Linear mapping/Dimension formula/Rank/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

Set ${}n=\dim _{K}{\left(V\right)}$. Let ${}U=\operatorname {kern} \varphi \subseteq V$ denote the kernel of the mapping and let ${}k=\dim _{K}{\left(U\right)}$ denote its dimension (${}k\leq n$). Let

$u_{1},\ldots ,u_{k}$

be a basis of ${}U$. Due to fact, there exist vectors

$v_{1},\ldots ,v_{n-k}$

such that

$u_{1},\ldots ,u_{k},\,v_{1},\ldots ,v_{n-k}$

is a basis of ${}V$. We claim that

$w_{j}=\varphi (v_{j}),\,j=1,\ldots ,n-k,$

is a basis of the image. Let ${}w\in W$ be an element of the image ${}\varphi (V)$. Then there exists a vector ${}v\in V$ such that ${}\varphi (v)=w$. We can write ${}v$ with the basis as

${}v=\sum _{i=1}^{k}s_{i}u_{i}+\sum _{j=1}^{n-k}t_{j}v_{j}\,.$

Then we have

${}{\begin{aligned}w&=\varphi (v)\\&=\varphi {\left(\sum _{i=1}^{k}s_{i}u_{i}+\sum _{j=1}^{n-k}t_{j}v_{j}\right)}\\&=\sum _{i=1}^{k}s_{i}\varphi (u_{i})+\sum _{j=1}^{n-k}t_{j}\varphi (v_{j})\\&=\sum _{j=1}^{n-k}t_{j}w_{j},\end{aligned}}$

which means that ${}w$ is a linear combination in terms of the ${}w_{j}$. In order to prove that the family ${}w_{j}$, ${}j=1,\ldots ,n-k$, is linearly independent, let a representation of zero be given,

${}0=\sum _{j=1}^{n-k}t_{j}w_{j}\,.$

Then

${}\varphi {\left(\sum _{j=1}^{n-k}t_{j}v_{j}\right)}=\sum _{j=1}^{n-k}t_{j}\varphi {\left(v_{j}\right)}=0\,.$

Therefore, ${}\sum _{j=1}^{n-k}t_{j}v_{j}$ belongs to the kernel of the mapping. Hence, we can write

${}\sum _{j=1}^{n-k}t_{j}v_{j}=\sum _{i=1}^{k}s_{i}u_{i}\,.$

Since this is altogether a basis of ${}V$, we can infer that all coefficients are ${}0$, in particular, ${}t_{j}=0$.

$\Box$

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

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

Remark

Let ${}\varphi \colon V\rightarrow W$ be a linear mapping, where ${}V$ has finite dimension. The dimension formula can be illustrated with the following special cases. If ${}\varphi$ is the zero mapping, then ${}\operatorname {kern} \varphi =V$ and

${}\operatorname {rk} \,\varphi =0\,.$

If ${}\varphi$ is injective, then ${}\operatorname {kern} \varphi =0$ and

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

The rank is always between ${}0$ and the dimension of the source space ${}V$. If ${}\varphi$ is surjective, then

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

and

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

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.

$\Box$