The rank-nullity theorem is a fundamental theorem in linear algebra which relates the dimensions of a linear map's kernel and image with the dimension of its domain.
Stating the Theorem
Let , be vector spaces, of which must be finite dimensional. Let . Then it holds true that
- and .
One can refine this theorem via the splitting lemma to be a statement about an isomorphism of spaces, not just dimensions.
As we know that , matrices immediately come to mind when discussing linear maps. In the case of a matrix, the dimension of the domain is given by , the number of columns in the matrix. Thusly the Rank-Nullity Theorem for a given matrix rather immediately becomes
Here we provide two proofs. The first operates in the general case, using linear maps. The second proof looks at the homogeneous system for with rank and shows explicitly that there exist a set of linearly independent solutions that span the kernel of .
It should be noted that the theorem holds true for any linear map with a finite dimensional domain, but a potentially infinitely dimensional codomain. This means that matrices are only a subset of linear maps that the theorem is true for, leaving the first proof more rigorous. Of course when the codomain is finitely dimensional, the proofs are equivalent.
Let be vector spaces over some field and defined as in the statment of the theorem with .
As is a subspace, there exists a basis for it.
Suppose and let
be such a basis.
We may now, by the Steiniz exchange lemma, extend with linearly independent vectors to form a full basis of .
is a basis for .
From this, we know that
We now claim that is a basis for .
By definition, is generating; it remains to be shown that it is also linearly independent to conlcude that it is a basis.
- for some .
- Thus, owing to the linearity of , it follows that
- This is a contradiction to being a basis, unless all are equal to zero. This shows that is linearly independent, and more specifically that it is a basis for .
To summarise, we have , a basis for , and , a basis for .
Finally we may state that
This concludes our proof.
Let with linearly independent columns (i.e. ). We will show that:
- There exists a set of linearly independent solutions to the homogeneous system .
- That every other solution is a linear combination of these solutions.
To do this, we will produce a matrix whose columns form a basis of the null space of .
Without loss of generality, assume that the first columns of are linearly independent. So, we can write
- with linearly independent column vectors, and
- , each of whose columns are linear combinations of the columns of .
This means that for some (see rank factorization) and, hence,
where is the identity matrix. We note that satisfies
Therefore, each of the columns of are particular solutions of .
Furthermore, the columns of are linearly independent because will imply for :
Therefore, the column vectors of constitute a set of linearly independent solutions for .
We next prove that any solution of must be a linear combination of the columns of .
For this, let
be any vector such that . Note that since the columns of are linearly independent, implies .
This proves that any vector that is a solution of must be a linear combination of the special solutions given by the columns of . And we have already seen that the columns of are linearly independent. Hence, the columns of constitute a basis for the null space of . Therefore, the nullity of is . Since equals rank of , it follows that . This concludes our proof.
Reformulations and generalizations
This theorem is a statement of the first isomorphism theorem of algebra for the case of vector spaces; it generalizes to the splitting lemma.
In more modern language, the theorem can also be phrased as follows: if
- 0 → U → V → R → 0
is a short exact sequence of vector spaces, then
Here R plays the role of im T and U is ker T, i.e.
In the finite-dimensional case, this formulation is susceptible to a generalization: if
- 0 → V1 → V2 → ... → Vr → 0
is an exact sequence of finite-dimensional vector spaces, then
The rank–nullity theorem for finite-dimensional vector spaces may also be formulated in terms of the index of a linear map. The index of a linear map , where and are finite-dimensional, is defined by
Intuitively, is the number of independent solutions of the equation , and is the number of independent restrictions that have to be put on to make solvable. The rank–nullity theorem for finite-dimensional vector spaces is equivalent to the statement
We see that we can easily read off the index of the linear map from the involved spaces, without any need to analyze in detail. This effect also occurs in a much deeper result: the Atiyah–Singer index theorem states that the index of certain differential operators can be read off the geometry of the involved spaces.
- Banerjee, Sudipto; Roy, Anindya (2014), Linear Algebra and Matrix Analysis for Statistics, Texts in Statistical Science (1st ed.), Chapman and Hall/CRC, ISBN 978-1420095388
- Meyer, Carl D. (2000), Matrix Analysis and Applied Linear Algebra, SIAM, ISBN 978-0-89871-454-8.