TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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Complexification preserves kernels, images, finite rank, nullity, and short exact sequences

Statement

Let $T : V \to W$ be a real-linear map. Under the canonical isomorphism of The tensor and direct-sum models of complexification are canonically complex-linearly isomorphic, the complexification acts componentwise: $T_{C} (v + i w) = T v + i T w$ . Consequently

$\ker (T_{C}) = (\ker T) \oplus i (\ker T), im (T_{C}) = (im T) \oplus i (im T),$

the complexifications of the real subspaces. If $V$ and $W$ are finite-dimensional, then ${rank}_{C} (T_{C}) = {rank}_{R} (T)$ and ${nullity}_{C} (T_{C}) = {nullity}_{R} (T)$ . If $0 \to U \overset{i}{\to} V \overset{p}{\to} W \to 0$ is a short exact sequence of real vector spaces, then

$0 ⟶ U_{C} \overset{i_{C}}{\to} V_{C} \overset{p_{C}}{\to} W_{C} ⟶ 0$

is a short exact sequence of complex vector spaces.

Facts & Assumptions

Given: A real-linear map $T : V \to W$ , and in the exactness clause real-linear maps $i : U \to V$ and $p : V \to W$ with $p \circ i = 0$ .

[L1]

The complexification of a real-linear map is $T_{C} (z \otimes v) = z \otimes T (v)$ (Complexification of a real-linear map).

[L2]

The canonical isomorphism $Φ : C \otimes_{R} V \to V \oplus i V$ satisfies $Φ (z \otimes v) = z (v, 0)$ , with inverse $Ψ (v + i w) = 1 \otimes v + i \otimes w$ (The tensor and direct-sum models of complexification are canonically complex-linearly isomorphic).

[L3]

Kernel and image of a linear map are linear subspaces, and a linear map is injective exactly when its kernel is zero (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial).

[L4]

For a linear map from a finite-dimensional space, $\dim V = nullity T + rank T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L5]

A short exact sequence is exact at every displayed module, so $im i = \ker p$ at the middle term (Exact sequences and short exact sequences of modules).

[L6]

Complexification of maps respects composition: $(p \circ i)_{C} = p_{C} \circ i_{C}$ (Complexification is a functor on real vector spaces and real-linear maps).

Proof

technique · direct

1.1

In the direct-sum model, $T_{C} (v + i w) = T v + i T w$ : applying $Φ$ to the formula of [L1], $Φ (T_{C} (1 \otimes v + i \otimes w)) = Φ (1 \otimes T v + i \otimes T w) = T v + i T w$ by [L2].

L1L2algebra

1.2

If $U \leq V$ is a real subspace with real basis $u_{1}, \dots, u_{m}$ , then every element of $U \oplus i U$ is $\sum_{j \leq m} (a_{j} + i b_{j}) u_{j}$ with $a_{j}, b_{j} \in R$ : the real and imaginary components are real combinations of the $u_{j}$ .

L2algebra

1.3

The list $u_{1}, \dots, u_{m}$ is complex-linearly independent in $U \oplus i U$ : $\sum_{j} (a_{j} + i b_{j}) u_{j} = 0$ means $(\sum_{j} a_{j} u_{j}, \sum_{j} b_{j} u_{j}) = (0, 0)$ in $V \oplus i V$ , and the real independence of the $u_{j}$ forces every $a_{j} = b_{j} = 0$ .

L2algebra

1.4

By [L6] and the hypothesis $p \circ i = 0$ , one has $p_{C} \circ i_{C} = 0$ , hence $im i_{C} \subseteq \ker p_{C}$ .

L6givenalgebra

2.1

$\ker (T_{C}) = {v + i w : T v = T w = 0} = (\ker T) \oplus i (\ker T)$ : the equality of the two descriptions is step 1.1, and by [L3] the kernel of the componentwise map is the complexification of $\ker T$ .

step 1.1L2L3

2.2

$im (T_{C}) = {T v + i T w} = (im T) \oplus i (im T)$ , again directly from step 1.1.

step 1.1L2

2.3

For a finite-dimensional real subspace $U$ , steps 1.2 and 1.3 exhibit $u_{1}, \dots, u_{m}$ as a complex basis of $U \oplus i U$ , so $\dim_{C} (U \oplus i U) = \dim_{R} U$ .

step 1.2step 1.3

3.1

Combining steps 2.1, 2.2 and 2.3 gives ${nullity}_{C} (T_{C}) = \dim_{C} \ker (T_{C}) = \dim_{R} \ker T = {nullity}_{R} (T)$ and the matching rank identity, with rank and nullity as in [L4].

step 2.1step 2.2step 2.3L4

3.2

At the middle term, $\ker p_{C} = (\ker p) \oplus i (\ker p) = (im i) \oplus i (im i) = im i_{C}$ by [L5] and steps 2.1 and 2.2 applied to $i$ and $p$ .

step 2.1step 2.2L5

3.3

The map $i_{C}$ is injective because $\ker i_{C} = (\ker i) \oplus i (\ker i) = 0$ by [L3], and $p_{C}$ is surjective because $im p_{C} = (im p) \oplus i (im p) = W_{C}$ .

step 2.1step 2.2L3

4.1

By [L5], exactness of the complexified sequence is: $0 = \ker i_{C}$ at $U_{C}$ , $im i_{C} = \ker p_{C}$ at $V_{C}$ , and $im p_{C} = W_{C}$ at $W_{C}$ ; these are step 3.3, step 3.2 and step 3.3 respectively, with the containment of step 1.4 absorbed into the equality.

step 1.4step 3.2step 3.3L5

5.1

Steps 2.1 and 2.2 prove the kernel and image formulas, step 3.1 the rank and nullity preservation, and step 4.1 the short-exact-sequence clause.

step 3.1step 4.1∎

Depends on

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Sources

Keith Conrad, Complexification (notes) (standard reference, not scraped)