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| In [[linear algebra]], the '''rank''' of a [[matrix (mathematics)|matrix]] ''A'' is a measure of the "[[Degenerate form|nondegenerateness]]" of the [[system of linear equations]] and [[linear transformation]] encoded by ''A''. There are many possible definitions of rank, including the size of the largest collection of [[linear independence|linearly independent]] columns of ''A''. Others are listed in the following section. The rank is one of the fundamental pieces of data associated with a matrix.
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| The rank is commonly denoted by either rk(''A'') or rank(''A''); sometimes the parentheses are unwritten, as in rank ''A''.
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| == Main definitions ==
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| In this section we give three definitions of the rank of a matrix. Many other definitions are possible; see [[#Alternative_definitions|below]] for a list of several of these.
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| The '''column rank''' of a matrix ''A'' is the maximum number of linearly independent column vectors of ''A''. The '''row rank''' of ''A'' is the maximum number of linearly independent row vectors of ''A''. Equivalently, the column rank of ''A'' is the [[dimension (linear algebra)|dimension]] of the [[column space]] of ''A'', while the row rank of ''A'' is the dimension of the [[row space]] of ''A''.
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| A result of fundamental importance in linear algebra is that the column rank and the row rank are always equal. (Two proofs of this result are given [[#Proofs_that_column_rank_.3D_row_rank|below]].) This number (i.e., the number of linearly independent rows or columns) is simply called the '''rank''' of ''A''.
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| The rank is also the dimension of the [[image (matrix)|image]] of the [[linear transformation]] that is given by multiplication by ''A''. More generally, if a [[linear operator]] on a [[vector space]] (possibly infinite-dimensional) has finite-dimensional image (e.g., a [[finite-rank operator]]), then the rank of the operator is defined as the dimension of the image.
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| == Examples ==
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| The matrix
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| :<math>\begin{bmatrix}1&2&1\\-2&-3&1\\3&5&0\end{bmatrix}</math>
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| has rank 2: the first two rows are linearly independent, so the rank is at least 2, but all three rows are linearly dependent (the first is equal to the sum of the second and third) so the rank must be less than 3.
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| The matrix
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| :<math>A=\begin{bmatrix}1&1&0&2\\-1&-1&0&-2\end{bmatrix}</math>
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| has rank 1: there are nonzero columns, so the rank is positive, but any pair of columns is linearly dependent. Similarly, the [[transpose]]
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| :<math>A^T = \begin{bmatrix}1&-1\\1&-1\\0&0\\2&-2\end{bmatrix}</math>
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| of ''A'' has rank 1. Indeed, since the column vectors of ''A'' are the row vectors of the [[transpose]] of ''A'', 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., rk(''A'') = rk(''A''<sup>T</sup>).
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| ==Computing the rank of a matrix==
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| === Rank from row echelon forms ===
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| {{main|Gaussian elimination|Gauss-Jordan elimination}}
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| 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.
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| For example, the matrix ''A'' given by
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| :<math>A=\begin{bmatrix}1&2&1\\-2&-3&1\\3&5&0\end{bmatrix}</math>
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| can be put in reduced row-echelon form by using the following elementary row operations:
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| :<math>\begin{bmatrix}1&2&1\\-2&-3&1\\3&5&0\end{bmatrix}R_2\rightarrow 2r_1 + r_2 \begin{bmatrix}1&2&1\\0&1&3\\3&5&0\end{bmatrix} R_3 \rightarrow -3r_1 + r_3 \begin{bmatrix}1&2&1\\0&1&3\\0&-1&-3\end{bmatrix} R_3 \rightarrow r_2 + r_3 \begin{bmatrix}1&2&1\\0&1&3\\0&0&0\end{bmatrix} R_1 \rightarrow -2r_2 + r_1 \begin{bmatrix}1&0&-5\\0&1&3\\0&0&0\end{bmatrix}</math>.
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| The final matrix (in reduced row echelon form) has two non-zero rows and thus the rank of matrix ''A'' is 2.
