## The Rank-Nullity Theorem

Today we start considering how a given linear transformation acts on a vector space. And we start with the “rank-nullity” theorem. This sounds really fancy, but it’s actually almost trivial.

We already said that (also ) is a abelian category. Now in any abelian category we have the first isomorphism theorem.

So let’s take this and consider a linear transformation . The first isomorphism theorem says we can factor as a surjection followed by an injection . We’ll just regard the latter as the inclusion of the image of as a subspace of . As for the surjection, it must be the linear map , just as in any abelian category. Then we can set up the short exact sequence

and the isomorphism theorem allows us to replace the last term

Now since every short exact sequence splits we have an isomorphism . This is the content of the rank-nullity theorem.

So where do “rank” and “nullity” come in? Well, these are just jargon terms. The “rank” of a linear transformation is the dimension of its image — not the target vector space, mind you, but the subspace of vectors of the form . That is, it’s the size of the largest set of linearly independent vectors in the image of the transformation. The “nullity” is the dimension of the kernel — the largest number of linearly independent vectors that sends to the zero vector in .

So what does the direct sum decomposition above mean? It tells us that there is a basis of which is in bijection with the disjoint union of a basis for and a basis for . In the finite-dimensional case we can take cardinalities and say that the dimension of is the sum of the dimensions of and . Or, to use our new words, the dimension of is the sum of the rank and the nullity of . Thus: the rank-nullity theorem.

[...] Let’s go back and consider a linear map . Remember that we defined its rank to be the dimension of its image. Let’s consider this a little more [...]

Pingback by Column Rank « The Unapologetic Mathematician | July 1, 2008 |

[...] be the dimension of this subspace, which we called the nullity of the linear transformation . The rank-nullity theorem then tells us that we have a relationship between the number of independent solutions to the system [...]

Pingback by Homogenous Linear Systems « The Unapologetic Mathematician | July 14, 2008 |

[...] system has no solutions at all. What’s the problem? Well, we’ve got a linear map . The rank-nullity theorem tells us that the dimension of the image (the rank) plus the dimension of the kernel (the nullity) [...]

Pingback by Unsolvable Inhomogenous Systems « The Unapologetic Mathematician | July 18, 2008 |

[...] the dimension of the cokernel add up to the dimension of the target space. But notice also that the rank-nullity theorem tells us that the dimension of the kernel and the dimension of the image add up to the dimension of [...]

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[...] the rank-nullity theorem says that , and similarly for all other linear maps. So we get — which expresses as the [...]

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[...] the time we get to the th power, where . Instead of repeating everything, let’s just use the rank-nullity theorem, which says for each power that . Now if then we [...]

Pingback by Images of Powers of Transformations « The Unapologetic Mathematician | February 18, 2009 |

[...] is, the kernel of is trivial. Since is a transformation from the vector space to itself, the rank-nullity theorem tells us that the image of is all of . That is, must be an invertible [...]

Pingback by Nondegenerate Forms II « The Unapologetic Mathematician | July 17, 2009 |

[...] this means that we can read off the rank of from the number of rows in , while the nullity is the number of zero columns in [...]

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[...] count the dimensions — if has dimension then each space has dimension — and use the rank-nullity theorem to see that they must be isomorphic. That is, every -multilinear functional is a linear combination [...]

Pingback by Multilinear Functionals « The Unapologetic Mathematician | October 22, 2009 |

[...] this were a linear system, the rank-nullity theorem would tell us that our solution space is (generically) dimensional. Indeed, we could use [...]

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[...] we already know that their dimensions are equal, so the rank-nullity theorem tells us all we need is to find an injective linear map from one to the [...]

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[...] that this is onto, so the dimension of the image is . The dimension of the source is , and so the rank-nullity theorem tells us that the dimension of the kernel — the dimension of the space that sends back to [...]

Pingback by Tensor Products over Group Algebras « The Unapologetic Mathematician | November 9, 2010 |

[...] to a constant function is automatically zero. Thus we conclude that . In fact, we can say more. The rank-nullity theorem tells us that the dimension of and the dimension of add up to the dimension of , which of course [...]

Pingback by Tangent Spaces and Regular Values « The Unapologetic Mathematician | April 26, 2011 |