Chapter 5: Problem 5
Evaluate the quadratic form \(f(\mathbf{x})=\mathbf{x}^{T} A \mathbf{x}\) for the given A and x. $$A=\left[\begin{array}{rrr} 1 & 0 & -3 \\ 0 & 2 & 1 \\ -3 & 1 & 3 \end{array}\right], \mathbf{x}=\left[\begin{array}{r} 2 \\ -1 \\ 1 \end{array}\right]$$
Short Answer
Expert verified
The value of the quadratic form is \(-5\).
Step by step solution
01
Understanding the Problem
The problem asks us to evaluate the quadratic form \( f(\mathbf{x}) = \mathbf{x}^{T} A \mathbf{x} \), where \( A \) is a symmetric matrix, and \( \mathbf{x} \) is a vector. Quadratic forms are expressions involving symmetric matrices and vectors.
02
Transposing Vector x
First, transpose the vector \( \mathbf{x} \) so that it becomes a row vector. Given \( \mathbf{x} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix} \), its transpose is \( \mathbf{x}^T = \begin{bmatrix} 2 & -1 & 1 \end{bmatrix} \).
03
Matrix-Vector Multiplication
Next, multiply \( A \) with \( \mathbf{x} \). Use the matrix-vector multiplication formula: \[ A \mathbf{x} = \begin{bmatrix} 1 & 0 & -3 \ 0 & 2 & 1 \ -3 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix} \]. The result is computed as follows: 1. First row: \( 1 \times 2 + 0 \times (-1) + (-3) \times 1 = -1 \) 2. Second row: \( 0 \times 2 + 2 \times (-1) + 1 \times 1 = -1 \) 3. Third row: \( -3 \times 2 + 1 \times (-1) + 3 \times 1 = -4 \) The result is \( A\mathbf{x} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix} \).
04
Vector Multiplication for Quadratic Form
Finally, multiply \( \mathbf{x}^T \) by \( A\mathbf{x} \). This operation is: \[ \mathbf{x}^{T} A \mathbf{x} = \begin{bmatrix} 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix} \]. Calculate the result: \( 2 \times (-1) + (-1) \times (-1) + 1 \times (-4) = -2 + 1 - 4 = -5 \) Hence, the value of the quadratic form is \( -5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix-Vector Multiplication
Matrix-vector multiplication is an essential concept in linear algebra and is widely used in various applications including physics, computer science, and statistics. In this process, a matrix is multiplied by a vector to yield another vector. To perform matrix-vector multiplication, multiply each element of the vector by the corresponding elements in each row of the matrix. The resulting products in each row are then summed to produce a component of the resulting vector. Here's how you can think about it:
- The number of columns in the matrix should match the number of rows in the vector for multiplication to be valid.
- Each entry in the resulting vector is a dot product of a row from the matrix and the input vector. ​
- In our example, multiplying matrix \(A\) (with dimensions 3x3) by the vector \(\mathbf{x}\) (with dimensions 3x1) gives us a new vector: \(\begin{bmatrix} -1 \ -1 \ -4 \ \end{bmatrix}\).
Symmetric Matrix
A symmetric matrix is a special type of square matrix that is equal to its transpose. This means the entries of the matrix are mirrored along its principal diagonal. In other words:
- If matrix \(A\) is symmetric, then \(A[i][j] = A[j][i]\) for all i and j.
- Such matrices are aesthetically pleasing because of their inherent symmetry and have various mathematical properties and applications.
- For the matrix given in the exercise, notice how the symmetric property is reflected across the diagonal: elements at \((1,3)\) and \((3,1)\) are both \(-3\), elements at \((2,3)\) and \((3,2)\) are both \(1\).
Vector Transposition
Vector transposition refers to the conversion of a vector from a column to a row arrangement or vice versa. This skill is essential when performing operations like the calculation of quadratic forms or dot products. When you transpose a vector:
- You flip its configuration. A vector \(\mathbf{x}\) with elements listed in a column becomes a row when transposed, noted as \(\mathbf{x}^T\).
- This is especially important in quadratic form evaluations or when aligning matrix dimensions for multiplicative operations, ensuring that dimensions agree to perform the operation correctly.
- In our problem, transposing \(\mathbf{x} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}\) yields the row vector \(\mathbf{x}^T = \begin{bmatrix} 2 & -1 & 1 \end{bmatrix}\).