The matrix A=[1101]A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}A=[1011] has characteristic polynomial (1−λ)2=0(1-\lambda)^2 = 0(1−λ)2=0. Which statement is true?
AAA is diagonalizable because the eigenvalue 1 has algebraic multiplicity 2.
AAA is not diagonalizable because the geometric multiplicity (1) is less than the algebraic multiplicity (2).
AAA is diagonalizable with eigenvectors [10]\begin{bmatrix} 1 \\ 0 \end{bmatrix}[10] and [11]\begin{bmatrix} 1 \\ 1 \end{bmatrix}[11].
AAA is diagonalizable because AAA is invertible.