A problem in Hoffman's Linear Algebra (1 Viewer)

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A problem in Hoffman's Linear Algebra.
Page 243

18. If T is a diagonalizable linear operator, then every T-invariant subspace has a complementary T-invariant subspace. And vice versa.

In fact, the answer lies on Pages 263~265,but I try not to use the conception T-admissible to prove this proposition.
Could someone help me out?
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There are a couple of different ways to do that:

The fact that T is "diagonalizable" means that there exist a basis for the vector space consisting entirely of eigenvectors of T (so that the matrix for T in that basis is diagonal). Using that, clearly any T-invariant subspace is spanned by some subset of those eigenvectors and it's orthogonal complement is spanned by the remaining eigenvectors- and so is T-invariant itself.

Or you could use the fact that, since T is diagonalizable, it is "self adjoint": for any vector u,v <Tu, v>= <u, Tv> where <u, v> is the inner product of u and v. That should make it easy to show that if a subspaced is T-invariant, then so is its orthogonal complement.
How about the Inversion of the proposition?

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