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| === Computation ===
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| 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 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.
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| == Proofs that column rank = row rank ==
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| The fact that the column and row ranks of any matrix are equal forms an important part of the [[fundamental theorem of linear algebra]]. We present two proofs of this result. The first is short, uses only basic properties of [[linear combination]]s of vectors, and is valid over any [[field (mathematics)|field]]. The proof is based upon {{harvtxt|Wardlaw|2005}}. The second is an elegant argument using [[orthogonality]] and is valid for matrices over the [[real numbers]]; it is based upon {{harvtxt|Mackiw|1995}}. <!--The third proof establishes a related result for [[complex number|complex]] matrices that reduces to rk(''A'') = rk(''A''<sup>T</sup>) in the case that ''A'' has only real entries.-->
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| ===First proof===
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| Let ''A'' be a matrix of size ''m × n'' (with ''m'' rows and ''n'' columns). Let the column rank of ''A'' be ''r'' and let
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| ''c<sub>1</sub>'',...,''c<sub>r</sub>'' be any basis for the column space of ''A''. Place these as the columns of an ''m × r'' matrix ''C''. Every column of ''A'' can be expressed as a linear combination of the ''r'' columns in ''C''. This means that there is an ''r × n'' matrix ''R'' such that ''A = CR''. ''R'' is the matrix whose ''i''-th column is formed from the coefficients giving the ''i''-th column of ''A'' as a linear combination of the ''r'' columns of ''C''. Now, each row of ''A'' is given by a linear combination of the ''r'' rows of ''R''. Therefore, the rows of ''R'' form a spanning set of the row space of ''A'' and, hence, the row rank of ''A'' cannot exceed ''r''. This proves that the row rank of ''A'' is less than or equal to the column rank of ''A''. This result can be applied to any matrix, so apply the result to the transpose of ''A''. Since the row rank of the transpose of ''A'' is the column rank of ''A'' and the column rank of the transpose of ''A'' is the row rank of ''A'', this establishes the reverse inequality and we obtain the equality of the row rank and the column rank of ''A''. (Also see [[rank factorization]].)
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| ===Second proof===
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| Let ''A'' be an ''m'' × ''n'' matrix with entries in the [[real number]]s whose row rank is ''r''. Therefore, the dimension of the row space of ''A'' is ''r''. Let <math>x_1, x_2,\ldots, x_r</math> be a [[basis (linear algebra)|basis]] of the row space of ''A''. We claim that the vectors <math>Ax_1, Ax_2,\ldots, Ax_r</math> are [[linearly independent]]. To see why, consider a linear homogeneous relation involving these vectors with scalar coefficients <math>c_1,c_2,\ldots,c_r</math>:
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| :<math>0 = c_1 Ax_1 + c_2 Ax_2 + \cdots + c_r Ax_r = A(c_1x_1 + c_2x_2 + \cdots + c_rx_r) = Av, </math>
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| where <math>v = c_1x_1 + c_2x_2 + \cdots + c_r x_r</math>. We make two observations: (a) ''v'' is a linear combination of vectors in the row space of ''A'', which implies that ''v'' belongs to the row space of ''A'', and (b) since ''A'' ''v'' = 0, the vector ''v'' is [[orthogonal]] to every row vector of ''A'' and, hence, is orthogonal to every vector in the row space of ''A''. The facts (a) and (b) together imply that ''v'' is orthogonal to itself, which proves that ''v'' = 0 or, by the definition of ''v'',
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| :<math>c_1x_1 + c_2x_2 + \cdots + c_r x_r = 0.</math>
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| But recall that the <math>x_i</math> were chosen as a basis of the row space of ''A'' and so are linearly independent. This implies that <math>c_1 = c_2 = \cdots = c_r = 0</math>. It follows that <math>Ax_1, Ax_2,\ldots, Ax_r</math> are linearly independent.
