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# Rank of a matrix

In linear algebra, the rank of a matrix A with entries in some field is defined to be the maximal number of columns of A which are linearly independent. It is usually denoted by rk(A).

### Alternative definitions

The maximal number of linearly independent columns of the m-by-n matrix A with entries in the field F is equal to the dimension of the column space of A (the column space being the subspace of Fm generated by the columns of A).

Alternatively and equivalently, we can define the rank of A as the maximal number of linearly independent rows of A, which equals the dimension of the row space of A.

If one considers the matrix A as a linear map

f : Fn -> Fm
with the rule
f(x) = Ax
then the rank of A can also be defined as the dimension of the image of f, or as n minus the dimension of the kernel of f (see linear map for a discussion of image and kernel). These definitions have the advantage that they can be applied to any linear map without need for a specific matrix.

The Rank Theorem can be expressed as: rank A + dim(Nul A) = n, given an m by n matrix A.

### Properties

We assume that A is an m-by-n matrix over the field F and describes a linear map f as above.

• rk(A) ≤ min(m,n)
• f is injective if and only if rk(A) = n.
• f is surjective if and only if rk(A) = m.
• In the case of a square matrix A (m = n), then A is invertible if and only if rk(A) = n.
• If B is any n-by-k matrix, then rk(AB) ≤ min(rk(A), rk(B).
• If B is an n-by-k matrix with rk(B) = n, then rk(AB) = rk(A)
• If C is an l-by-m matrix with rk(C) = m, then rk(CA) = rk(A)
• The rank of A is equal to r if and only if there exists an invertible m-by-m matrix X and an invertible n-by-n matrix Y such that

$ XAY = \begin{bmatrix} I_r & 0 \\ 0 & 0 \\ \end{bmatrix} $

where Ir denotes the r-by-r identity matrix.
• The rank of a matrix plus the nullity of the matrix equals the number of columns of the matrix.

### Computation

The easiest way to compute the rank of a matrix A is given by the Gauss elimination method. The row-echelon form of A produced by the Gauss algorithm has the same rank as A, and its rank can be read off as the number of non-zero rows.

Consider for example the 4-by-4 matrix

$ A = \begin{bmatrix} 2 & 4 & 1 & 3 \\ -1 & -2 & 1 & 0 \\ 0 & 0 & 2 & 2 \\ 3 & 6 & 2 & 5 \\ \end{bmatrix} $

We see that the second column is twice the first column, and that the fourth column equals the sum of the first and the third. The first and the third columns are linearly independent, so the rank of A is two. This can be confirmed with the Gauss algorithm. It produces the following row echelon form of A:

$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $

which has two non-zero rows.

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