Rank–nullity theorem

The rank–nullity theorem is a theorem in linear algebra, which asserts that the dimension of the domain of a linear map is the sum of its rank (the dimension of its image) and its nullity (the dimension of its kernel) .

Rank–nullity theorem

Stating the theoremEdit

Let  ,   be vector spaces, where   is finite dimensional. Let   be a linear transformation. Then[1]

 ,

where

  and  

One can refine this theorem 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 immediately.

MatricesEdit

Since  ,[2] matrices immediately come to mind when discussing linear maps. In the case of an   matrix, the dimension of the domain is  , the number of columns in the matrix. Thus the Rank-Nullity theorem for a given matrix   immediately becomes

 .

ProofsEdit

Here we provide two proofs. The first[3] operates in the general case, using linear maps. The second proof[4] looks at the homogeneous system   for   with rank   and shows explicitly that there exists a set of   linearly independent solutions that span the kernel of  .

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 proofEdit

Let   be vector spaces over some field   and   defined as in the statement 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 Steinitz exchange lemma, extend   with   linearly independent vectors   to form a full basis of  .

Let

 

such that

 

is a basis for  . 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  , 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.

Second proofEdit

Let   with   linearly independent columns (i.e.  ). We will show that:

  1. There exists a set of   linearly independent solutions to the homogeneous system  .
  2. 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

 ,

where

  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,

 .

Let

 ,

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  .

Therefore,

 
 

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 generalizationsEdit

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 im T and U is ker T, i.e.

 

In the finite-dimensional case, this formulation is susceptible to a generalization: if

0 → V1V2 → ... → Vr → 0

is an exact sequence of finite-dimensional vector spaces, then

 [5]

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.

NotesEdit

  1. ^ Friedberg; Insel; Spence. Linear Algebra. Pearson. p. 70. ISBN 9780321998897.
  2. ^ Friedberg; Insel; Spence. Linear Algebra. pp. 103–104. ISBN 9780321998897.
  3. ^ Friedberg; Insel; Spence. Linear Algebra. Pearson. p. 70. ISBN 9780321998897.
  4. ^ 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
  5. ^ Zaman, Ragib. "Dimensions of vector spaces in an exact sequence". Mathematics Stack Exchange. Retrieved 27 October 2015.

ReferencesEdit