a) Let U 1 , U 2 ⊂ V be subspaces of V satisfying U 1 ∩ U 2 = {0}. Then any basis of U 2 can be extended to a basis of a complementary space of U 1 inside V .

STRUCTURE OF NILPOTENT ENDOMORPHISMS

RONALD VAN LUIJK

This is an alternative proof of Theorem 3.3 in Michael Stoll’s “Linear Algebra II” (2007).

Lemma 1. Let V be a finite-dimensional vector space over a field F .

a) Let U 1 , U 2 ⊂ V be subspaces of V satisfying U ₁ ∩ U ₂ = {0}. Then any basis of U 2 can be extended to a basis of a complementary space of U ₁ inside V .

b) Let U ₁ , U ₂ , U ₃ ⊂ V be subspaces of V such that U ₃ is a complementary space of U ₁ + U ₂ inside V , and U 2 is a complementary space of U 1 inside U 1 + U 2 . Then U 2 + U 3 is a complementary space of U ₁ inside V and the union of any bases for U ₂ and U ₃ is a basis for U 2 + U 3 .

Proof. Exercise. 

Theorem 2. Let V be a finite-dimensional vector space over a field F and set n = dim V . Let f : V → V be a nilpotent endomorphism. Then V has a basis (v 1 , . . . , v n ) such that for all i ∈ {1, . . . , n} we have f (v _i ) = v _i+1 or f (v _i ) = 0.

Proof. Let m be an integer such that f ^m = 0. Note that we have a chain of inclusions {0} = ker f ⁰ ⊂ ker f ¹ ⊂ ker f ² ⊂ · · · ⊂ ker f ^m−1 ⊂ ker f ^m = V.

We prove by descending induction that for all j ∈ {0, 1, . . . , m} there are elements w 1 , . . . , w s ∈ V and non-negative integers d ₁ , . . . , d _s , such that the sequence

(1) w ₁ , f (w ₁ ), . . . , f ^d

(w ₁ ), w ₂ , f (w ₂ ), . . . , f ^d

(w ₂ ), . . . , w _s , f (w _s ), . . . , f ^d

(w _s )
is a basis of a complementary space X j of ker f ^j inside V and, if j > 0, the sequence (2) f ^d

⁺¹ (w 1 ), . . . , f ^d

⁺¹ (w s )

is a basis of a subspace Y _j ⁰ of ker f ^j satisfying Y _j ⁰ ∩ ker f ^j−1 = {0}.

For j = m this is true because we can take ` = 0 and X _j = Y _j ⁰ = 0 (the zero space is a complementary space of V inside V ). Suppose 0 ≤ j < m and suppose we have elements w ₁ , . . . , w _s ∈ V and integers d ₁ , . . . , d _s , such that the sequence A of (1) is a basis for a complementary space X j+1 of ker f ^j+1 inside V and the sequence of (2) is a basis of a subspace Y _j+1 ⁰ of ker f ^j+1 with Y _j+1 ⁰ ∩ ker f ^j = {0}. Using part (a) of Lemma 1, we extend the sequence (2) to a basis

B = f ^d

⁺¹ (w 1 ), . . . , f ^d

⁺¹ (w s ), w s+1 , w s+2 , . . . , w t

of a complementary space Y j+1 of ker f ^j inside ker f ^j+1 . We set X j = X j+1 + Y j+1 . Then by part (b) of Lemma 1, the space X j is a complementary space of ker f ^j inside V , which, after reordering the elements of A and B, has a basis

w ₁ , f (w ₁ ), . . . , f ^e

(w ₁ ), w ₂ , f (w ₂ ), . . . , f ^e

(w ₂ ), . . . , w _t , f (w _t ), . . . , f ^e

(w _t )
,

1

where e _k = d _k + 1 for 1 ≤ k ≤ s and e _k = 0 for s < k ≤ t. Note that this is exactly (1), with w 1 , . . . , w s replaced by w 1 , . . . , w t and d 1 , . . . , d s replaced by e 1 , . . . , e t . Suppose j > 0, and set Y _j ⁰ = f (Y _j+1 ). The sequence

C = f ^e

⁺¹ (w 1 ), . . . , f ^e

⁺¹ (w t )

equals f (B) and therefore generates Y _j ⁰ . We show that the elements in C are linearly inde- pendent. Suppose λ ₁ , . . . , λ _t ∈ F are such that

(3)

t

X

k=1

λ _k f ^e

⁺¹ (w _k ) = 0,

and set x = P t

k=1 λ _k f ^e

(w _k ) ∈ X _j . Then (3) says f (x) = 0, so x ∈ X _j ∩ ker f ⊂ X _j ∩ ker f ^j = {0}, so x = 0. Since the elements of B are linearly independent, we get λ ₁ = · · · = λ k = 0, so the elements of C are also independent. Since Y j+1 is contained in ker f ^j+1 , its image Y _j ⁰ is contained in ker f ^j . For any y ∈ Y _j ⁰ ∩ ker f ^j−1 there is a y ⁰ ∈ Y _j+1 with y = f (y ⁰ ), which satisfies f ^j (y ⁰ ) = f ^j−1 (y) = 0, which implies y ⁰ ∈ Y _j+1 ∩ ker f ^j = {0}, so we have y ⁰ = 0 and hence y = 0. We obtain Y _j ⁰ ∩ ker f ^j−1 = 0. This finishes the induction argument.

The statement of the theorem follows, as for j = 0, the only complementary space of ker f ^j = ker id V = {0} is V , so we can take (v 1 , . . . , v n ) to be the sequence (1) associated to

j = 0.