Correct option is B
Given:
- A is a 3 × 3 real matrix.
- Its characteristic polynomial is divisible by T², meaning 0 is an eigenvalue with algebraic multiplicity at least 2.
We now analyze each option step by step:
Option (a):
"The eigenspace of A for the eigenvalue 0 is two-dimensional."
- The algebraic multiplicity of the eigenvalue 0 is at least 2, but the geometric multiplicity (dimension of the eigenspace) could be less than 2.
- For example, if A is not diagonalizable, the eigenspace of 0 might only have dimension 1.
- This statement is not always true.
Option (b):
"All the eigenvalues of A are real."
- A is a 3 × 3 real matrix, so its characteristic polynomial has real coefficients.
- When the polynomial is divisible by T², 0 is an eigenvalue. The remaining root of the characteristic polynomial must also be real because the polynomial’s degree is 3 and complex roots occur in conjugate pairs.
- Therefore, all eigenvalues of A are guaranteed to be real.
- This statement is true.
Option (c):
"A³ = 0."
- For A³ = 0, A must be nilpotent. While 0 is an eigenvalue with algebraic multiplicity at least 2, this does not imply that A³ = 0.
- There is no evidence to suggest that A satisfies this condition.
- This statement is false.
Option (d):
"A is diagonalizable."
- For A to be diagonalizable, the geometric multiplicity of every eigenvalue must equal its algebraic multiplicity.
- Since 0 is an eigenvalue with algebraic multiplicity 2, but the geometric multiplicity might be smaller (if A is not diagonalizable), A is not guaranteed to be diagonalizable.
- This statement is false.
Conclusion:
The correct answer is (b): "All the eigenvalues of A are real."
