Question

Consider the matrix
\[ A=\begin{pmatrix} \lambda_1 & 1 & 0 \\ 0 & \lambda_2 & 1 \\ 0 & 0 & \lambda_3 \end{pmatrix}, \qquad \lambda_1,\lambda_2,\lambda_3\in\mathbb{R}. \]
Which of the following statements is/are correct?

• A. \(A\) is never diagonalizable
• B. \(A\) is diagonalizable if and only if \(\lambda_1, \lambda_2, \lambda_3\) are distinct
• C. The dimension of eigenspaces corresponding to each \(\lambda_i\) is \(1, i=1,2,3\)
• D. For any \(A\), there exist real numbers \(\alpha\) and \(\beta\) such that \(A^2+\alpha A+\beta I_3=0\)

Hint:

The 1's on the superdiagonal keep rank(A - lambda I) equal to 2 for every eigenvalue lambda, distinct or repeated, so each eigenspace always has dimension 1; use that to test diagonalizability and the minimal polynomial degree.

Answer 1

The Correct Option is B, C

Step 1: Eigenvalues.
\(A\) upper triangular, eigenvalues \(\lambda_1,\lambda_2,\lambda_3\).

Step 2: (A) false.
Distinct eigenvalues always diagonalizable.

Step 3: Repeated eigenvalue case.
\(A-\lambda I\) has rank 2, eigenspace dim \(3-2=1\), less than multiplicity if repeated.

Step 4: (B) TRUE.
Diagonalizable iff distinct.

Step 5: (C) TRUE.
Eigenspace dim always 1 (non-derogatory).

Step 6: (D) false.
With distinct eigenvalues, minimal polynomial has degree 3, cannot satisfy degree-2 polynomial.

Final Answer: \[ \boxed{\text{B, C}} \]