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| Now, each <math>Ax_i</math> is obviously a vector in the column space of ''A''. So, <math>Ax_1, Ax_2,\ldots, Ax_r</math> is a set of ''r'' linearly independent vectors in the column space of ''A'' and, hence, the dimension of the column space of ''A'' (i.e., the column rank of ''A'') must be at least as big as ''r''. This proves that row rank of ''A'' is no larger than the column rank of ''A''. Now apply this result to the transpose of ''A'' to get the reverse inequality and conclude as in the previous proof.
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| <!--
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| ===Third proof===
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| Finally, we provide a proof of the related result, rk(''A'') = rk(''A''<sup>*</sup>), where ''A''<sup>*</sup> is the [[conjugate transpose]] or [[hermitian transpose]] of ''A''. When the elements of ''A'' are real numbers, this result becomes rk(''A'') = rk(''A''<sup>T</sup>) and can constitute another proof that row rank is equal to column rank. Otherwise, for complex matrices, rk(''A'') = rk(''A''<sup>*</sup>) is not equivalent to row rank = column rank, and the first proof above should be used. This proof is short, elegant and makes use of the [[null space]].
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| Let ''A'' be an ''m'' × ''n'' matrix. Define rk(''A'') to mean the column rank of ''A'' and let ''A''<sup>*</sup> denote the [[conjugate transpose]] or [[hermitian transpose]] of ''A''. First note that <math>A^*Ax = 0</math> if and only if <math>Ax = 0</math>. Indeed, one direction is trivial and the other follows from the following chain of reasoning:
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| <math>A^*Ax = 0 \Rightarrow x^*A^*Ax = 0 \Rightarrow (Ax)^*(Ax)= 0 \Rightarrow \|Ax\|^2 = 0 \Rightarrow Ax = 0 </math>
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| where <math>\|\cdot\|</math> is the [[Euclidean norm]]. This proves that the [[null space]] of <math>A</math> is equal to the [[null space]] of <math>A^*A</math>. From the [[rank–nullity theorem]], we obtain <math>rk(A) = rk(A^*A)</math>. (Alternate argument: Since <math>A^*Ax = 0 </math> if and only if <math>Ax = 0</math>, the columns of <math>A^*A</math> satisfy the same linear relationships as the columns of <math>A</math>. In particular, they must have the same number of linearly independent columns and, hence, the same column rank.) Each column of <math>A^*A</math> is a linear combination of the columns of <math>A^*</math>. Therefore, the column space of <math>A^*A</math> is a subspace of the column space of <math>A^*</math>. This implies that <math>rk(A^*A) \leq rk(A^*)</math>. We have proved: <math>rk(A) = rk(A^*A) \leq rk(A^*)</math>. Now apply this result to <math>A^*</math> to obtain the reverse inequality: since (<math>A^*)^* = A</math>, we can write <math>rk(A^*) \leq rk((A^*)^*) = rk(A)</math>. This proves <math>rk(A) = rk(A^*)</math>. When the elements of <math>A</math> are real, the [[conjugate transpose]] is the [[transpose]] and we obtain <math>rk(A) = rk(A^T)</math>, as desired. --> | |
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| == Alternative definitions ==
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| In all the definitions in this section, the matrix ''A'' is taken to be an ''m'' × ''n'' matrix over an arbitrary [[field (mathematics)|field]] ''F''.
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| ;dimension of image:
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| Given the matrix ''A'', there is an associated [[linear mapping]]
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| : ''f'' : ''F''<sup>''n''</sup> → ''F''<sup>''m''</sup>
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| defined by
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| :''f''('''x''') = ''A'''''x'''.
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| The rank of ''A'' is the dimension of the image of ''f''. This definition has the advantage that it can be applied to any linear map without need for a specific matrix.
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| ;rank in terms of nullity:
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| Given the same linear mapping ''f'' as above, the rank is ''n'' minus the dimension of the [[kernel (algebra)|kernel]] of ''f''. The [[rank–nullity theorem]] states that this definition is equivalent to the preceding one.
