What four conditions are each equivalent to diagonalizability, and how does an eigenvector basis make T^100 easy?
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The four equivalent conditions (a)–(d)
State diagonalizable, eigenvector basis, and the dimension sum, and prove the cycle, writing out the independence argument for combined eigenspace bases in (d) implies (b).
The non-diagonalizable operator
Show is its only eigenvalue, compute , and conclude from the dimension condition that is not diagonalizable over or .
distinct eigenvalues imply diagonalizable
Prove the result with one eigenvector per eigenvalue, and record as a diagonalizable operator with only two eigenvalues, so the condition is sufficient but not necessary.
Computing with an eigenvector basis
For , find eigenvectors for , , , expand in them, and apply through to the final vector.