Question

Show that the Rayleigh quotient of a real matrix \(A\) and vector \(\mathbf{v}, \mu(\mathbf{v})=\frac{\mathbf{v}^{T} A \mathbf{v}}{\mathbf{v}^{T \mathbf{v}}}\), is the least squares solution of the problem $$ \min _{\mu}\|A \mathbf{v}-\mu \mathbf{v}\|, $$ where \(\mathbf{v}\) is given.

Short answer

Question: Prove that the Rayleigh quotient of a real matrix A and vector v is the least squares solution of the problem min(µ)||A v - µ v||. Answer: The Rayleigh quotient µ(v) minimizes the objective function ||A v - µ v||^2, as shown in the detailed step-by-step solution above. Therefore, it is the least squares solution of the problem.

Step-by-step solution

Define the objective function

Define the objective function for the least squares problem as: $$ f(\mu) = \|A \mathbf{v} - \mu \mathbf{v}\|^2. $$

Expand the objective function

Expand the objective function: \begin{align*} f(\mu) &= (A\mathbf{v}-\mu\mathbf{v})^T(A\mathbf{v}-\mu\mathbf{v}) \\ &= (\mathbf{v}^TA^T-\mu\mathbf{v}^T)(A\mathbf{v}-\mu\mathbf{v}) \\ &= \mathbf{v}^TA^TA\mathbf{v}-\mu\mathbf{v}^TA^T\mathbf{v} -\mu\mathbf{v}^TA\mathbf{v}+\mu^2\mathbf{v}^T\mathbf{v}. \end{align*}

Simplify the expression

Since \(A\) is a real matrix, we have \(A^T\mathbf{v}=(A\mathbf{v})^T\). Substitute this back into the expression: $$ f(\mu) = \mathbf{v}^T A^T A \mathbf{v} - 2\mu \mathbf{v}^T A\mathbf{v} + \mu^2 \mathbf{v}^T\mathbf{v}. $$

Find the minimum of the objective function

Find the minimum of the objective function by taking the derivative with respect to \(\mu\) and setting it to zero: $$ \frac{df(\mu)}{d\mu} = -2\mathbf{v}^TA\mathbf{v} + 2\mu\mathbf{v}^T\mathbf{v} = 0. $$ Solving for \(\mu\), we get: $$ \mu = \frac{\mathbf{v}^TA\mathbf{v}}{\mathbf{v}^T\mathbf{v}} = \mu(\mathbf{v}). $$

Conclusion

We have shown that the Rayleigh quotient is the least squares solution of the given problem, i.e., it minimizes the objective function \(\|A \mathbf{v}-\mu \mathbf{v}\|\).

Key concepts

Least Squares Solution

In mathematics, the least squares solution is a method used to find the best-fit solution of an overdetermined system. This is particularly useful when the system doesn't have an exact solution. The least squares solution minimizes the sum of the squares of the residuals, which are the differences between observed and calculated values.
\[\text{Residual} = A\mathbf{v} - \mu\mathbf{v}\]Here, we aim to find the optimal \(\mu\) that minimizes the expression \(\|A\mathbf{v} - \mu\mathbf{v}\|\). This approach is widely utilized in data fitting, where noise or errors in data lead to a need for an approximate solution.
The beauty of the least squares method lies in its ability to provide a unique and optimal solution, especially when dealing with large datasets. It finds the parameter values (here, \(\mu\)) that minimize the difference between the given observations (\(A\mathbf{v}\)) and the model predictions (\(\mu\mathbf{v}\)).
When applied to the Rayleigh quotient, the least squares method ensures that the quotient truly reflects the optimal value \(\mu\) for reducing the discrepancy between \(A\mathbf{v}\) and \(\mu\mathbf{v}\).

Matrix Calculations

Matrix calculations are fundamental in various math and engineering fields, and understanding them is crucial for leveraging the power of matrices in problem-solving. With matrix operations, we can transform and handle multiple, complex equations in a structured manner.
In this exercise, the matrix \(A\) transforms the vector \(\mathbf{v}\) and the task involves expanding and manipulating these transformations to find \(\mu\). A crucial aspect of matrix calculations is the ability to multiply matrices and vectors, which is evident when we compute expressions like \(\mathbf{v}^TA^TA\mathbf{v}\).
Matrix calculations like these are essential when simplifying expressions involving Rayleigh quotients. By organizing terms such as \(\mathbf{v}^TA^T\mathbf{v}\), we can reduce an expression to its simplest form, making further manipulation convenient.
Remember, the key to mastering matrix calculations is understanding how to apply operations like addition, subtraction, and multiplication to matrices and vectors. These skills will help you solve a wide range of optimization and computational problems.

Optimization Problem

The concept of optimization is central to mathematics and computer science. At its core, an optimization problem aims to find the best solution from a set of feasible solutions. In this exercise, we are tasked with finding the optimal \(\mu\) that minimizes the residual \(\|A\mathbf{v} - \mu\mathbf{v}\|\).
Here is the process:

• Define your objective function, which in this case is \(f(\mu) = \|A\mathbf{v} - \mu\mathbf{v}\|^2\).
• Expand and simplify the problem using matrix operations to obtain a handleable form.
• Find the minimum by setting the derivative \(\frac{df(\mu)}{d\mu}\) to zero, which provides the value of \(\mu\) that minimizes the function.

Optimization problems often involve various constraints and objective functions, determined by real-world scenarios. Understanding these principles allows for applications across different fields, from machine learning to logistics.
In summary, the exercise showcases the elegance and utility of optimizing matrix equations to find solutions tailored perfectly to the problem, emphasizing the power of these mathematical concepts.