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Independent initial vectors make a family of nilpotent Jordan strings independent

Statement

Let $N : V \to V$ be an endomorphism and let $(v_{i, 1}, \dots, v_{i, m_{i}})$ be finitely many Jordan strings for $N$ at $0$ . If their initial vectors $v_{i, 1}$ are linearly independent, then the union of all vectors in the strings is linearly independent. The empty family is allowed.

Facts & Assumptions

Given: A finite family of nilpotent Jordan strings whose initial vectors are linearly independent.

[L1]

In each string, $N v_{i, 1} = 0$ and $N v_{i, j} = v_{i, j - 1}$ for $j \geq 2$ (Jordan blocks, Jordan strings, and their endpoints).

[L2]

A finite family is linearly independent when its only vanishing finite linear combination has every coefficient zero (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

Proof

technique · induction

1.1

Induct on a natural number $r$ bounding all the string lengths, taking $r = 0$ for the empty family, which has no strings and no maximum length. At $r = 0$ the family is empty and the assertion is immediate, the empty union being independent by [L2].

baseL2

1.2

Assume the result for families whose lengths are bounded by $r - 1$ , and let the present family have lengths bounded by $r \geq 1$ with some string of length exactly $r$ . In a relation $\sum_{i, j} a_{i, j} v_{i, j} = 0$ , applying $N^{r - 1}$ and using [L1] gives $\sum_{i : m_{i} = r} a_{i, r} v_{i, 1} = 0$ .

L1ihalgebra

2.1

Independence of the initial vectors forces every coefficient $a_{i, r}$ to vanish. Removing the terminal vectors at position $r$ leaves a relation among truncated strings of maximum length at most $r - 1$ , so the induction hypothesis makes all remaining coefficients zero; [L2] gives independence of the whole union.

step 1.2ihL2discharge-induction∎

Depends on

Used by

Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 46 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

S. Treil, Linear Algebra Done Wrong, Chapter 9, Theorem 4.1 (standard reference, not scraped)