Let A=(1201)A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}A=(1021). Which statement about AAA is true?
It is diagonalizable because it has distinct eigenvalues.
It is not diagonalizable because the geometric multiplicity of λ=1\lambda = 1λ=1 is less than its algebraic multiplicity.
It is not diagonalizable because the determinant is 111.
It has two linearly independent eigenvectors for λ=1\lambda = 1λ=1.