Question

Consider the matrices 
\( M = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} \)

\( N = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 0 \end{pmatrix} \)

Which one of the following is true?

• A. \( M \text{ is not diagonalizable but } N \text{ is diagonalizable} \)
• B. \( \text{Both } M \text{ and } N \text{ are not diagonalizable} \)
• C. \( \text{Both } M \text{ and } N \text{ are diagonalizable} \)
• D. \( M \text{ is diagonalizable but } N \text{ is not diagonalizable} \)

Hint:

A matrix is diagonalizable if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity. Check the eigenvalues and compute eigenvectors to determine this property.

Answer 1

The Correct Option is A

Step 1: Analyze the diagonalizability of \( M \).
Matrix \( M \) is \[ \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}. \] The eigenvalues of \( M \) are \( \lambda = 2 \) with algebraic multiplicity 2. To check diagonalizability, compute the eigenvectors: \[ M - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. \] The rank of \( M - 2I \) is 1, so the geometric multiplicity of eigenvalue \( \lambda = 2 \) is 1. Since the geometric multiplicity is less than the algebraic multiplicity, \( M \) is not diagonalizable. 

Step 2: Analyze the diagonalizability of \( N \).
Matrix \( N \) is \[ \begin{pmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 0 \end{pmatrix}. \] The eigenvalues of \( N \) are obtained by solving \( \det(N - \lambda I) = 0 \): \[ \det \begin{pmatrix} 1 - \lambda & 0 & 0 \\ 1 & 2 - \lambda & 0 \\ 1 & 1 & -\lambda \end{pmatrix} = 0. \] Expanding the determinant: \[ (1 - \lambda)\left((2 - \lambda)(-\lambda)\right) = 0. \] This gives \( \lambda = 1, \lambda = 2, \lambda = 0 \). All eigenvalues of \( N \) have linearly independent eigenvectors (verified through eigenvector computation). Therefore, \( N \) is diagonalizable. 

Step 3: Conclusion.
Matrix \( M \) is not diagonalizable due to insufficient independent eigenvectors, while \( N \) is diagonalizable.