/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Without multiplying \(S=\left[\b... [FREE SOLUTION] | 91Ó°ÊÓ

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Without multiplying \(S=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\\ \sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 0 & 5\end{array}\right]\left[\begin{array}{rr}\cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta\end{array}\right]\), find (a) the determinant of \(S\) (b) the eigenvalues of \(S\) (c) the eigenvectors of \(S\) (d) a reason why \(S\) is symmetric positive definite.

Short Answer

Expert verified
(a) 10, (b) 2, 5, (c) Rotated standard basis vectors, (d) Symmetric with positive eigenvalues.

Step by step solution

01

Understanding the Matrices

To solve the problem, we first need to understand the components involved. The given matrix product \( S = M_1 \, D \, M_2 \), where:- \( M_1 = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \) is a rotation matrix.- \( D = \begin{bmatrix} 2 & 0 \ 0 & 5 \end{bmatrix} \) is a diagonal matrix.- \( M_2 = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix} \) is another rotation matrix, which is the inverse of \( M_1 \).
02

Determinant of S

The determinant of a product of matrices is the product of their determinants. The determinant of a rotation matrix is 1, and the determinant of a diagonal matrix is the product of its diagonal elements.\[\det(S) = \det(M_1) \cdot \det(D) \cdot \det(M_2) = 1 \cdot (2 \cdot 5) \cdot 1 = 10\]
03

Eigenvalues of S

Since the matrix \( S \) is similar to the diagonal matrix \( D \) through rotation matrices, its eigenvalues are the same as those of \( D \). Thus, the eigenvalues of \( S \) are 2 and 5.
04

Eigenvectors of S

The eigenvectors of a diagonal matrix \( D \) are the standard basis vectors, \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \), for eigenvalues 2 and 5, respectively. After applying the rotation matrix \( M_1 \), the eigenvectors of \( S \) become:\[\begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} 1 \ 0 \end{bmatrix} = \begin{bmatrix} \cos \theta \ \sin \theta \end{bmatrix},\begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} 0 \ 1 \end{bmatrix} = \begin{bmatrix} -\sin \theta \ \cos \theta \end{bmatrix} \]
05

Symmetric Positive Definite Verification

A matrix is symmetric positive definite if it is symmetric and all its eigenvalues are positive. Since \( S\) can be expressed as \( M_1 D M_1^T \) and both \( M_1 \) and \( D \) have non-negative entries, \( S \) is symmetric due to the structure of the matrices involved. The eigenvalues, 2 and 5, are positive, proving \( S \) is symmetric positive definite.

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Key Concepts

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

Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors play a crucial role in understanding matrices. An eigenvalue is a number that describes the factor by which a linear transformation alters a vector. The corresponding eigenvector is the vector that undergoes this transformation while remaining in the same direction.
Consider the matrix transformation represented by matrix \( S \). In linear algebra, if \( A \) is our transformation matrix, the equation \( Av = lambda v \) holds, where \(v\) is an eigenvector and \( lambda \) is the corresponding eigenvalue.
In the context of the exercise, matrix \( S \) is expressed similar to matrix \( D \), implying it preserves eigenvalues this way. \( D \), being diagonal, has straightforward eigenvalues, which are just the entries on its diagonal: 2 and 5. Therefore, these are also eigenvalues of \( S \).
Finding eigenvectors involves solving \( (S - lambda I)v = 0 \), where \( I \) is the identity matrix. For diagonal matrices, eigenvectors correspond to each axis: \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \) correspond to eigenvalues 2 and 5. Since \( S \) is constructed by rotating \( D \), the eigenvectors of \( S \) are rotated versions of these standard basis vectors. Thus, they are \( \begin{bmatrix} \cos \theta \ \sin \theta \end{bmatrix} \) and \( \begin{bmatrix} -\sin \theta \ \cos \theta \end{bmatrix} \).
Matrix Theory
Matrix theory provides a fundamental understanding of linear transformations in math. A matrix is a rectangular arrangement of numbers or functions used for transformations. Through manipulation, matrices can illustrate systems of equations simultaneously.
  • Rotation Matrix: One interesting type of matrix is the rotation matrix. It rotates points in a plane. Rotation matrices are orthogonal, meaning their inverse is their transpose.
  • Diagonal Matrix: Another key player in matrix theory is the diagonal matrix. It simplifies operations as its non-diagonal entries are zero. The ease of eigenvalue computation is due to the direct representation of values on the diagonal.
The given problem utilizes a rotation matrix and a diagonal matrix. By expressing \( S \) as \( M_1 \, D \, M_2 \), where both \( M_1 \) and \( M_2 \) are rotation matrices and \( D \) is diagonal, we simplify our understanding of the problem.
The manipulation stems from realizing the multiplication of these structured matrices doesn't alter the rotation, allowing us to derive properties without matrix multiplication.
Symmetric Positive Definite Matrices
A matrix is symmetric if it equals its transpose, meaning \( A = A^T \). Symmetric matrices have nicer properties, especially in computations and applications like optimization and physics.
A matrix \( A \) is also considered positive definite if all its eigenvalues are positive, which generally implies specific beneficial properties like ensuring unique solutions and stability in systems.
  • **Symmetric Nature of S:** In our problem, matrix \( S \) maintains symmetry due to the construction with symmetric matrices, ensuring the eigenvectors remain orthogonal.
  • **Positive Definiteness:** It means matrix \( S \) will map any non-zero vector \( x \) to a positive dot product \( x^T S x > 0 \).
Since the eigenvalues of \( S \) are 2 and 5, both positive, plus the symmetry derived from the rotation matrix and diagonal matrix combination, \( S \) is confirmed to be symmetric positive definite.

