Abstract Algebra?

Theorem: Let T be a linear operator on a finite-dimensional vector space such that the characteristic polynomial of T splits. Suppose that b is an eigenvalue of T with multiplicity m. Then dim(K_b) %26lt;= m.





Proof: Let W = K_b, and let h(t) be the characteristic polynomial of T_w. By theorem h(t) divides the characteristic polynomial of T, and by Theorem **, b is the only eigenvalue of T_w. Hence, h(t) = (-1)^d(t-b)^b, where d = dim(W), and d%26lt;=m.





Theorem ** : Let T be a linear operator on a vector space V, and let b be an eignevalue of T. Then





For any scalar c not equal to b, the restriction of T- cI to K_b is one-to-one.





Note: 'I' is the identity transformation. Also K_b is the generalized eigenspace of T corresponding to b.





What I need help on is on the the proof, I don't understand why is it that it can be concluded from theorem** that b is the ONLY eigenvalue of T_w, or even that it is an eigenvalue for that matter. If someone could explain, it would be helpful.

Abstract Algebra?
Suppose that b' ≠ b is another eigenvalue of T_w. Theorem (**) then asserts that T - b'I restricted to K_b is injective, which means that its kernel is {0}, so b' cannot be an eigenvalue (remember that an eigenspace is always ≠ {0}).





Regarding the second part of your question, b is, by hypothesis, an eigenvalue of T, so it's also an eigenvalue of T_w, because K_b is an invariant subspace for T.





Hope this helps.