Linear mapping/Dimension formula/No proof/Section

The following statement is called dimension formula.

Theorem

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

$\varphi \colon V\longrightarrow W$

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

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

holds.

Proof

This proof was not presented in the lecture.

$\Box$

Definition

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

$\varphi \colon V\longrightarrow W$

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

${}\operatorname {rk} \,\varphi :=\dim _{K}{\left(\operatorname {Im} \varphi \right)}\,$

the rank of ${}\varphi$.

The dimension formula can also be expressed as

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

Example

We consider the linear mapping

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

given by the matrix

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

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

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

The solution space is

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

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

Corollary

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

$\varphi \colon V\longrightarrow W$

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

surjective.

Proof

This follows from the dimension formula and fact.

$\Box$