The maximal number of linearly independent columns of the mbyn 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 F^{m} 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
The Rank Theorem can be expressed as: rank A + dim(Nul A) = n, given an m by n matrix A.
We assume that A is an mbyn matrix over the field F and describes a linear map f as above.
XAY = \begin{bmatrix} I_r & 0 \\ 0 & 0 \\ \end{bmatrix}</math>
The easiest way to compute the rank of a matrix A is given by the Gauss elimination method. The rowechelon 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 nonzero rows.
Consider for example the 4by4 matrix
A = \begin{bmatrix} 2 & 4 & 1 & 3 \\ 1 & 2 & 1 & 0 \\ 0 & 0 & 2 & 2 \\ 3 & 6 & 2 & 5 \\ \end{bmatrix}</math>
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}</math>
which has two nonzero rows.
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