Question

Let \(M = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\)The maximum number of linearly independent eigenvectors of \( M \) is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

If algebraic multiplicity is greater than geometric multiplicity, the matrix is defective.

Answer 1

The Correct Option is C

Step 1: Finding characteristic equation. 

\[\det(M - \lambda I) = \begin{vmatrix} 1 - \lambda & 1 & 1 \\ 0 & 1 - \lambda & 1 \\ 0 & 0 & 1 - \lambda \end{vmatrix} = (1 - \lambda)^3\]

Step 2: Finding eigenvalues. - The only eigenvalue is \( \lambda = 1 \) with algebraic multiplicity 3. - Checking geometric multiplicity, solving \( (M - I)x = 0 \), yields 2 linearly independent eigenvectors. 
Step 3: Selecting the correct option. Since geometric multiplicity is 2, the correct answer is (C) 2.