A problem in Hoffman's Linear Algebra

1. Oct 15, 2006

tghg

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?

Last edited: Oct 15, 2006
2. Oct 15, 2006

HallsofIvy

Staff Emeritus
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.

3. Oct 15, 2006

tghg

How about the Inversion of the proposition?