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A nilpotent shift has minimal polynomial $x^{n}$ and, for $n > 0$ , a single primary component

Example

For $n > 0$ , let $T : F^{n} \to F^{n}$ satisfy $T (e_{0}) = 0$ and $T (e_{i}) = e_{i - 1}$ for $0 < i < n$ , where $(e_{i})_{i < n}$ is the standard basis. Then $μ_{T} = x^{n}$ , and the only primary component is $\ker T^{n} = F^{n}$ . For $n = 0$ , $μ_{T} = 1$ and the primary decomposition is empty.

Facts & Assumptions

Given: The standard basis and shift operator in the Example.

[L1]

The list $(e_{i})_{i < n}$ is an ordered basis of $F^{n}$ , including the empty case (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

The minimal polynomial is the least-degree monic annihilator (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

[L3]

Irreducible-power factors of the minimal polynomial give the primary components (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

Verification

technique · direct

1.1

For $n > 0$ , repeated application of $T$ gives $T^{n} = 0$ , while $T^{n - 1} e_{n - 1} = e_{0} \neq 0$ by [L1]. Hence [L2] gives $μ_{T} = x^{n}$ .

L1L2

2.1

The sole irreducible factor is $x$ , so [L3] gives the one primary component $\ker T^{n} = F^{n}$ . At $n = 0$ , [L1] and [L2] give $μ_{T} = 1$ .

step 1.1L1L2L3∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 86 results over 24 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

Sheldon Axler, Linear Algebra Done Right, 4th ed., §§8A–8B (standard reference, not scraped)

A nilpotent shift has minimal polynomial xnxn and, for n>0n>0, a single primary component