Question

In each case, show that \(\langle\mathbf{v}, \mathbf{w}\rangle=\mathbf{v}^{T} A \mathbf{w}\) defines an inner product on \(\mathbb{R}^{2}\) and hence show that \(A\) is positive definite. a. \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right]\) b. \(A=\left[\begin{array}{rr}5 & -3 \\ -3 & 2\end{array}\right]\) c. \(A=\left[\begin{array}{ll}3 & 2 \\ 2 & 3\end{array}\right]\) d. \(A=\left[\begin{array}{ll}3 & 4 \\ 4 & 6\end{array}\right]\)

Short answer

Each matrix defines an inner product on \(\mathbb{R}^{2}\) and is positive definite.

Step-by-step solution

Understand Inner Product Properties

To verify that \( \langle\mathbf{v}, \mathbf{w}\rangle=\mathbf{v}^{T} A \mathbf{w} \) defines an inner product, ensure it satisfies these properties: **1. Linearity**, **2. Symmetry**, and **3. Positive Definiteness**. This means \( \langle\mathbf{v}, \mathbf{w}\rangle = \langle\mathbf{w}, \mathbf{v}\rangle \), and \( \langle\mathbf{v}, \mathbf{v}\rangle \geq 0 \) with equality if \( \mathbf{v} = \mathbf{0} \).

Analyze Case (a) with Symmetry

Given matrix \(A=\begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix}\), check if \(A\) is symmetric. Since \(A = A^T\), this matrix is symmetric, satisfying the symmetry property of an inner product.

Check Positive Definiteness for (a)

Compute \( \mathbf{v}^T A \mathbf{v}\) where \( \mathbf{v} = \begin{bmatrix} x \ y \end{bmatrix} \):\[ \mathbf{v}^T A \mathbf{v} = \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 2x^2 + 2xy + y^2 \]For positivity, set \(2x^2 + 2xy + y^2 > 0\). This is always positive as its corresponding quadratic form has no real roots unless \(\mathbf{v} = \mathbf{0}\). Hence, \(A\) is positive definite.

Use Similar Approach for Cases (b)-(d)

Repeat Steps 2 and 3 for each given matrix:**b.** \(A=\begin{bmatrix} 5 & -3 \ -3 & 2 \end{bmatrix}\): it's symmetric. Test positive definiteness:\[ \mathbf{v}^T A \mathbf{v} = 5x^2 - 6xy + 2y^2 > 0 \]. Confirm positivity over all non-zero \(\mathbf{v}\).**c.**\(A=\begin{bmatrix} 3 & 2 \ 2 & 3 \end{bmatrix}\): symmetric. For positive definiteness, \[ \mathbf{v}^T A \mathbf{v} = 3x^2 + 4xy + 3y^2 > 0 \]. This remains positive.**d.** \(A=\begin{bmatrix} 3 & 4 \ 4 & 6 \end{bmatrix}\): symmetric. Check\[ \mathbf{v}^T A \mathbf{v}= 3x^2 + 8xy + 6y^2 > 0 \]. Also remains positive.All matrices are symmetric and positive definite.

Conclude Results

In each case from (a) to (d), \( \mathbf{v}^T A \mathbf{w} \) serves as an inner product on \(\mathbb{R}^{2}\) since the required properties are met, especially symmetry and positive definiteness. Thus, all matrices are confirmed as positive definite.

Key concepts

Positive Definite Matrix

A matrix is considered positive definite if, for any non-zero vector \( \mathbf{v} \), the product \( \mathbf{v}^T A \mathbf{v} > 0 \). This ensures that the associated quadratic form is always positive. In mathematical terms, positive definiteness indicates that the matrix reflects all vectors onto positive values of a quadratic expression.
This property is crucial, as it guarantees there is no instance where a vector gets mapped to a negative value.

• A necessary condition for a matrix to be positive definite is that all its eigenvalues are positive.
• For symmetry matrices, determinants of all leading principal minors must be positive.

Positive definite matrices play a significant role in optimization and ensure that cost functions acquire a minimum value at the solution point.

Matrix Symmetry

Matrix symmetry is a property where the matrix remains unchanged if it is transposed, meaning \( A = A^T \). This property is vital for defining an inner product because it ensures the term \( \langle \mathbf{v}, \mathbf{w} \rangle = \mathbf{v}^T A \mathbf{w} \) equals \( \langle \mathbf{w}, \mathbf{v} \rangle \).

Symmetric matrices have several outstanding features:

• All eigenvalues of a symmetric matrix are real numbers.
• Symmetric matrices can be diagonalized using an orthogonal basis.

These properties simplify problems and calculations in linear algebra, making it easier to extract meaningful insights from data and solve complex equations.

Properties of Inner Products

Inner products are essential in vector spaces, providing a method to measure angles, lengths, and projections. For a function \( \langle \mathbf{v}, \mathbf{w} \rangle = \mathbf{v}^T A \mathbf{w} \) to be an inner product, it must satisfy the following properties:

• Linearity: It is linear in both arguments, meaning scalars can factor out, and addition of vectors inside the expression distributes.
• Symmetry: As noted earlier, \( \langle \mathbf{v}, \mathbf{w} \rangle = \langle \mathbf{w}, \mathbf{v} \rangle \).
• Positive Definiteness: \( \langle \mathbf{v}, \mathbf{v} \rangle \) is always greater than or equal to zero, and equals zero only when \( \mathbf{v} \) is a zero vector.

When these conditions are met, the function properly acts as an inner product, allowing for consistent and reliable measurements in a vector space.

Linear Algebra

Linear algebra serves as the backbone of many mathematical computations, especially in dealing with vectors and matrices. It addresses solving systems of linear equations, and the properties and operations of matrices.
In the context of inner products and matrix properties, linear algebra offers tools to:

• Understand matrix behaviors through eigenvectors and eigenvalues.
• Identify characteristics like positive definiteness and symmetry.

These tools enable engineers and scientists to model and solve real-world problems efficiently, from computer graphics to machine learning algorithms. Linear algebra's application in these areas underscores its fundamental importance for quantitatively analyzing data and structures.