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| ;column rank – dimension of column space:
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| The rank of ''A'' is the maximal number of linearly independent columns <math>c_1,c_2,\dots,c_k</math> of ''A''; this is the [[dimension of a vector space|dimension]] of the [[column space]] of ''A'' (the column space being the subspace of ''F''<sup>''m''</sup> generated by the columns of ''A'', which is in fact just the image of the linear map ''f'' associated to ''A'').
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| ;row rank – dimension of row space:
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| The rank of ''A'' is the maximal number of linearly independent rows of ''A''; this is the dimension of the [[row space]] of ''A''.
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| ;decomposition rank:
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| The rank of ''A'' is the smallest integer ''k'' such that ''A'' can be factored as <math>A=CR</math>, where ''C'' is an ''m'' × ''k'' matrix and ''R'' is a ''k'' × ''n'' matrix. In fact, for all integers ''k'', the following are equivalent:
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| # the column rank of ''A'' is less than or equal to ''k'',
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| # there exist ''k'' columns <math>c_1,\ldots,c_k</math> of size ''m'' such that every column of ''A'' is a linear combination of <math>c_1,\ldots,c_k</math>,
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| # there exist an <math>m \times k</math> matrix ''C'' and a <math>k \times n</math> matrix ''R'' such that <math>A = CR</math> (when ''k'' is the rank, this is a [[rank factorization]] of ''A''),
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| # there exist ''k'' rows <math>r_1,\ldots,r_k</math> of size ''n'' such that every row of ''A'' is a linear combination of <math>r_1,\ldots,r_k</math>,
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| # the row rank of ''A'' is less than or equal to ''k''.
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| Indeed, the following equivalences are obvious: <math>(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Leftrightarrow(5)</math>.
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| For example, to prove (3) from (2), take ''C'' to be the matrix whose columns are <math>c_1,\ldots,c_k</math> from (2).
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| To prove (2) from (3), take <math>c_1,\ldots,c_k</math> to be the columns of ''C''.
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| It follows from the equivalence <math>(1)\Leftrightarrow(5)</math> that the row rank is equal to the column rank.
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| 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 ''f'' : ''V'' → ''W'' is the minimal dimension ''k'' of an intermediate space ''X'' such that ''f'' can be written as the composition of a map ''V'' → ''X'' and a map ''X'' → ''W''. 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.
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| ;determinantal rank – size of largest non-vanishing minor:
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| The rank of ''A'' is the largest order of any non-zero [[Minor (linear algebra)|minor]] in ''A''. (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.
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| A non-vanishing ''p''-minor (''p'' × ''p'' 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 ''n'' vectors has dimension ''p,'' then ''p'' 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 ''n'' vectors has dimension ''p,'' then ''p'' of these vectors span the space ''and'' there is a set of ''p'' coordinates on which they are linearly independent).
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| ;tensor rank – minimum number of simple tensors:
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| The rank of ''A'' is the smallest number ''k'' such that ''A'' can be written as a sum of ''k'' rank 1 matrices, where a matrix is defined to have rank 1 if and only if it can be written as a nonzero product <math>c \cdot r</math> of a column vector ''c'' and a row vector ''r''. This notion of rank is called [[tensor rank]]; it can be generalized in the [[Singular_value_decomposition#Separable_models|separable models]] interpretation of the [[singular value decomposition]].
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| == Properties ==
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| We assume that ''A'' is an ''m'' × ''n'' matrix, and we define the linear map ''f'' by ''f''('''x''') = ''A'''''x''' as above.
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| * The rank of an ''m'' × ''n'' matrix is a [[nonnegative]] [[integer]] and cannot be greater than either ''m'' or ''n''. That is, rk(''A'') ≤ min(''m'', ''n''). A matrix that has a rank as large as possible is said to have '''full rank'''; otherwise, the matrix is '''rank deficient'''.
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| * Only a [[zero matrix]] has rank zero.
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| * ''f'' is [[injective]] if and only if ''A'' has rank ''n'' (in this case, we say that ''A'' has ''full column rank'').
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| * ''f'' is [[surjective]] if and only if ''A'' has rank ''m'' (in this case, we say that ''A'' has ''full row rank'').
