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 because AAA is invertible.
AAA is diagonalizable with eigenvectors [10]\begin{bmatrix} 1 \\ 0 \end{bmatrix}[10] and [11]\begin{bmatrix} 1 \\ 1 \end{bmatrix}[11].