# Linear mapping/Dimension formula/No proof/Section

The following statement is called dimension formula.

## Theorem

Let ${\displaystyle {}K}$ denote a field, let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote ${\displaystyle {}K}$-vector spaces, and let

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

denote a ${\displaystyle {}K}$-linear mapping. Suppose that ${\displaystyle {}V}$ has finite dimension. Then

${\displaystyle {}\dim _{}{\left(V\right)}=\dim _{}{\left(\operatorname {kern} \varphi \right)}+\dim _{}{\left(\operatorname {Im} \varphi \right)}\,}$

holds.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

## Definition

Let ${\displaystyle {}K}$ denote a field, let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote ${\displaystyle {}K}$-vector spaces, and let

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

denote a ${\displaystyle {}K}$-linear mapping. Suppose that ${\displaystyle {}V}$ has finite dimension. Then we call

${\displaystyle {}\operatorname {rk} \,\varphi :=\dim _{}{\left(\operatorname {Im} \varphi \right)}\,}$
the rank of ${\displaystyle {}\varphi }$.

The dimension formula can also be expressed as

${\displaystyle {}\dim _{}{\left(V\right)}=\dim _{}{\left(\operatorname {kern} \varphi \right)}+\operatorname {rk} \,\varphi \,.}$

## Example

We consider the linear mapping

${\displaystyle \varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{4},{\begin{pmatrix}x\\y\\z\end{pmatrix}}\longmapsto M{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}y+z\\2y+2z\\x+3y+4z\\2x+4y+6z\end{pmatrix}},}$

given by the matrix

${\displaystyle {}M={\begin{pmatrix}0&1&1\\0&2&2\\1&3&4\\2&4&6\end{pmatrix}}\,.}$

To determine the kernel, we have to solve the homogeneous linear system

${\displaystyle {}{\begin{pmatrix}y+z\\2y+2z\\x+3y+4z\\2x+4y+6z\end{pmatrix}}={\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}\,.}$

The solution space is

${\displaystyle {}L={\left\{s{\begin{pmatrix}1\\1\\-1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}\,,}$

and this is the kernel of ${\displaystyle {}\varphi }$. The kernel has dimension one, therefore the dimension of the image is ${\displaystyle {}2}$, due to the dimension formula.

## Corollary

Let ${\displaystyle {}K}$ denote a field, let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote ${\displaystyle {}K}$-vector spaces with the same dimension ${\displaystyle {}n}$. Let

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

denote a linear mapping. Then ${\displaystyle {}\varphi }$ is injective if and only if ${\displaystyle {}\varphi }$ is surjective.

### Proof

This follows from the dimension formula and fact.

${\displaystyle \Box }$