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| * If ''A'' is a square matrix (i.e., ''m'' = ''n''), then ''A'' is [[invertible matrix|invertible]] if and only if ''A'' has rank ''n'' (that is, ''A'' has full rank).
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| * If ''B'' is any ''n'' × ''k'' matrix, then
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| ::<math>\operatorname{rank}(AB) \leq \min(\operatorname{rank}\ A, \operatorname{rank}\ B).</math>
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| * If ''B'' is an ''n'' × ''k'' matrix of rank ''n'', then
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| ::<math>\operatorname{rank}(AB) = \operatorname{rank}(A).</math>
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| * If ''C'' is an ''l'' × ''m'' matrix of rank ''m'', then
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| ::<math>\operatorname{rank}(CA) = \operatorname{rank}(A).</math>
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| * The rank of ''A'' is equal to ''r'' if and only if there exists an invertible ''m'' × ''m'' matrix ''X'' and an invertible ''n'' × ''n'' matrix ''Y'' such that
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| ::<math>
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| XAY =
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| \begin{bmatrix}
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| I_r & 0 \\
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| 0 & 0 \\
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| \end{bmatrix},
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| </math>
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| :where ''I''<sub>''r''</sub> denotes the ''r'' × ''r'' [[identity matrix]].
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| * [[Sylvester]]’s rank inequality: if ''A'' is an ''m'' × ''n'' matrix and ''B'' is ''n'' × ''k'', then
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| ::<math>\operatorname{rank}(A) + \operatorname{rank}(B) - n \leq \operatorname{rank}(A B).</math><ref>Proof: Apply the rank–nullity theorem to the inequality
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| ::<math>\dim \operatorname{ker}(AB) \le \dim \operatorname{ker}(A) + \dim \operatorname{ker}(B)</math>.</ref>
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| :This is a special case of the next inequality.
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| * The inequality due to [[Frobenius]]: if ''AB'', ''ABC'' and ''BC'' are defined, then
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| ::<math>\operatorname{rank}(AB) + \operatorname{rank}(BC) \le \operatorname{rank}(B) + \operatorname{rank}(ABC).</math><ref>Proof: The map
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| :<math>C: \operatorname{ker}(ABC) / \operatorname{ker}(BC) \to \operatorname{ker}(AB) / \operatorname{ker}(B)</math>
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| is well-defined and injective. We thus obtain the inequality in terms of dimensions of kernel, which can then be converted to the inequality in terms of ranks by the rank–nullity theorem. Alternatively, if ''M'' is a linear subspace then dim(''AM'') ≤ dim(''M''); apply this inequality to the subspace defined by the (orthogonal) complement of the image of ''BC'' in the image of ''B'', whose dimension is rk(''B'') – rk(''BC''); its image under ''A'' has dimension rk(''AB'') – rk(''ABC'')</ref>
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| * Subadditivity: rank(''A'' + ''B'') ≤ rank(''A'') + rank(''B'') when ''A'' and ''B'' are of the same dimension. As a consequence, a rank-''k'' matrix can be written as the sum of ''k'' rank-1 matrices, but not fewer.
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| * The rank of a matrix plus the [[Kernel (matrix)|nullity]] of the matrix equals the number of columns of the matrix. (This is the [[rank–nullity theorem]].)
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| * If ''A'' is a matrix over the [[real numbers]] then the rank of ''A'' and the rank of its corresponding [[Gram matrix]] are equal. Thus, for real matrices
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| ::<math>\operatorname{rank}(A^T A) = \operatorname{rank}(A A^T) = \operatorname{rank}(A) = \operatorname{rank}(A^T)</math>.
