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Rank-nullity: $\dim_F V=\operatorname{nullity}T+\operatorname{rank}T$

Statement

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

$\dim_F V=\operatorname{nullity}T+\operatorname{rank}T.$

Equivalently,

$\dim_F V=\dim_F(\ker T)+\dim_F(\operatorname{im}T).$

Facts & Assumptions

Given: A linear map $T:V\to W$ 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 $K\subseteq B$ ; for $C=B\setminus K$ , the set $C$ is finite, $T[C]$ is a basis of $\operatorname{im}T$ , and $T|_C:C\to T[C]$ 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 $\dim_F V$ ; infinite-dimensional means having no finite basis).

[L4]

If two finite sets are disjoint, the cardinality of their union is the sum of their cardinalities (The sum rule: a finite disjoint union is finite with $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , and a sum over a finite index set splits along a partition, clause 1).

[L5]

A bijection between finite sets transports their cardinality (The cardinality $\lvert A\rvert$ of a finite set, consequence (c)).

Proof

technique · direct

1.1

Choose $K,B,C$ as in [L2]. Since $B=K\mathbin{\dot\cup}C$ , [L4] gives $|B|=|K|+|C|$ . The bijection $T|_C:C\to T[C]$ gives $|C|=|T[C]|$ .

L2L4L5given

2.1

Since $B$ , $K$ , and $T[C]$ are bases of $V$ , $\ker T$ , and $\operatorname{im}T$ , respectively, [L1] and [L3] turn step 1.1 into $\dim_F V=\operatorname{nullity}T+\operatorname{rank}T$ .

step 1.1L1L2L3

3.1

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

step 2.1L1∎

Remarks

If $V=\{0_V\}$ , all three dimensions are zero and the formula reads $0=0+0$ ; no positive-dimension hypothesis is hidden.
No finite-dimensionality assumption is made on $W$ . The image is finite-dimensional for the reason isolated in Extending a basis of the kernel to a basis of the domain gives a basis of the image.

Depends on

Used by

For an m× n matrix A, rank(A)+dim N(A)=n Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 100 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

UCLA Algebra Notes, rank-nullity theorem (standard reference, not scraped)
Axler, Linear Algebra Done Right, 4th ed., Chapter 3 (standard reference, not scraped)

Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T\dim_F V=\operatorname{nullity}T+\operatorname{rank}TdimF​V=nullityT+rankT