# Matrix/Rank/Columns and rows/Invertible/Introduction/Section

Let be a
field,
and let denote an
-matrix
over . Then the
dimension
of the
linear subspace
of , generated by the columns, is called the *column rank* of the matrix, written

Let denote a field, and let and denote -vector spaces of dimensions and . Let

be a linear mapping which is described by the matrix , with respect to bases of the spaces. Then

holds.

### Proof

To formulate the next statement, we introduce *row rank* of an -matrix to be the dimension of the linear subspace of generated by the rows.

Let be a field, and let denote an -matrix over . Then the column rank coincides with the row rank. If is transformed with elementary row manipulations to a matrix in the sense of fact, then the rank equals the number of relevant rows of .

Let denote the number of the relevant rows in the matrix in echelon form, gained by elementary row manipulations. We have to show that this number is the column rank, and the row rank of and of . In an elementary row manipulation, the linear subspace generated by the rows is not changed, therefore the row rank is not changed. So the row rank of equals the row rank of . This matrix has row rank , since the first rows are linearly independent, and beside this, there are only zero rows. But has also column rank , since the columns, where there is a new step, are linearly independent, and the other columns are linear combinations of these columns. By exercise, the column rank is preserved by elementary row manipulations.

Both ranks coincide, so we only talk about the *rank of a matrix*.

Let be a field, and let denote an -matrix

over . Then the following statements are equivalent.- is invertible.
- The rank of is .
- The rows of are linearly independent.
- The columns of are linearly independent.

The equivalence of (2), (3) and (4) follows from the definition and from
fact.

For the equivalence of (1) and (2), let's consider the
linear mapping

defined by . The property that the column rank equals , is equivalent with the map being surjective, and this is, due to
fact,
equivalent with the map being bijective. Because of
fact,
bijectivity is equivalent with the matrix being
invertible.