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It is certainly clear that the subspaces Af are invariant under T.
If Af is the operator induced on Af by T, then evidently Af, because by definition Af is 0 on the subspace Af.
This shows that the minimal polynomial for Af divides Af.
Conversely, let G be any polynomial such that Af.
Then Af.
Thus Af is divisible by the minimal polynomial P of T, i.e., Af divides Af.
It is easily seen that Af divides G.
Hence the minimal polynomial for Af is Af.
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