Question

Let \( A \) be a square matrix such that \[ \text{det}(xI - A) = x^4(x - 1)^2(x - 2)^3, \] where \( \text{det}(M) \) denotes the determinant of a square matrix \( M \). If \[ \text{rank}(A^2) < \text{rank}(A^3) = \text{rank}(A^4), \] then the geometric multiplicity of the eigenvalue 0 of \( A \) is \(\underline{\hspace{1cm}}\) .

Hint:

The geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors corresponding to it.

Answer 1

Correct Answer: 4

From the characteristic polynomial of \( A \), we see that the eigenvalue 0 has multiplicity 4. Since the rank of \( A^2 \) is less than the rank of \( A^3 \), this indicates that the geometric multiplicity of eigenvalue 0 is 4. Thus, the geometric multiplicity of eigenvalue 0 is \( \boxed{4} \).