The most we can do is decompose a linear operator into a smaller, simpler collection of operators which tell us how the linear operator works. More formally, for α:V→V, where V is a finite dimensional vector space over any field, the aim is to decompose V as a direct sum of α-invariant subspaces. (a subspace W is α-invariant if for any w∈W, α(w)∈W.)

The primary decomposition theorem states that this decomposition is determined by the minimal polynomial of α :
*Let α:V→V be a linear operator whose minimal polynomial factorises into monic, coprime polynomials-
*

*m*

_{α}(t)=p_{1}(t)p_{2}(t)

Then,V= W

Where the W

Then,

_{1}⊕W

_{2}

Where the W

_{i}are α-invariant subspaces such that p_{i}is the minimum polynomial of α|_{Wi}Repeated application of this result (i.e., fully factorising m_{α} into pairwise coprime factors) gives a more general version: if m_{α}(t)=p_{1}(t)...p_{k}(t) as described, then

_{1}⊕...⊕W

_{k}

With α-invariant W

_{i}with corresponding minimal polynomal p

_{i}.

It may now be apparent that diagonalisation is a special case in which each W_{i} has a minimal polynomial which consists of a single factor, (t-λ_{i}); i.e. if m_{α}=(t-λ_{1})...(t-λ_{k}) for distinct λ_{i} then α is diagonalisable with the λ_{i} as the diagonal entries.

**Proof of the Primary Decomposition Theorem**

The theorem makes two assertions; that we can construct α-invariant subspaces W_{i} based on the p_{i}; and that the direct sum of these W_{i} constructs V.

For the first, a result about invariant subspaces is needed:
*Lemma*: if α,β:V→V are linear maps such that αβ = βα, then ker β is α-invariant.
*Proof*: Take w∈Ker β - we need to show that α(w) is also in ker β. Now β(α(w))=α(β(w)) by assumption, so β(α(w))=α(** 0**) since w∈ker β , =

**since α is a linear map. But if β(α(w))=0 then α(w) ∈ ker β, hence ker β is α-invariant.**

__0__Given this result, we now take the W

_{i}as Ker p

_{i}(α). Then since p

_{i}(α)α = αp

_{i}(α), it follows that ker p

_{i}= W

_{i}is α-invariant.

We now seek to show that (i) V = W

_{1}+ W

_{2}, and (ii) W

_{1}∩W

_{2}= {

**} (that is, V decomposes as a direct sum of the W**

__0___{i}'s.)

Using Euclid's Algorithm for polynomials, since the p

_{i}are coprime there are polynomials q

_{i}such that p

_{1}(t)q

_{1}(t) + p

_{2}(t)q

_{2}(t) =1.

So for any v ∈V, consider w

_{1}=p

_{2}(α)q

_{2}(α)v and w

_{2}=p

_{1}(α)q

_{1}(α)v. Then v= w

_{1}+ w

_{2}by the above identity. We can confirm that w

_{1}∈W

_{1}: p

_{1}(α)w

_{1}=m

_{α}(α)q

_{2}(α)v = 0. Similarly, w

_{2}∈W

_{2}. So we have (i).

For (ii), let v ∈ W

_{1}∩W

_{2}. Then

v = id(v) = q

_{1}(α)p

_{1}(α)v + q

_{2}(α)p

_{2}(α)v =

**. So W**

__0___{1}∩W

_{2}= {

**}.**

__0__Finally, for the claimed minimum polynomials. Let m_{i} be the min.poly. of α|_{Wi} . We have that p_{i}(α|_{Wi})=0, so the degree of p_{i} is at least that of m_{i}. This holds for each i. However,
p_{1}(t)p_{2}(t)=m_{α}(t)=lcm{m_{1}(t), m_{2}(t)} so we obtain

deg p_{1} + deg p_{2} = deg m_{α} ≤ deg m_{1} + deg m_{2}.

It follows that deg p_{i} = deg m_{i} for each i, and given monic p_{i} it must be that m_{i}=p_{i}. The proof is complete.