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

Statement

Every nilpotent endomorphism $N$ of a finite-dimensional vector space $V$ has an ordered basis that is the concatenation of Jordan strings for $N$ at $0$ . For $V = 0$ , this is the empty basis and the empty family of strings.

Facts & Assumptions

Given: A nilpotent endomorphism $N$ of a finite-dimensional vector space $V$ .

[L1]

Jordan strings with linearly independent initial vectors have linearly independent union (Independent initial vectors make a family of nilpotent Jordan strings independent).

[L2]

Rank-nullity gives $\dim V = \dim \ker N + \dim im N$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L3]

In a finite-dimensional vector space, every linearly independent subset extends to a basis without a choice principle; a subspace of the same dimension as the whole space equals it (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · induction on $\dim\operatorname{im}N$

1.1

If $im N = 0$ , extend the empty independent set to a basis of $V$ by [L3]; every basis vector is then a length-one Jordan string. This includes $V = 0$ .

baseL3

1.2

Put $W = im N$ . If $W \neq 0$ , its restriction $N ∣_{W}$ is nilpotent and $\dim im (N ∣_{W}) < \dim W$ : equality would make the restriction surjective, hence all its powers surjective, contradicting nilpotence on nonzero $W$ . The induction hypothesis therefore gives a Jordan-string basis $(w_{i, 1}, \dots, w_{i, m_{i}})$ of $W$ .

ihalgebra

2.1

For every $i$ , choose $v_{i, m_{i} + 1} \in V$ with $N v_{i, m_{i} + 1} = w_{i, m_{i}}$ ; adjoining it extends the $i$ th string by one. The vectors $w_{i, 1}$ form a basis of $\ker N \cap W$ , because in the Jordan-string basis of $W$ the kernel of $N ∣_{W}$ consists exactly of the initial-vector combinations.

step 1.2choosealgebra

3.1

Use the finite-dimensional extension clause [L3] to extend the independent family $(w_{i, 1})$ to a basis of $\ker N$ ; regard each added vector as a length-one string. The initial vectors of all resulting strings are independent, so [L1] makes their union independent.

step 2.1L1L3

4.1

The union has $\dim W$ vectors inherited from the strings in $W$ , one lift for each old string, and $\dim \ker N$ minus that same number of added kernel vectors; hence it has $\dim W + \dim \ker N = \dim V$ vectors by [L2]. Its span therefore has dimension $\dim V$ , so [L3] makes it all of $V$ .

step 3.1L2L3

5.1

Thus the union is a basis of Jordan strings, completing the induction.

step 1.1step 1.2step 4.1discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 80 results over 23 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 4.2 (standard reference, not scraped)