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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03
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Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T

Statement

Let T:V→W be a linear map of vector spaces over F, with V finite-dimensional. Then

dim⁡FV=nullity⁡T+rank⁡T.

Equivalently,

dim⁡FV=dim⁡F(ker⁡T)+dim⁡F(im⁡T).

Facts & Assumptions

Given: A linear map T:V→W with V finite-dimensional over F.

[L1]

Nullity and rank are the dimensions of the kernel and image (Rank and nullity of a linear map with finite-dimensional domain).

[L2]

There are finite bases K of ker⁡T and B of V with K⊆B; for C=B∖K, the set C is finite, T[C] is a basis of im⁡T, and T∣C:C→T[C] is bijective (Extending a basis of the kernel to a basis of the domain gives a basis of the image).

[L3]

The dimension of a finite-dimensional vector space is the number of elements in any finite basis (Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis).

[L5]

A bijection between finite sets transports their cardinality (The cardinality ∣A∣ of a finite set, consequence (c)).

Proof

technique · direct
1.1

Choose K,B,C as in [L2]. Since B=K∪˙C, [L4] gives ∣B∣=∣K∣+∣C∣. The bijection T∣C:C→T[C] gives ∣C∣=∣T[C]∣.

L2L4L5given
2.1

Since B, K, and T[C] are bases of V, ker⁡T, and im⁡T, respectively, [L1] and [L3] turn step 1.1 into dim⁡FV=nullity⁡T+rank⁡T.

step 1.1L1L2L3
3.1

The displayed equivalent form follows by unfolding the definitions of rank and nullity.

step 2.1L1∎

Remarks

Depends on

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Dependency tree · two levels

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Sources