Let v= [1,2] be a vector in the plane and let A = 2[[1/sqrt(2), - 1/sqrt(2)], [1/sqrt(2), 1/sqrt(2)]]. What is (A^8)v?

Let's solve the problem step by step.

We are given the vector \( v = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \) and the matrix \( A = 2 \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \). We need to find \( A^8v \).

### Step 1: Simplify Matrix A

Start by calculating the simplified form of matrix \( A \):
\[
A = 2 \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \sqrt{2} & -\sqrt{2} \\ \sqrt{2} & \sqrt{2} \end{bmatrix}.
\]

### Step 2: Analyze the Matrix \( A \)

To make computing higher powers of the matrix easier, we should check if the matrix can be diagonalized or if it has a special form like a rotation or reflection matrix. Let's calculate the eigenvalues and eigenvectors.

Calculate the characteristic equation:
\[
\det(A - \lambda I) = \det \begin{bmatrix} \sqrt{2} - \lambda & -\sqrt{2} \\ \sqrt{2} & \sqrt{2} - \lambda \end{bmatrix} = (\sqrt{2} - \lambda)^2 - (-\sqrt{2})(\sqrt{2}) = \lambda^2 - 2\sqrt{2}\lambda.
\]
Simplifying this will provide insights about the matrix's behavior.

Since the matrix relates to rotation, we can infer that its power to even numbers will yield cyclic results.

Solve the matrix exponentiation problem where A is a 2x2 rotation matrix, and v is a vector in the plane. Learn key concepts like eigenvalues, matrix diagonalization, and matrix powers. Suitable for undergraduate linear algebra students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation