In linear algebra, the rank of a matrix is the dimension of the vector space generated (or spanned) by its columns. pp. 111-112, §§ 3.115, 3.119 p. 48, § 1.16Bourbaki, Algebra, ch. II, §10.12, p. 359 This corresponds to the maximal number of linearly independent columns of . This, in turn, is identical to the dimension of the vector space spanned by its rows. Rank is thus a measure of the "nondegenerateness" of the system of linear equations and linear transformation encoded by . There are multiple equivalent definitions of rank. A matrix's rank is one of its most fundamental characteristics.
The rank is commonly denoted by or ; sometimes the parentheses are not written, as in .Alternative notation includes from and . The rank can also be denoted by , from German Rang.
More generally, the rank of a linear map between two vector spaces is the dimension of its image.
Main definitions
In this section, we give some definitions of the rank of a matrix. Many definitions are possible; see Alternative definitions for several of these.
The column rank of is the dimension of the column space of , while the row rank of is the dimension of the row space of .
A fundamental result in linear algebra is that the column rank and the row rank are always equal. (Three proofs of this result are given in , below.) This number (i.e., the number of linearly independent rows or columns) is simply called the rank of .
A matrix is said to have full rank if its rank equals the largest possible for a matrix of the same dimensions, which is the lesser of the number of rows and columns. A matrix is said to be rank-deficient if it does not have full rank. The rank deficiency of a matrix is the difference between the lesser of the number of rows and columns, and the rank.
The rank of a linear map or operator is defined as the dimension of its image: p. 200, ch. 3, Definition 2.1 p. 52, § 2.5.1 p. 71, § 4.3 p. 90, § 50where is the dimension of a vector space, and is the image of a map.
Examples The matrix
has rank 2: the first two columns are linearly independent, so the rank is at least 2, but since the third is a linear combination of the first two (the first column plus the second), the three columns are linearly dependent so the rank must be less than 3.
The matrix
has rank 1: there are nonzero columns, so the rank is positive, but any pair of columns is linearly dependent. Similarly, the transpose
of has rank 1. Indeed, since the column vectors of are the row vectors of the transpose of , the statement that the column rank of a matrix equals its row rank is equivalent to the statement that the rank of a matrix is equal to the rank of its transpose, i.e., .
Computing the rank of a matrix
Rank from row echelon forms
A common approach to finding the rank of a matrix is to reduce it to a simpler form, generally row echelon form, by elementary row operations. Row operations do not change the row space (hence do not change the row rank), and, being invertible, map the column space to an isomorphic space (hence do not change the column rank). Once in row echelon form, the rank is clearly the same for both row rank and column rank, and equals the number of pivots (or basic columns) and also the number of non-zero rows.
For example, the matrix given by
can be put in reduced row-echelon form by using the following elementary row operations:
The final matrix (in reduced row echelon form) has two non-zero rows and thus the rank of matrix is 2.
Computation When applied to floating point computations on computers, basic Gaussian elimination (LU decomposition) can be unreliable, and a rank-revealing decomposition should be used instead. An effective alternative is the singular value decomposition (SVD), but there are other less computationally expensive choices, such as QR decomposition with pivoting (so-called rank-revealing QR factorization), which are still more numerically robust than Gaussian elimination. Numerical determination of rank requires a criterion for deciding when a value, such as a singular value from the SVD, should be treated as zero, a practical choice which depends on both the matrix and the application.
Proofs that column rank = row rank
Proof using row reduction The fact that the column and row ranks of any matrix are equal is fundamental in linear algebra. Many proofs have been given. One of the most elementary ones has been sketched in . Here is a variant of this proof:
It is straightforward to show that neither the row rank nor the column rank are changed by an elementary row operation. As Gaussian elimination proceeds by elementary row operations, the reduced row echelon form of a matrix has the same row rank and the same column rank as the original matrix. Further elementary column operations allow putting the matrix in the form of an identity matrix possibly bordered by rows and columns of zeros. Again, this changes neither the row rank nor the column rank. It is immediate that both the row and column ranks of this resulting matrix is the number of its nonzero entries.
We present two other proofs of this result. The first uses only basic properties of linear combinations of vectors, and is valid over any field. The proof is based upon Wardlaw (2005). The second uses orthogonality and is valid for matrices over the real numbers; it is based upon Mackiw (1995). Both proofs can be found in the book by Banerjee and Roy (2014).