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| :This can be shown by proving equality of their [[kernel (matrix)|null spaces]]. Null space of the Gram matrix is given by vectors ''x'' for which <math>A^T A x = 0</math>. If this condition is fulfilled, also holds <math>0 = x^T A^T A x = |A x|^2</math>. <ref>{{cite book| last = Mirsky| first = Leonid| title = An introduction to linear algebra| year = 1955| publisher = Dover Publications| isbn = 978-0-486-66434-7 }}</ref>
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| * If ''A'' is a matrix over the [[complex numbers]] and ''A''* denotes the conjugate transpose of ''A'' (i.e., the [[Hermitian adjoint|adjoint]] of ''A''), then
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| ::<math>\operatorname{rank}(A) = \operatorname{rank}(\overline{A}) = \operatorname{rank}(A^T) = \operatorname{rank}(A^*) = \operatorname{rank}(A^*A).</math>
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| == Applications ==
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| One useful application of calculating the rank of a matrix is the computation of the number of solutions of a [[system of linear equations]]. According to the [[Rouché–Capelli theorem]], the system is inconsistent if the rank of the [[augmented matrix]] is greater than the rank of the [[coefficient matrix]]. If, on the other hand, the ranks of these two matrices are equal, then the system must have at least one solution. The solution is unique if and only if the rank equals the number of variables. Otherwise the general solution has ''k'' free parameters where ''k'' is the difference between the number of variables and the rank. In this case (and assuming the system of equations is in the real or complex numbers) the system of equations has infinitely many solutions.
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| In [[control theory]], the rank of a matrix can be used to determine whether a [[linear system]] is [[controllability|controllable]], or [[observability|observable]].
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| ==Generalization==
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| There are different generalisations of the concept of rank to matrices over arbitrary [[ring (mathematics)|ring]]s. In those generalisations, column rank, row rank, dimension of column space and dimension of row space of a matrix may be different from the others or may not exist.
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| Thinking of matrices as [[tensors]], the [[tensor rank]] generalizes to arbitrary tensors; note that for tensors of order greater than 2 (matrices are order 2 tensors), rank is very hard to compute, unlike for matrices.
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| There is a notion of [[rank (differential topology)|rank]] for [[smooth map]]s between [[smooth manifold]]s. It is equal to the linear rank of the [[pushforward (differential)|derivative]].
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| ==Matrices as tensors==
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| Matrix rank should not be confused with [[tensor order]], which is called tensor rank. Tensor order is the number of indices required to write a [[tensor]], and thus matrices all have tensor order 2. More precisely, matrices are tensors of type (1,1), having one row index and one column index, also called covariant order 1 and contravariant order 1; see [[Tensor (intrinsic definition)]] for details.
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| Note that the tensor rank of a matrix can also mean the minimum number of [[simple tensor]]s necessary to express the matrix as a linear combination, and that this definition does agree with matrix rank as here discussed.
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| == See also ==
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| * [[Matroid rank]]
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| * [[Nonnegative rank (linear algebra)]]
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| * [[Rank (differential topology)]]
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| ==References==
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| {{Citation
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| | last=Mackiw
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| | first=G.
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| | title=A Note on the Equality of the Column and Row Rank of a Matrix
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| | year=1995
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| | journal=[[Mathematics Magazine]]
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| | volume=68
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| | issue=4}}
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| {{Citation
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| | last=Wardlaw
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| | first=William P.
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| | title=Row Rank Equals Column Rank
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| | year=2005
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| | journal=[[Mathematics Magazine]]
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| | volume=78
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| | issue=4}}
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| <references/>
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| ==Further reading==
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| * {{cite book| author = Roger A. Horn and Charles R. Johnson| title = Matrix Analysis| year = 1985| isbn = 978-0-521-38632-6 }}
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| * Kaw, Autar K. Two Chapters from the book Introduction to Matrix Algebra: 1. Vectors [http://numericalmethods.eng.usf.edu/mws/che/04sle/mws_che_sle_bck_vectors.pdf] and System of Equations [http://numericalmethods.eng.usf.edu/mws/che/04sle/mws_che_sle_bck_system.pdf]
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| * Mike Brookes: Matrix Reference Manual. [http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/property.html#rank]
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| {{linear algebra}}
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| {{DEFAULTSORT:Rank (Linear Algebra)}}
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| [[Category:Linear algebra]]
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