Case of a general linear transformation

MA1112: Linear Algebra II

Dr. Vladimir Dotsenko (Vlad) Lecture 9

Case of a general linear transformation

Now, suppose that ϕ is an arbitrary linear transformation ofV (no assumptionϕ^k =0 anymore). Let us nevertheless consider the sequence of subspacesN₁=ker(ϕ),N₂=ker(ϕ²), . . . , N_m=ker(ϕ^m), . . . .

Note that this sequence is increasing:

N₁⊂N₂⊂. . . N_m⊂. . .

Indeed, ifv∈N_s, that isϕ^s(v) =0, then we haveϕ^s+1(v) =ϕ(ϕ^s(v)) =0.

Since we only work with finite-dimensional vector spaces, this sequence of subspaces cannot be strictly increasing; if Ni 6= Ni+1, then, obviously, dimNi+1 > 1+dimNi. It follows that for some k we have Nk =Nk+1.

Lemma 1. In this case we have Nk+l=Nk for all l > 0.

Proof. We shall prove that Nk+l = Nk+l−1 by induction on l. The induction basis (case l = 1) follows immediately from our notation. Suppose that Nk+l= Nk+l−1; let us prove that Nk+l+1 =Nk+l. Let us take a vectorv∈Nk+l+1, soϕ^k+l+1(v) =0. We haveϕ^k+l+1(v) =ϕ^k+l(ϕ(v)), soϕ(v)∈Nk+l. But by the induction hypothesisNk+l=Nk+l−1, so ϕ^k+l−1(ϕ(v)) =0, or ϕ^k+l(v) =0, so v∈Nk+l, as required.

Lemma 2. For the indexk that we found, we haveV=ker(ϕ^k)⊕Im(ϕ^k).

Proof. Let us first show that ker(ϕ^k)∩Im(ϕ^k) = {0}. Indeed, suppose that v ∈ ker(ϕ^k)∩Im(ϕ^k). This means thatϕ^k(v) =0and thatv=ϕ^k(w)for some vectorw. It follows thatϕ^2k(w) =0, sow∈N_2k. But from the previous lemma we know thatN_2k =N_k, so w ∈N_k. Thus, v = ϕ^k(w) =0, which is what we claimed.

Now we consider the sum of these two subspaces which we just proved to be direct. It is a subspace of V of dimension dim ker(ϕ^k) +dim Im(ϕ^k) =dim(V),so it has to coincide withV.

Note that the result we just proved explains the difference between the case ϕ² = ϕ and ϕ² = 0. In the caseϕ²=ϕ we of course have Ker(ϕ) =Ker(ϕ²), so V= Ker(ϕ)⊕Im(ϕ), while in the case ϕ²=0, usually Ker(ϕ)6=Ker(ϕ²)but Ker(ϕ²) =Ker(ϕ³)always, so we cannot expect that V=Ker(ϕ)⊕Im(ϕ), but we only have the trivial decompositionV=V⊕0=Ker(ϕ²)⊕Im(ϕ²).

Lemma 3. For the indexk that we found,

1. bothKer(ϕ^k)and Im(ϕ^k)are invariant subspaces ofϕ,

2. on the first subspace, the linear transformationϕ has just the zero eigenvalue, 3. on the second subspace, all eigenvalues ofϕare different from zero.

Proof. 1. The invariance is straightforward: ifv∈Ker(ϕ^k), so thatϕ^k(v) =0, then of course ϕ^k(ϕ(v)) =ϕ^k+1(v) =0,

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soϕ(v)∈Ker(ϕ^k), and similarly, ifv∈Im(ϕ^k), so thatv=ϕ^k(w), then of course ϕ(v) =ϕ(ϕ^k(w)) =ϕ^k+1(w) =ϕ^k(ϕ(w)),

soϕ(v)∈Im(ϕ^k).

2. If ϕ(v) =µvfor some 06=v∈Ker(ϕ^k), then 0=ϕ^k(v) =µ^kv, so µ=0.

3. If ϕ(v) = 0 for some0 6= v∈ Im(ϕ^k), thenϕ^k(v) = 0, but we know that Im(ϕ^k)∩Ker(ϕ^k) = {0}, which is a contradiction.

We shall conclude the proof by induction, using the results we obtained. The induction parameter would be slightly unconventional: the number of distinct eigenvalues ofϕ. Our strategy would be to decomposeV into a direct sum of invariant subspaces for each of which ϕhas only one eigenvalue, proving the following result.

Theorem 1. For every linear transformationϕ:V→V whose (different) eigenvalues areλ1, . . . ,λk, there exist invariant subsspaces U1, . . . ,Uk such that on each subspaceUi the only eigenvalue ofϕ isλi, and

V=U1⊕U2⊕ · · · ⊕Uk.

Proof. We shall prove this result by induction on the number of distinct eigenvalues ofϕ.

Let us consider the transformation ϕ_λ1 =ϕ−λ₁I. Considering kernels of its powers, we find the first placek₁ where they stabilise, so that Ker(ϕ^k_λ¹

1) =Ker(ϕ^k_λ¹⁺¹

1 ) =. . ..

Note that the subspaces Ker(ϕ^k_λ¹

1) and Im(ϕ^k_λ¹

1) are invariant subspaces of ϕ. (Indeed, we al- ready know that these are invariant subspaces of ϕλ₁, and ϕ = ϕλ₁ +λ1I). Note also that we have V=Ker(ϕ^k_λ¹

1)⊕Im(ϕ^k_λ¹

1).

On the invariant subspace Ker(ϕ^k_λ¹

1), the transformationϕ_λ1 has only the eigenvalue0, soϕ=ϕ_λ1+λ₁I has only the eigenvalueλ1. Also, on the invariant subspace Im(ϕ^k_λ¹

1), ϕλ₁ has no zero eigenvalues, hence ϕ has no eigenvalues equal toλ1. Hence, we may putU1:=Ker(ϕ^k_λ

1)and then apply the induction hypothesis to the linear transformationϕ on the subspaceV⁰=Im(ϕ^k_λ

1)where it has fewer eigenvalues.

Let us remark that the claim that the eigenvalues ofϕon the subspaceV⁰ =Im(ϕ^k_λ

1)will beλ2, . . . ,λk. This follows from the fact that ifV =V1⊕V2is a decomposition ofVinto a sum of two invariant subspaces, the characteristic polynomial ofϕ onV is the product of the characteristic polynomials onV1andV2: the determinant of a block matrix is equal to the product of determinant of blocks.

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