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Let \\[ A=\left(\begin{array}{rr} 2 & 0 \\ 0 & -2 \end{array}\right) \quad \text { and } \quad \mathbf{x}=\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \\] and set \\[ f\left(x_{1}, x_{2}\right)=\|A \mathbf{x}\|_{2} /\|\mathbf{x}\|_{2} \\] Determine the value of \(\|A\|_{2}\) by finding the maximum value of \(f\) for all \(\left(x_{1}, x_{2}\right) \neq(0,0)\)

Short Answer

Expert verified
The 2-norm of the given matrix \(A\) is \(\|A\|_2 = 2\).

Step by step solution

01

Define the function f

Given the matrix \(A\) and the vector \(\mathbf{x}\), the function \(f\) is defined as: \[f\left(x_{1}, x_{2}\right)=\frac{\|A \mathbf{x}\|_{2}}{\|\mathbf{x}\|_{2}}\]
02

Rewrite the function f in terms of x1 and x2

First, compute \(A\mathbf{x}\) by multiplying the matrix \(A\) and the vector \(\mathbf{x}\): \[A\mathbf{x} = \begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 2x_1 \\ -2x_2 \end{pmatrix}\] Now, calculate the 2-norm of \(A\mathbf{x}\): \[\|A\mathbf{x}\|_2 = \sqrt{(2x_1)^2 + (-2x_2)^2} = 2\sqrt{x_1^2 + x_2^2} = 2\|\mathbf{x}\|_2\] Then, compute the 2-norm of \(\mathbf{x}\): \[\|\mathbf{x}\|_2 = \sqrt{x_1^2 + x_2^2}\] Finally, substitute the norms into the function \(f\): \[f\left(x_{1}, x_{2}\right) = \frac{2\|\mathbf{x}\|_2}{\|\mathbf{x}\|_2} = 2\]
03

Use the Rayleigh quotient properties to find the maximum value of f

Using the properties of the Rayleigh quotient, \(f(x_1, x_2)\) will achieve its maximum value when \(A\mathbf{x}\) and \(\mathbf{x}\) are parallel, and this maximum value will be the 2-norm of matrix \(A\). Since the result of our calculations in Step 2 does not depend on the values of \(x_1\) and \(x_2\), it's always equal to 2.
04

Determine the 2-norm of matrix A

As we found the maximum value of the function \(f\), the 2-norm of matrix \(A\) is: \[\|A\|_2 = 2\] Hence, the 2-norm of the given matrix \(A\) is 2.

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