Proof using linear combinations Let be an matrix. Let the column rank of be , and let be any basis for the column space of . Place these as the columns of an matrix . Every column of can be expressed as a linear combination of the columns in . This means that there is an matrix such that . is the matrix whose th column is formed from the coefficients giving the th column of as a linear combination of the columns of . In other words, is the matrix which contains the multiples for the bases of the column space of (which is ), which are then used to form as a whole. Now, each row of is given by a linear combination of the rows of . Therefore, the rows of form a spanning set of the row space of and, by the Steinitz exchange lemma, the row rank of cannot exceed . This proves that the row rank of is less than or equal to the column rank of . This result can be applied to any matrix, so apply the result to the transpose of . Since the row rank of the transpose of is the column rank of and the column rank of the transpose of is the row rank of , this establishes the reverse inequality and we obtain the equality of the row rank and the column rank of . (Also see Rank factorization.)
Proof using orthogonality Let be an matrix with entries in the real numbers whose row rank is . Therefore, the dimension of the row space of is . Let be a basis of the row space of . We claim that the vectors are linearly independent. To see why, consider a linear homogeneous relation involving these vectors with scalar coefficients :
where . We make two observations: (a) is a linear combination of vectors in the row space of , which implies that belongs to the row space of , and (b) since , the vector is orthogonal to every row vector of and, hence, is orthogonal to every vector in the row space of . The facts (a) and (b) together imply that is orthogonal to itself, which proves that or, by the definition of ,
But recall that the were chosen as a basis of the row space of and so are linearly independent. This implies that . It follows that are linearly independent.
Every is in the column space of . So, is a set of linearly independent vectors in the column space of and, hence, the dimension of the column space of (i.e., the column rank of ) must be at least as big as . This proves that row rank of is no larger than the column rank of . Now apply this result to the transpose of to get the reverse inequality and conclude as in the previous proof.
Alternative definitions In all the definitions in this section, the matrix is taken to be an matrix over an arbitrary field .
Dimension of image Given the matrix , there is an associated linear mapping
defined by
The rank of is the dimension of the image of . This definition has the advantage that it can be applied to any linear map without need for a specific matrix.
Rank in terms of nullity Given the same linear mapping as above, the rank is minus the dimension of the kernel of . The rank–nullity theorem states that this definition is equivalent to the preceding one.
Column rank – dimension of column space The rank of is the maximal number of linearly independent columns of ; this is the dimension of the column space of (the column space being the subspace of generated by the columns of , which is in fact just the image of the linear map associated to ).
Row rank – dimension of row space The rank of is the maximal number of linearly independent rows of ; this is the dimension of the row space of .
Decomposition rank The rank of is the smallest positive integer such that can be factored as , where is an matrix and is a matrix. In fact, for all integers , the following are equivalent:
the column rank of is less than or equal to , there exist columns of size such that every column of is a linear combination of , there exist an matrix and a matrix such that (when is the rank, this is a rank factorization of ), there exist rows of size such that every row of is a linear combination of , the row rank of is less than or equal to .
Indeed, the following equivalences are obvious: . For example, to prove (3) from (2), take to be the matrix whose columns are from (2). To prove (2) from (3), take to be the columns of .
It follows from the equivalence that the row rank is equal to the column rank.
As in the case of the "dimension of image" characterization, this can be generalized to a definition of the rank of any linear map: the rank of a linear map is the minimal dimension of an intermediate space such that can be written as the composition of a map and a map . Unfortunately, this definition does not suggest an efficient manner to compute the rank (for which it is better to use one of the alternative definitions). See rank factorization for details.
Rank in terms of singular values The rank of equals the number of non-zero singular values, which is the same as the number of non-zero diagonal elements in Σ in the singular value decomposition
Determinantal rank – size of largest non-vanishing minor The rank of is the largest order of any non-zero minor in . (The order of a minor is the side-length of the square sub-matrix of which it is the determinant.) Like the decomposition rank characterization, this does not give an efficient way of computing the rank, but it is useful theoretically: a single non-zero minor witnesses a lower bound (namely its order) for the rank of the matrix, which can be useful (for example) to prove that certain operations do not lower the rank of a matrix.
A non-vanishing -minor ( submatrix with non-zero determinant) shows that the rows and columns of that submatrix are linearly independent, and thus those rows and columns of the full matrix are linearly independent (in the full matrix), so the row and column rank are at least as large as the determinantal rank; however, the converse is less straightforward. The equivalence of determinantal rank and column rank is a strengthening of the statement that if the span of vectors has dimension , then of those vectors span the space (equivalently, that one can choose a spanning set that is a subset of the vectors): the equivalence implies that a subset of the rows and a subset of the columns simultaneously define an invertible submatrix (equivalently, if the span of vectors has dimension , then of these vectors span the space and there is a set of coordinates on which they are linearly independent).
Tensor rank – minimum number of simple tensors