Discussion
Let T be a linear operator on \(\mathbf{R}^{4}\) having the minimal polynomial \((x + 1)^{3}\). Then which of the following statements is true ?
Answer: C
The true statement is T is triangulable over \(\mathbf{R}\).
An operator is triangulable exactly when its minimal polynomial splits into linear factors over the field. Here \((x + 1)^{3}\) splits over \(\mathbf{R}\), so T has an upper triangular matrix.
The others are false. T is diagonalizable only if the minimal polynomial has distinct linear factors, but \((x + 1)\) is repeated. T is not nilpotent, since its only eigenvalue is -1, not 0. The characteristic polynomial has degree 4 on \(\mathbf{R}^{4}\), and with the same roots as the minimal polynomial it must be \((x + 1)^{4}\), not \((x + 1)^{3}\).
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