and \(y\), then by writing \(z=t\) the solutions become parametric equations for a line in space (Section 1.3). b) Assume that the two equations represent two planes in space. Interpret geometrically the case in which the equations have no solution and the case in which elimination leads to a second equation \(0=0\).
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