Question

Find the eigenvalues of the matrix \(A = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\).

• A. \( 0, 1, 1 \)
• B. \( 0, -1, -1 \)
• C. \( 1, -1, 0 \)
• D. \( -1, -1, -1 \)

Hint:

If a matrix is upper or lower triangular, its eigenvalues are simply the elements on the main diagonal.
While this matrix is not strictly triangular due to the \( -1 \) in the top right, it is effectively block diagonal, making the diagonal elements the roots.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Eigenvalues of a matrix \( A \) are the roots of the characteristic equation \( \det(A - \lambda I) = 0 \).

Step 2: Key Formula or Approach:
Set up the characteristic determinant and solve for \( \lambda \).

Step 3: Detailed Explanation:
The characteristic equation is:
\[ \det \begin{pmatrix} 0 - \lambda & 0 & -1
0 & -1 - \lambda & 0
0 & 0 & -1 - \lambda \end{pmatrix} = 0 \]
Expanding the determinant along the first column:
\[ (-\lambda) \left[ (-1 - \lambda)(-1 - \lambda) - (0)(0) \right] - 0 + 0 = 0 \]
\[ -\lambda (-1 - \lambda)^2 = 0 \]
\[ -\lambda (1 + \lambda)^2 = 0 \]
This gives:
\( \lambda = 0 \) or \( (1 + \lambda)^2 = 0 \).
Solving for \( \lambda \), we get \( \lambda = 0, -1, -1 \).

Step 4: Final Answer:
The eigenvalues are \( 0, -1, -1 \).