# Ranking

The rank set or dimension set is a set from the mathematical branch of linear algebra . It shows a connection between the dimensions of the definition set , the core and the image of a linear mapping between two vector spaces .

## sentence

If there is a linear mapping from a vector space into a vector space , then the equation applies to the dimensions of the definition set , the kernel and the image of the mapping${\ displaystyle f \ colon V \ to W}$ ${\ displaystyle V}$${\ displaystyle W}$ ${\ displaystyle V}$ ${\ displaystyle \ mathrm {ker} (f)}$ ${\ displaystyle \ mathrm {im} (f)}$${\ displaystyle f}$

${\ displaystyle \ dim V = \ dim \ mathrm {ker} (f) + \ dim \ mathrm {im} (f)}$.

If one uses the terms defect for the dimension of the core and rank for the dimension of the image of the figure , the ranking is: ${\ displaystyle \ mathrm {def} (f)}$ ${\ displaystyle \ mathrm {rg} (f)}$${\ displaystyle f}$

${\ displaystyle \ dim V = \ mathrm {def} (f) + \ mathrm {rg} (f)}$.

## proofs

### Proof of the homomorphism theorem

The theorem follows directly from the homomorphism theorem

${\ displaystyle \ mathrm {im} (f) \ cong V / \ mathrm {ker} (f)}$.

Since the factor space is isomorphic to a complementary space of in , the following applies ${\ displaystyle V / \ mathrm {ker} (f)}$ ${\ displaystyle U}$${\ displaystyle \ mathrm {ker} (f)}$${\ displaystyle V}$

${\ displaystyle \ mathrm {im} (f) \ cong U}$.

After now

${\ displaystyle V = \ mathrm {ker} (f) \ oplus U}$

is follows from the equivalence of isomorphism and equality of dimension

${\ displaystyle \ dim V = \ dim \ mathrm {ker} (f) + \ dim U = \ dim \ mathrm {ker} (f) + \ dim \ mathrm {im} (f)}$.

If a set is a basis of , which is supplemented by a set with to a basis of ( is then a basis of a complementary space of ), then is ${\ displaystyle B \ subset \ mathrm {ker} (f)}$${\ displaystyle \ mathrm {ker} (f)}$${\ displaystyle A}$${\ displaystyle A \ cap B = \ emptyset}$${\ displaystyle A \ cup B}$${\ displaystyle V}$${\ displaystyle A}$${\ displaystyle \ mathrm {ker} (f)}$

${\ displaystyle f (A) = \ left \ {f (a) \ mid a \ in A \ right \}}$

a base of the image . Consider now the restriction of to the linear envelope${\ displaystyle \ mathrm {im} (f)}$ ${\ displaystyle f ^ {\ prime}}$${\ displaystyle f}$

${\ displaystyle U = \ mathrm {span} (A)}$,

then is injective and ${\ displaystyle f ^ {\ prime}}$

${\ displaystyle \ mathrm {im} (f ^ {\ prime}) = \ mathrm {im} (f)}$.

Thus there is an isomorphism between and the image of . Therefore applies ${\ displaystyle f ^ {\ prime}}$${\ displaystyle U}$${\ displaystyle f}$

${\ displaystyle \ dim V = \ left | A \ right | + \ left | B \ right | = \ dim U + \ dim \ mathrm {ker} (f) = \ dim \ mathrm {im} (f) + \ dim \ mathrm {ker} (f)}$.

The homomorphism theorem also follows - through the transition from the complementary space to the factor space.

## reversal

The theorem is valid for vector spaces of any (also infinite) dimension. In the finite-dimensional case, the dimension of the image space can be derived from the dimension of the core as

${\ displaystyle \ dim \ mathrm {im} (f) = \ dim V- \ dim \ mathrm {ker} (f)}$

to calculate. The reverse also applies accordingly

${\ displaystyle \ dim \ mathrm {ker} (f) = \ dim V- \ dim \ mathrm {im} (f)}$.

In the infinite-dimensional case, the dimension of the image space cannot be calculated from the dimension of the core (or vice versa) using the rank theorem if the core has the same dimension as the entire space. Otherwise the dimension of the image space is equal to the dimension of . ${\ displaystyle W}$${\ displaystyle V}$

## generalization

A far-reaching generalization of the ranking is the statement that the alternating sum of the dimensions of the individual components of a chain complex is equal to the alternating sum of the dimensions of its homology groups. See the Euler characteristic of a chain complex .