Question

Let \( I \) denote the identity matrix of order 7, and \( A \) be a \( 7 \times 7 \) real matrix having characteristic polynomial \( C_A(\lambda) = \lambda^2 (\lambda - 1)^\alpha (\lambda + 2)^\beta \), where \( \alpha \) and \( \beta \) are positive integers. If \( A \) is diagonalizable and \( \text{rank}(A) = \text{rank}(A + 2I) \), then \( \text{rank}(A - I) \) is \(\underline{\hspace{2cm}}\) (in integer).

Hint:

For a diagonalizable matrix, the rank is the number of nonzero eigenvalues. When subtracting a scalar from the matrix, shift the eigenvalues by that scalar.

Answer 1

Correct Answer: 4

Given that \( A \) is diagonalizable, we can deduce the eigenvalues of \( A \) from its characteristic polynomial:
\[ C_A(\lambda) = \lambda^2 (\lambda - 1)^\alpha (\lambda + 2)^\beta \] The eigenvalues of \( A \) are:
- \( \lambda = 0 \) (with multiplicity 2),
- \( \lambda = 1 \) (with multiplicity \( \alpha \)),
- \( \lambda = -2 \) (with multiplicity \( \beta \)).
Since \( A \) is diagonalizable, the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity. The rank of a matrix is the number of nonzero eigenvalues. The matrix \( A - I \) will have eigenvalues \( 0, \lambda - 1 \) for each eigenvalue \( \lambda \) of \( A \). Thus, the eigenvalues of \( A - I \) are:
- \( \lambda = 0 \rightarrow -1 \),
- \( \lambda = 1 \rightarrow 0 \),
- \( \lambda = -2 \rightarrow -3 \).
Thus, \( A - I \) has 2 zero eigenvalues (from \( \lambda = 1 \)) and \( 2 + \beta \) nonzero eigenvalues. Therefore, the rank of \( A - I \) is \( 7 - 2 = 4 \). Thus, \( \text{rank}(A - I) = 4 \).