Rank階数かいすう and nullity退化次数たいかじすう of a linear map線型写像せんけいしゃぞう

1Introduction

The previous rank lecture read the rank of a matrix as the dimension of its column space. This lecture generalizes that definition to linear maps: rank is the dimension of the image of the map. This does not create a new quantity; after choosing bases and writing a representation matrix, the two definitions agree.

2Terms and definitions

Let $T:V\to W$T:V\to W be a linear map線型写像せんけいしゃぞう.

The image像ぞう is

The kernel核かく is

The generalized definition of rank is

and the nullity退化次数たいかじすう is

Do not confuse the two levels of definition. For a matrix, rank was introduced as the dimension of the column space. For a linear map, rank is the dimension of the image. Their agreement is a theorem.

3Plan

The proof strategy is to choose bases and compare the image of a linear map with the column space of its representation matrix. Then we prove rank-nullity and connect it to invertibility.

4Intuitive explanation

A linear map sends input vectors to output vectors. Some input directions survive as output directions; others collapse to zero. The image is the space of surviving outputs. The kernel is the space of collapsed inputs. Rank counts the surviving dimension, and nullity counts the collapsed dimension.

5Precise explanation

5.11. Image and kernel are subspaces

If $y_1=T\left(v_1\right)$y_1=T(v_1) and $y_2=T\left(v_2\right)$y_2=T(v_2) are in $\mathrm{im}T$\operatorname{im}T, then for scalars $a,b$a,b,

so the image is a subspace. If $u_1,u_2\in \ker T$u_1,u_2\in\ker T, then

so the kernel is also a subspace.

5.22. Agreement with matrix rank

Let $A=[T{]}_{C\leftarrow B}$A=[T]_{C\leftarrow B} be the representation matrix of $T:V\to W$T:V\to W. The columns of $A$A are the coordinate vectors of $T\left(v_1\right),\dots ,T\left(v_n\right)$T(v_1),\ldots,T(v_n) in the basis $C$C. Therefore the column space of $A$A is the coordinate representation of $\mathrm{im}T$\operatorname{im}T. Since coordinate maps are dimension-preserving isomorphisms,

5.33. Rank-nullity theorem

Let $u_1,\dots ,u_k$u_1,\ldots,u_k be a basis of $\ker T$\ker T, and extend it to a basis

of $V$V. Then $T\left(v_1\right),\dots ,T\left(v_r\right)$T(v_1),\ldots,T(v_r) is a basis of $\mathrm{im}T$\operatorname{im}T. Spanning follows by writing any $x\in V$x\in V in this basis and using $T\left(u_i\right)=0$T(u_i)=0. Independence follows because any relation among $T\left(v_i\right)$T(v_i) gives an element of $\ker T$\ker T, and uniqueness of the basis expression forces all coefficients to be zero. Hence

5.44. Relation to invertibility

For a linear map $T:V\to V$T:V\to V on a finite-dimensional vector space, the following are equivalent: $T$T is invertible, $\ker T=\left\{0\right\}$\ker T=\{0\}, $\mathrm{rank}\left(T\right)=\dim V$\operatorname{rank}(T)=\dim V, and $\mathrm{im}T=V$\operatorname{im}T=V. Rank-nullity supplies the bridge between kernel and image.

6Worked example

Define $T:R^3\to R^2$T:\mathbb R^3\to\mathbb R^2 by

In standard bases,

Its columns span $R^2$\mathbb R^2, so $\mathrm{rank}\left(T\right)=2$\operatorname{rank}(T)=2. The kernel is given by

so $(x,y,z)=t(-1,1,-1)$(x,y,z)=t(-1,1,-1) and $\mathrm{nullity}\left(T\right)=1$\operatorname{nullity}(T)=1. Thus

7How to recognize the right definition

Use column-space rank when a matrix is the main object. Use image dimension when a linear map is the main object. After bases are chosen, the two definitions agree. For invertibility, check kernel zero, full image, or full rank.

8Scope

Rank-nullity is used here for finite-dimensional vector spaces. Determinants enter only for square matrices or maps $T:V\to V$T:V\to V. Rank is defined for rectangular matrices and maps $V\to W$V\to W, but the same determinant criterion is not.

9Final forms

10In one sentence

• Matrix rank is the dimension of the column space.
• Linear-map rank is the dimension of the image.
• Rank-nullity says input dimension splits into surviving dimension and collapsed dimension.

11Exercise links

12Related links