Question

Consider a vector \( p \) in 2-dimensional space. Let its direction (counter-clockwise angle with the positive x-axis) be \( \theta \). Let \( p \) be an eigenvector of a 2 \( \times \) 2 matrix \( A \) with corresponding eigenvalue \( \lambda \), where \( \lambda>0 \). If we denote the magnitude of a vector \( v \) by \( \|v\| \), identify the VALID statement regarding \( p' \), where \( p' = Ap \).

• A. Direction of \( p' = \lambda \theta, \|p'\| = \|p\| \)
• B. Direction of \( p' = \theta, \|p'\| = \lambda \|p\| \)
• C. Direction of \( p' = \lambda \theta, \|p'\| = \lambda \|p\| \)
• D. Direction of \( p' = \theta, \|p'\| = \frac{\|p\|}{\lambda} \)

Hint:

For an eigenvector \( p \) with eigenvalue \( \lambda \), the direction of the transformed vector remains the same, but the magnitude is scaled by \( \lambda \).

Answer 1

The Correct Option is B

Let the vector \( p \) be an eigenvector of the matrix \( A \), and its corresponding eigenvalue is \( \lambda \). The definition of an eigenvector and eigenvalue tells us that: \[ A p = \lambda p. \] This implies that the vector \( p' = A p \) is simply the vector \( p \) scaled by \( \lambda \). Step 1: Direction of \( p' \)
Since \( p' = A p \), and the matrix \( A \) is a linear transformation, the direction of \( p' \) remains the same as that of \( p \) because eigenvectors are aligned along the same direction as the transformation applied to them. Thus, the direction of \( p' \) is \( \theta \).
Step 2: Magnitude of \( p' \)
The magnitude of \( p' \) is simply the magnitude of \( p \) multiplied by the eigenvalue \( \lambda \): \[ \|p'\| = \lambda \|p\|. \] Therefore, the correct statement is Option (B): Direction of \( p' = \theta \), and \( \|p'\| = \lambda \|p\| \).
Final Answer: (B)