Lecture 23 - Decomposition of an Operator

The main result of this section is as follows.

Theorem

Given an operator TL(V) where V is a complex finite-dimensional vector space, then there is a basis of generalized eigenvectors of T.

This is big! It says that any operator can be turned into a diagonal-matrix as a direct corollary:

Corollary

V is a complex finite-dimensional vector space. Let λ1,...,λm be the distinct eigenvalues of TL(V). Then:

  • V=G(λ1,T)G(λ2,T)G(λm,T)
  • Each G(λi,T) is T-invariant.
  • Each (TλiI)|G(λi,T) is nilpotent.

In trying to prove this, we'll have to use one of these facts:

Fact

The null space and the range of a polynomial operator in T are both T-invariant.

We first prove the Lecture 23 - Decomposition of an Operator#^3c4724 proof:
Proof
(a) By induction on dim(V).

The base case is dim(V)=1. Then every non-zero vector is an eigenvector, so every basis chosen is a generalized eigenbasis.

For the inductive case, suppose dim(V)=n+1 and that the theorem holds for all dim(V)=n. Reminder that T has an eigenvalue, because every operator on a complex space has an eigenvalue. Call it λ1.

Recall from yesterday:

Proposition

V=null(Tdim(V))range(Tdim(V))

Thus:

V=null((Tλ1I)dim(V))range((Tλ1I)dim(V))

The left part null((Tλ1I)dim(V))=G(λ1,T) and let U=range((Tλ1I)dim(V)). Notice that:

  1. U is invariant
  2. dim(U)<dim(V)
  3. λ1 is not an eigenvalues of T|U because the eigenvalues of T|U are λ2,...,λm

Using the inductive hypothesis on the vector space U and operator T|U:

U=G(λ2,T|U)G(λm,T|U)

If we can show that G(λi,T|U)=G(λi,T) for all 2im then we are done.

We know because if you're in the left set, you're in U so applying T to those items is equivalent to T|U and thus we are done.

For , suppose v=vn1+vr where vn1G(λ1,T) and vrU. Because U has it's own decomposition. Also (from inductive hypothesis):

vr=vn2++vnm

where vniG(λi,T|U) and thus vniG(λi,T) (the case). We have:

v=vn1+vn2++vnm

where now each vniG(λi,T) (notice the change from T|U). All the vni's are linearly independent since they are all eigenvectors of different λi values. Therefore, vn1,...,vnm is LI.

Thus for all ji we have vni=v and vnj=0. This suggests vG(λi,T|U) and thus we are done.

(b) Using the definition of G:

G(λi,T)=null(TλiI)dim(V)

Where the right side is a polynomial operator p(T), so then it's T-invariant.

(c) Clear by definition, because vG(λi,T) then (Tλi)dim(V)v=0 so then:

(TλiI)|G(λi,T)=0the zero operator

showing nilpotency. Notice that it doesn't mean that TλiI is nilpotent. The restriction is important.

multiplicity of an eigenvalue

For TL(V), the multiplicity of an eigenvalue is dim(G(λ,T))=dim(null(TλI)dim(V)).

Note

We've talked about algebraic and geometric multiplicity of an eigenvalue in Linear I. Here the definition above is for the algebraic version. For the geometric one, it's defined as:

dim(E(λ,T))
Fact

i=1kG(λi,T)=dim(V)

The proof from this comes from (a) of the main theorem we looked at, using the fact that direct sums add up in dimension.