The rank–nullity theorem is a theorem in linear algebra, which asserts: M is the sum of the rank of M and the nullity of M It follows that for linear transformations of vector spaces of equal finite dimension, either injectivity or surjectivity implies bijectivity.

Stating the theorem

Linear transformations

Let T : V \to W be a linear transformation between two vector spaces where T's domain V is finite dimensional. Then where is the rank of T (the dimension of its image) and is the nullity of T (the dimension of its kernel). In other words, This theorem can be refined via the splitting lemma to be a statement about an isomorphism of spaces, not just dimensions. Explicitly, since T induces an isomorphism from to the existence of a basis for V that extends any given basis of implies, via the splitting lemma, that Taking dimensions, the rank–nullity theorem follows.

Matrices

Linear maps can be represented with matrices. More precisely, an m\times n matrix M represents a linear map where F is the underlying field. So, the dimension of the domain of f is n, the number of columns of M, and the rank–nullity theorem for an m\times n matrix M is

Proofs

Here we provide two proofs. The first operates in the general case, using linear maps. The second proof looks at the homogeneous system where \mathbf{A} is a m\times n with rank r, and shows explicitly that there exists a set of n-r linearly independent solutions that span the null space of \mathbf{A}. While the theorem requires that the domain of the linear map be finite-dimensional, there is no such assumption on the codomain. This means that there are linear maps not given by matrices for which the theorem applies. Despite this, the first proof is not actually more general than the second: since the image of the linear map is finite-dimensional, we can represent the map from its domain to its image by a matrix, prove the theorem for that matrix, then compose with the inclusion of the image into the full codomain.

First proof

Let V, W be vector spaces over some field F, and T defined as in the statement of the theorem with \dim V = n. As is a subspace, there exists a basis for it. Suppose and let be such a basis. We may now, by the Steinitz exchange lemma, extend \mathcal{K} with n-k linearly independent vectors to form a full basis of V. Let such that is a basis for V. From this, we know that We now claim that is a basis for. The above equality already states that is a generating set for ; it remains to be shown that it is also linearly independent to conclude that it is a basis. Suppose is not linearly independent, and let for some. Thus, owing to the linearity of T, it follows that This is a contradiction to \mathcal{B} being a basis, unless all \alpha _j are equal to zero. This shows that is linearly independent, and more specifically that it is a basis for. To summarize, we have \mathcal{K}, a basis for, and , a basis for. Finally we may state that This concludes our proof.

Second proof

Let \mathbf{A} be an m\times n matrix with r linearly independent columns (i.e. ). We will show that: To do this, we will produce an matrix \mathbf{X} whose columns form a basis of the null space of \mathbf{A}. Without loss of generality, assume that the first r columns of \mathbf{A} are linearly independent. So, we can write where This means that for some matrix \mathbf{B} (see rank factorization) and, hence, Let where is the identity matrix. So, \mathbf{X} is an matrix such that Therefore, each of the n-r columns of \mathbf{X} are particular solutions of. Furthermore, the n-r columns of \mathbf{X} are linearly independent because will imply for : Therefore, the column vectors of \mathbf{X} constitute a set of n-r linearly independent solutions for. We next prove that any solution of must be a linear combination of the columns of \mathbf{X}. For this, let be any vector such that. Since the columns of are linearly independent, implies. Therefore, This proves that any vector \mathbf{u} that is a solution of must be a linear combination of the n-r special solutions given by the columns of \mathbf{X}. And we have already seen that the columns of \mathbf{X} are linearly independent. Hence, the columns of \mathbf{X} constitute a basis for the null space of \mathbf{A}. Therefore, the nullity of \mathbf{A} is n - r. Since r equals rank of \mathbf{A}, it follows that. This concludes our proof.

A third fundamental subspace

When T: V \to W is a linear transformation between two finite-dimensional subspaces, with n = \dim(V) and m = \dim(W) (so can be represented by an m \times n matrix M), the rank–nullity theorem asserts that if T has rank r, then n - r is the dimension of the null space of M, which represents the kernel of T. In some texts, a third fundamental subspace associated to T is considered alongside its image and kernel: the cokernel of T is the quotient space, and its dimension is m - r. This dimension formula (which might also be rendered ) together with the rank–nullity theorem is sometimes called the fundamental theorem of linear algebra.

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 saying that each short exact sequence of vector spaces splits. Explicitly, given that is a short exact sequence of vector spaces, then, hence Here R plays the role of and U is, i.e. In the finite-dimensional case, this formulation is susceptible to a generalization: if 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 V and W are finite-dimensional, is defined by Intuitively, is the number of independent solutions v of the equation Tv = 0, and is the number of independent restrictions that have to be put on w to make Tv = w 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 T from the involved spaces, without any need to analyze T 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.

Citations