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

## Definition

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

${\displaystyle \operatorname {rk} \,M.}$

## Lemma

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

${\displaystyle \varphi \colon V\longrightarrow W}$

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

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

holds.

### Proof

${\displaystyle \Box }$

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

## Lemma

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

### Proof

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

${\displaystyle \Box }$

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

## Corollary

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

over ${\displaystyle {}K}$. Then the following statements are equivalent.
1. ${\displaystyle {}M}$ is invertible.
2. The rank of ${\displaystyle {}M}$ is ${\displaystyle {}n}$.
3. The rows of ${\displaystyle {}M}$ are linearly independent.
4. The columns of ${\displaystyle {}M}$ 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

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

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

${\displaystyle \Box }$