Rank–nullity theorem

The rank–nullity theorem is a theorem in linear algebra, which asserts:

Rank–nullity theorem

It follows that for linear transformations of vector spaces of equal finite dimension, either injectivity or surjectivity implies bijectivity.

Stating the theorem

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Linear transformations

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Let   be a linear transformation between two vector spaces where  's domain   is finite dimensional. Then   where   is the rank of   (the dimension of its image) and   is the nullity of   (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   induces an isomorphism from   to   the existence of a basis for   that extends any given basis of   implies, via the splitting lemma, that   Taking dimensions, the rank–nullity theorem follows.

Matrices

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Linear maps can be represented with matrices. More precisely, an   matrix M represents a linear map   where   is the underlying field.[5] So, the dimension of the domain of   is n, the number of columns of M, and the rank–nullity theorem for an   matrix M is  

Proofs

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Here we provide two proofs. The first[2] operates in the general case, using linear maps. The second proof[6] looks at the homogeneous system   where   is a   with rank   and shows explicitly that there exists a set of   linearly independent solutions that span the null space 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 proof

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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 summarize, we have  , a basis for  , and  , a basis for  .

Finally we may state that  

 

This concludes our proof.

Second proof

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Let   be an   matrix 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 an   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

  •   is an   matrix with   linearly independent column vectors, and
  •   is an   matrix such that each of its   columns is linear combinations of the columns of  .

This means that   for some   matrix   (see rank factorization) and, hence,  

Let   where   is the   identity matrix. So,   is an   matrix such that  

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  . 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.

A third fundamental subspace

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When   is a linear transformation between two finite-dimensional subspaces, with   and   (so can be represented by an   matrix  ), the rank–nullity theorem asserts that if   has rank  , then   is the dimension of the null space of  , which represents the kernel of  . In some texts, a third fundamental subspace associated to   is considered alongside its image and kernel: the cokernel of   is the quotient space  , and its dimension is  . This dimension formula (which might also be rendered  ) together with the rank–nullity theorem is sometimes called the fundamental theorem of linear algebra.[7][8]

Reformulations and generalizations

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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   plays the role of   and   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[9]   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.

Citations

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  1. ^ Axler (2015) p. 63, §3.22
  2. ^ a b Friedberg, Insel & Spence (2014) p. 70, §2.1, Theorem 2.3
  3. ^ Katznelson & Katznelson (2008) p. 52, §2.5.1
  4. ^ Valenza (1993) p. 71, §4.3
  5. ^ Friedberg, Insel & Spence (2014) pp. 103-104, §2.4, Theorem 2.20
  6. ^ 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
  7. ^ * Strang, Gilbert. Linear Algebra and Its Applications. 3rd ed. Orlando: Saunders, 1988.
  8. ^ Strang, Gilbert (1993), "The fundamental theorem of linear algebra" (PDF), American Mathematical Monthly, 100 (9): 848–855, CiteSeerX 10.1.1.384.2309, doi:10.2307/2324660, JSTOR 2324660
  9. ^ Zaman, Ragib. "Dimensions of vector spaces in an exact sequence". Mathematics Stack Exchange. Retrieved 27 October 2015.

References

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