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Chapter 7: Problem 5

Determine which of the matrices in Exercises \(1-6\) are symmetric. $$ \left[\begin{array}{rrr}{-6} & {2} & {0} \\ {2} & {-6} & {2} \\ {0} & {2} & {-6}\end{array}\right] $$

Short Answer

Expert verified
The matrix is symmetric as it is equal to its transpose.

Step by step solution

01

Understand the Definition of a Symmetric Matrix

A matrix is called symmetric if it is equal to its own transpose. In simpler terms, a matrix \( A \) is symmetric if \( A = A^T \). This means that for all elements \( a_{ij} \) in the matrix, \( a_{ij} \) must equal \( a_{ji} \).
02

Find the Transpose of the Given Matrix

Given the matrix, \( A = \begin{bmatrix} -6 & 2 & 0 \ 2 & -6 & 2 \ 0 & 2 & -6 \end{bmatrix} \). The transpose of \( A \), denoted \( A^T \), is obtained by switching its rows and columns. Therefore, \( A^T = \begin{bmatrix} -6 & 2 & 0 \ 2 & -6 & 2 \ 0 & 2 & -6 \end{bmatrix} \).
03

Compare the Matrix to Its Transpose

Compare the original matrix \( A = \begin{bmatrix} -6 & 2 & 0 \ 2 & -6 & 2 \ 0 & 2 & -6 \end{bmatrix} \) with its transpose \( A^T = \begin{bmatrix} -6 & 2 & 0 \ 2 & -6 & 2 \ 0 & 2 & -6 \end{bmatrix} \). Each corresponding element is the same in both matrices.
04

Conclude if the Matrix is Symmetric

Since every element of \( A \) matches the corresponding element of \( A^T \), \( A = A^T \). Therefore, the matrix is symmetric.

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Key Concepts

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

Matrix Transpose
In linear algebra, understanding the concept of a matrix transpose is crucial. To "transpose" a matrix means to swap its rows with its columns. If you have a matrix \( A \), its transpose, denoted as \( A^T \), is formed by converting the first row of \( A \) into the first column of \( A^T \), the second row into the second column, and so forth.
  • The process keeps the diagonal elements in their place.
  • Elements above the diagonal are moved symmetrically below the diagonal in \( A^T \), and vice versa.
For example, given a matrix \(\begin{bmatrix}1 & 2 \3 & 4\end{bmatrix}\), its transpose will be \( \begin{bmatrix}1 & 3 \2 & 4\end{bmatrix} \). This action is simple yet powerful, playing a pivotal role in various linear algebra applications.
Symmetric Matrix Properties
A symmetric matrix is one that is equal to its transpose. In mathematical terms, a matrix \( A \) is symmetric if \( A = A^T \). This means:
  • All elements across its main diagonal (from top left to bottom right) remain unchanged.
  • The elements of the matrix are mirrored symmetrically above and below this diagonal.
A notable property of symmetric matrices is that their eigenvectors are orthogonal, meaning they are at right angles to each other. Symmetric matrices feature prominently in optimization algorithms and physics, depicting relationships that are bidirectional, like forces between two interacting particles.
Linear Algebra Concepts
Linear algebra is the study of vectors, vector spaces, and linear transformations. Matrices are a key representation of these concepts.
  • Matrices allow for efficient solutions to systems of linear equations.
  • They encode information in various real-world applications, from computer graphics to quantum mechanics.
Understanding how to manipulate matrices鈥攕uch as taking transposes, identifying symmetry, and finding determinants鈥攊s fundamental. These skills enable us to explore solutions of linear systems, perform computer modeling, and understand theoretical properties like linear mappings. Linear algebra concepts form the bedrock of many modern technological advancements, making them indispensable in both theoretical and applied science domains.

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Most popular questions from this chapter

Let \(A=P D P^{-1},\) where \(P\) is orthogonal and \(D\) is diagonal, and let \(\lambda\) be an eigenvalue of \(A\) of multiplicity \(k\) . Then \(\lambda\) appears \(k\) times on the diagonal of \(D .\) Explain why the dimension of the eigenspace for \(\lambda\) is \(k\)

Find an SVD of each matrix [Hint: In Exercise 11, one choice for \(U\) is \(\left[\begin{array}{rrr}{-1 / 3} & {2 / 3} & {2 / 3} \\ {2 / 3} & {-1 / 3} & {2 / 3} \\ {2 / 3} & {2 / 3} & {-1 / 3}\end{array}\right]\) In Exercise \(12,\) one column of \(U\) can be \(\left[\begin{array}{c}{1 / \sqrt{6}} \\ {-2 / \sqrt{6}} \\ {1 / \sqrt{6}}\end{array}\right].\)] \(\left[\begin{array}{rr}{-3} & {1} \\ {6} & {-2} \\ {6} & {-2}\end{array}\right]\)

Find the singular values of the matrices. \(\left[\begin{array}{rr}{1} & {0} \\ {0} & {-3}\end{array}\right]\)

[M] Orthogonally diagonalize the matrices in Exercises \(37-40\) . To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and, for each eigenvalue \(\lambda,\) find an orthonormal basis for \(\mathrm{Nul}(A-\lambda I),\) as in Examples 2 and \(3 .\) $$ \left[\begin{array}{rrrr}{6} & {2} & {9} & {-6} \\ {2} & {6} & {-6} & {9} \\\ {9} & {-6} & {6} & {2} \\ {-6} & {9} & {2} & {6}\end{array}\right] $$

[M] A Landsat image with three spectral components was made of Homestead Air Force Base in Florida (after the base was hit by Hurricane Andrew in 1992 ). The covariance matrix of the data is shown below. Find the first principal component of the data, and compute the percentage of the total variance that is contained in this component. $$ S=\left[\begin{array}{rrr}{164.12} & {32.73} & {81.04} \\ {32.73} & {539.44} & {249.13} \\ {81.04} & {249.13} & {189.11}\end{array}\right] $$
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