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: Structure.
$A$ is non-derogatory Hessenberg; eigenspace always dim 1 per eigenvalue.

Step 2: (C) settled.
TRUE for all cases.

Step 3: (B) settled.
Diagonalizable iff eigenspace dim = multiplicity, so iff all distinct. TRUE.

Step 4: (A) false.
Distinct case is diagonalizable.

Step 5: (D) test with example.
$\lambda=0,1,2$: minimal poly degree 3, cannot satisfy quadratic relation. FALSE.
\[ \boxed{\text{B, C}} \]