Question

In each case, find a symmetric matrix \(A\) such that \(q=\mathbf{x}^{T} B \mathbf{x}\) takes the form \(q=\mathbf{x}^{T} A \mathbf{x}\). a. \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) b. \(\left[\begin{array}{rr}1 & 1 \\ -1 & 2\end{array}\right]\) c. \(\left[\begin{array}{lll}1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\) d. \(\left[\begin{array}{rrr}1 & 2 & -1 \\ 4 & 1 & 0 \\ 5 & -2 & 3\end{array}\right]\)

Short answer

Average the given matrix with its transpose to form a symmetric matrix A.

Step-by-step solution

Understanding Matrix Transposition

Recall that for any matrix \(B\), the transpose \(B^T\) is the matrix obtained by swapping rows and columns of \(B\). A symmetric matrix \(A\) is one which satisfies \(A = A^T\). The goal is to express \(q = \mathbf{x}^T B \mathbf{x}\) in a symmetric form \(q = \mathbf{x}^T A \mathbf{x}\).

Constructing a Symmetric Matrix from (a)

For part (a), given matrix \(B = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}\), we can construct a symmetric matrix by averaging \(B\) and its transpose \(B^T\). Compute \(B^T = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}\). Then, the symmetric matrix \(A\) is \(\frac{1}{2}(B + B^T) = \begin{bmatrix} 1 & \frac{1}{2} \ \frac{1}{2} & 1 \end{bmatrix}\).

Constructing a Symmetric Matrix from (b)

For part (b), given matrix \(B = \begin{bmatrix} 1 & 1 \ -1 & 2 \end{bmatrix}\), calculate its transpose: \(B^T = \begin{bmatrix} 1 & -1 \ 1 & 2 \end{bmatrix}\). Then, form the symmetric matrix \(A\) by averaging \(B\) and \(B^T\): \(\frac{1}{2}(B + B^T) = \begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix}\).

Constructing a Symmetric Matrix from (c)

For part (c), the matrix \(B = \begin{bmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ 0 & 1 & 1 \end{bmatrix}\) has a transpose \(B^T = \begin{bmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 \end{bmatrix}\). The symmetric matrix \(A\) is obtained as \(\frac{1}{2}(B + B^T) = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & 1 & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} & 1 \end{bmatrix}\).

Constructing a Symmetric Matrix from (d)

For part (d), start with \(B = \begin{bmatrix} 1 & 2 & -1 \ 4 & 1 & 0 \ 5 & -2 & 3 \end{bmatrix}\) and its transpose \(B^T = \begin{bmatrix} 1 & 4 & 5 \ 2 & 1 & -2 \ -1 & 0 & 3 \end{bmatrix}\). Then, the symmetric matrix \(A\) is found by \(\frac{1}{2}(B + B^T) = \begin{bmatrix} 1 & 3 & 2 \ 3 & 1 & -1 \ 2 & -1 & 3 \end{bmatrix}\).

Key concepts

Matrix Transposition

When we talk about matrix transposition, we are referring to the process of converting a matrix into another by simply swapping its rows with its columns. This operation is fundamental in linear algebra, and helps to perform various mathematical computations in a more structured form. Having a matrix \( B \), the transpose of this matrix, denoted as \( B^T \), is formed by changing the entry at row \( i \), column \( j \) of \( B \) to row \( j \), column \( i \) of \( B^T \). This operation is crucial when dealing with symmetric matrices, as a matrix \( A \) is symmetric if it is equal to its transpose, meaning \( A = A^T \). This characteristic is a key concept when translating quadratic forms into symmetric form.
To find a symmetric matrix from a given \( B \), you can average the matrix with its transpose: \( A = \frac{1}{2}(B + B^T) \). This results in \( A \) having the same elements along its main diagonal, and each element outside the diagonal being the average of its (transpose matricular) position in \( B \) and \( B^T \).

Quadratic Forms

Quadratic forms are expressions involving a symmetric matrix and a vector, often appearing in various fields of science and engineering such as physics and optimization. They usually have the form \( q = \mathbf{x}^T A \mathbf{x} \), where \( \mathbf{x} \) is a vector with components \( x_1, x_2, \ldots, x_n \), and \( A \) is a symmetric matrix. The quadratic form effectively describes a surface or shape in multidimensional space, such as an ellipse or paraboloid.
By expressing a non-symmetric quadratic form as \( q = \mathbf{x}^T B \mathbf{x} \), it can be rewritten to a symmetric form by replacing \( B \) with the symmetric \( A \). This conversion ensures the expression accurately retains its mathematical properties, facilitating easier analysis and solution of various algebraic problems. Understanding how to convert into symmetric form is critical in disciplines that involve calculating distance, optimization problems, and algorithms in computer science.

Linear Algebra

Linear algebra serves as the backbone of modern mathematics and many practical applications, providing tools to solve equations involving linear mappings and structures. It investigates vector spaces, and linear transformations between them, utilizing matrices and functions. Matrices operate as tools for simplifying multiple equations and performing operations on vector spaces.
Linear algebra concepts are imperative when dealing with symmetric matrices and quadratic forms. The understanding of operations like matrix transposition is vital, since they enable the manipulation of systems to derive solutions effectively. It also aids in understanding eigenvalues, eigenvectors, and the diagonalization of matrices, allowing for the development of more efficient computational methods. Moreover, these principles support machine learning algorithms, data analysis, and various engineering fields by modeling and solving large systems of equations. Mastery of linear algebra helps in interpreting results that are fundamental in scientific computations and design processes across multiple disciplines.