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

Definition

Let ${}K$ be a field, and let ${}M$ denote an ${}m\times n$-matrix over ${}K$. Then the dimension of the linear subspace of ${}K^{m}$, generated by the columns, is called the column rank of the matrix, written

$\operatorname {rk} \,M.$

Lemma

Let ${}K$ denote a field, and let ${}V$ and ${}W$ denote ${}K$-vector spaces of dimensions ${}n$ and ${}m$. Let

$\varphi \colon V\longrightarrow W$

be a linear mapping, which is described by the matrix ${}M\in \operatorname {Mat} _{m\times n}(K)$, with respect to bases of the spaces. Then

${}\operatorname {rk} \,\varphi =\operatorname {rk} \,M\,$

holds.

Proof

See exercise.

$\Box$

To formulate the next statement, we introduce row rank of an ${}m\times n$-matrix to be the dimension of the linear subspace of ${}K^{n}$ generated by the rows.

Lemma

Let ${}K$ be a field, and let ${}M$ denote an ${}m\times n$-matrix over ${}K$. Then the column rank coincides with the row rank. If ${}M$ is transformed with elementary row manipulations to a matrix ${}M'$ in the sense of fact,

then the rank equals the number of relevant rows of ${}M'$.

Proof

Let ${}r$ denote the number of the relevant rows in the matrix ${}M'$ in echelon form, gained by elementary row manipulations. We have to show that this number is the column rank, and the row rank of ${}M'$ and of ${}M$. 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 ${}M$ equals the row rank of ${}M'$. This matrix has row rank ${}r$, since the first ${}r$ rows are linearly independent, and beside this, there are only zero rows. But ${}M'$ has also column rank ${}r$, since the ${}r$ columns, where there is a new step, are linearly independent, and the other columns are linear combinations of these ${}r$ columns. By exercise, the column rank is preserved by elementary row manipulations.

$\Box$

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

Corollary

Let ${}K$ be a field, and let ${}M$ denote an ${}n\times n$-matrix

over ${}K$. Then the following statements are equivalent.

1. ${}M$ is invertible.
2. The rank of ${}M$ is ${}n$.
3. The rows of ${}M$ are linearly independent.
4. The columns of ${}M$ are linearly independent.

Proof

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

$\varphi \colon K^{n}\longrightarrow K^{n}$

defined by ${}M$. The property that the column rank equals ${}n$, 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.

$\Box$