Question

Let \( M \) be a 3 \(\times\) 3 real symmetric matrix with eigenvalues \(-1, 1, 2\) and the corresponding unit eigenvectors \( u, v, w \), respectively. Let \( x \) and \( y \) be two vectors in \( \mathbb{R}^3 \) such that \[ Mx = u + 2(v + w) \quad \text{and} \quad M^2 y = u - (v + 2w). \] Considering the usual inner product in \( \mathbb{R}^3 \), the value of \( |x + y|^2 \), where \( |x + y| \) is the length of the vector \( x + y \), is

• A. 1.25
• B. 0.25
• C. 0.75
• D. 1

Hint:

When working with eigenvalues and eigenvectors, remember that the matrix operation on an eigenvector results in a scalar multiple of the eigenvector itself, and the dot product of orthogonal vectors is zero.

Answer 1

The Correct Option is A

We are given that \(M\) is a symmetric matrix with eigenvalues \(-1, 1, 2\), and we are tasked with finding \( |x + y|^2 \). First, let’s break down the given equations for \( Mx \) and \( M^2 y \).
Step 1: Use the properties of eigenvectors and eigenvalues.
We know that:
- The eigenvectors corresponding to the eigenvalues \(-1, 1, 2\) of the matrix \( M \) are \( u, v, w \), respectively.
- Thus, \( M u = -u \), \( M v = v \), and \( M w = 2w \).
Step 2: Find \( x \) and \( y \) in terms of eigenvectors.
We are given that \( Mx = u + 2(v + w) \). Applying the matrix \( M \), we get: \[ M x = M u + 2(M v + M w) = -u + 2(v + 2w) = -u + 2v + 4w \] This gives us the expression for \( x \), so \( x = -u + 2v + 4w \). For \( M^2 y = u - (v + 2w) \), since \( M^2 = M \times M \), we apply \( M \) to both sides: \[ M^2 y = M(u - v - 2w) = M u - M v - 2M w = -u - v - 4w \] This gives us the expression for \( y \), so \( y = -u - v - 4w \).
Step 3: Compute \( |x + y|^2 \).
Now we compute the square of the length of \( x + y \): \[ x + y = (-u + 2v + 4w) + (-u - v - 4w) = -2u + v \] Thus, the squared length of \( x + y \) is: \[ |x + y|^2 = (-2u + v) \cdot (-2u + v) \] Using the properties of the inner product and the fact that \( u, v, w \) are unit vectors: \[ |x + y|^2 = 4u \cdot u - 4u \cdot v + v \cdot v = 4(1) - 4(0) + (1) = 4 + 1 = 5 \] So, \( |x + y|^2 = 5 \).