Discussion
Matrix \(A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix}\) is :
Answer: D
The matrix is diagonalizable over C.
The characteristic polynomial is
\[ (1 - \lambda)[(1 - \lambda)^{2} + 1] = 0 \]
giving eigenvalues \(1, 1 + i, 1 - i\). These are three distinct eigenvalues, so A is diagonalizable over C.
Over R it is neither diagonalizable nor triangulable, because triangulating over R requires all eigenvalues to be real, and \(1 \pm i\) are not. That rules out A and B, and C is wrong because distinct eigenvalues guarantee diagonalizability over C.
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