A matrix AAA has the Jordan canonical form J=(210021002)J = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}J=200120012. Which statement about AAA is true?
AAA is diagonalizable with three distinct eigenvalues
AAA is not diagonalizable; it has one eigenvalue λ=2\lambda = 2λ=2 with algebraic multiplicity 3 and geometric multiplicity 1
A2=J2A^2 = J^2A2=J2 (Jordan form remains unchanged when squaring)
AAA has eigenvalues 0, 1, and 2