Is the matrix A=(1101)A = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}A=(1011) diagonalizable? Why or why not?
No — it has a repeated eigenvalue λ=1\lambda=1λ=1 with only one linearly independent eigenvector
Yes — all 2×22\times 22×2 matrices are diagonalizable
Yes — because det(A)=1≠0\det(A) = 1 \neq 0det(A)=1=0
No — because tr(A)=2\text{tr}(A) = 2tr(A)=2