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Most popular questions from this chapter

Tall thin matrices \(Q\) with orthonormal columns: \(Q^{\mathrm{T}} Q=I\). $$ Q^{\mathrm{T}} Q=\left[\begin{array}{c} q_{1}^{\mathrm{T}} \\ : \\ q_{n}^{\mathrm{T}} \end{array}\right]\left[\begin{array}{l} q_{1} \cdots q_{n} \\ {\left[\begin{array}{c} 1 \\ n \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I} \end{array}\right] $$ If this \(Q\) multiplies any vector \(x\), the length of the vector does not change : $$ \|Q x\|=\|x\| \text { because }(Q x)^{\mathrm{T}}(Q x)=x^{\mathrm{T}} Q^{\mathrm{T}} Q x=x^{\mathrm{T}} x $$ If \(m>n\) the \(m\) rows cannot be orthogonal in \(\mathbf{R}^{n}\). Tall thin matrices haye \(Q Q^{\mathrm{T}} \neq I\).

Find the matrices \(C_{1}\) and \(C_{2}\) containing independent columns of \(A_{1}\) and \(A_{2}\) : $$ A_{1}=\left[\begin{array}{lll} 1 & 3 & -2 \\ 3 & 9 & -6 \\ 2 & 6 & -4 \end{array}\right] \quad A_{2}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right] $$

If \(a_{11}, \ldots, a_{1 n}\) is the first row of a rank-1 matrix \(A\) and \(a_{11}, \ldots, a_{m 1}\) is the first column, find a formula for \(a_{i j}\). Good to check when \(a_{11}=2, a_{12}=3, a_{21}=4\). When will your formula break down? Then rank 1 is impossible or not unique.

Given an \(m\) by \(n\) by \(p\) tensor \(T\), how do you decide if \(T\) has rank 1 ?

In some data science applications, the first pivot is the largest number \(\left|a_{i j}\right|\) in \(A\). Then row \(i\) becomes the first pivot row \(u_{1}^{*}\). Column \(j\) is the first pivot column. Divide that column by \(a_{i j}\) so \(\ell_{1}\) has 1 in row \(i\). Then remove that \(\ell_{1} u_{1}^{*}\) from \(A\). This example finds \(a_{22}=4\) as the first pivot \((i=j=2)\). Dividing by 4 gives \(\ell_{1}\) : $$ \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{c} 1 / 2 \\ 1 \end{array}\right]\left[\begin{array}{ll} 3 & 4 \end{array}\right]+\left[\begin{array}{cc} -1 / 2 & 0 \\ 0 & 0 \end{array}\right]=\boldsymbol{\ell}_{1} \boldsymbol{u}_{1}^{*}+\boldsymbol{\ell}_{2} \boldsymbol{u}_{2}^{*}=\left[\begin{array}{cc} 1 / 2 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{cc} 3 & 4 \\ -1 / 2 & 0 \end{array}\right] $$ For this \(A\), both \(L\) and \(U\) involve permutations. \(P_{1}\) exchanges the rows to give \(L\). \(P_{2}\) exchanges the columns to give an upper triangular \(U\). Then \(P_{1} A P_{2}=L U\). Permuted in advance \(\quad P_{1} A P_{2}=\left[\begin{array}{cc}1 & 0 \\ 1 / 2 & 1\end{array}\right]\left[\begin{array}{cc}4 & 3 \\ 0 & -1 / 2\end{array}\right]=\left[\begin{array}{ll}4 & 3 \\ 2 & 1\end{array}\right]\) Question for \(A=\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right]:\) Apply complete pivoting to produce \(P_{1} A P_{2}=L U\).

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