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Assuming choice, $\ker T^{} = (im T)^{\circ}$ and $im T^{} = (\ker T)^{\circ}$ ; in finite dimensions $rank T^{*} = rank T$

Statement

Assume the axiom of choice. For a linear map $T : V \to W$ ,

$\ker T^{*} = (im T)^{\circ}, im T^{*} = (\ker T)^{\circ} .$

If $V$ and $W$ are finite-dimensional, then $rank T^{*} = rank T$ .

Facts & Assumptions

Given: The axiom of choice and a linear map $T : V \to W$ .

[L1]

The transpose satisfies $T^{*} (g) = g \circ T$ (The transpose or algebraic adjoint $T^{*} : W^{*} \to V^{*}$ , $T^{*} (g) = g \circ T$ , of a linear map $T : V \to W$ ).

[L2]

An annihilator consists exactly of the functionals vanishing on the named subspace (The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ ).

[L3]

In finite dimension, $\dim U^{\circ} = \dim X - \dim U$ for $U \leq X$ (Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$ ).

[L4]

Assuming choice, every independent set extends to a basis (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ ).

[L5]

Rank-nullity gives $\dim V = \dim \ker T + rank T$ when $V$ is finite-dimensional (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

technique · direct

1.1

A functional $g \in W^{*}$ lies in $\ker T^{*}$ exactly when $g (T v) = 0$ for every $v \in V$ , which by [L1] and [L2] is exactly $g \in (im T)^{\circ}$ .

L1L2

1.2

Every $T^{*} (g) = g \circ T$ vanishes on $\ker T$ , so $im T^{*} \subseteq (\ker T)^{\circ}$ .

L1L2

1.3

Conversely let $f \in (\ker T)^{\circ}$ . Define $h_{0} : im T \to F$ by $h_{0} (T v) = f (v)$ . If $T v = T v^{'}$ , then $v - v^{'} \in \ker T$ and $f (v) = f (v^{'})$ , so $h_{0}$ is well defined and linear. Extend a basis of $im T$ to a basis of $W$ using [L4], and extend $h_{0}$ by value $0$ on the added basis vectors to obtain $h \in W^{*}$ . Then $T^{*} (h) = f$ .

L1L2L4choosealgebra

2.1

Steps 1.2 and 1.3 prove $im T^{*} = (\ker T)^{\circ}$ .

step 1.2step 1.3

3.1

If $V, W$ are finite-dimensional, [L3] and step 2.1 give $rank T^{*} = \dim (\ker T)^{\circ} = \dim V - \dim \ker T$ , which equals $rank T$ by [L5].

step 2.1L3L5

4.1

Steps 1.1, 2.1, and 3.1 give the two identities and the finite-dimensional rank equality, including zero source or target.

step 1.1step 2.1step 3.1∎

Depends on

Used by

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Sources

H. Pinkham, Linear Algebra, Chapter 6 (standard reference, not scraped)

Assuming choice, ker⁡T∗=(im⁡T)∘kerT∗=(imT)∘ and im⁡T∗=(ker⁡T)∘imT∗=(kerT)∘; in finite dimensions rank⁡T∗=rank⁡TrankT∗=rankT