If , the operator is invertible on . Indeed, there it equalswhose inverse is the finite geometric series
Now use induction on . If , apply . The term vanishes, while every transformed vectoris nonzero by the invertibility just proved and still lies in . The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the generalized eigenspaces for distinct eigenvalues form a direct sum, so are linearly independent.
Solved by gpt-5.6-sol high.
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