# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 11/refcontrol

    Linear subspace under a linear mapping
A typical property of a linear mapping is that it maps lines to lines (or a point). More general is following statement.

## LemmaCreate referencenumber

Let $K$ be a field,MDLD/field let $V$ and $W$ denote $K$-vector spacesMDLD/vector spaces and let
    $\varphi \colon V\longrightarrow W$
be a
$K$
* For a linear subspaceMDLD/linear subspace $S\subseteq V$, the imageMDLD/image (map) $\varphi (S)$ is a linear subspace of $W$.
* In particular, the image $\operatorname {Im} \varphi =\varphi (V)$ of the mapping is a linear subspace of $W$.
* For a linear subspace $T\subseteq W$, the preimageMDLD/preimage $\varphi ^{-1}(T)$ is a linear subspace of $V$.
* In particular, $\varphi ^{-1}(0)$ is a linear subspace of $V$.

### Proof

$\Box$

## Definition

Let $K$ denote a field,MDLD/field let $V$ and $W$ denote $K$-vector spaces,MDLD/vector spaces and let
    $\varphi \colon V\longrightarrow W$
denote a $K$-linear mapping.MDLD/linear mapping Then
    $\operatorname {kern} \varphi :=\varphi ^{-1}(0)={\left\{v\in V\mid \varphi (v)=0\right\}}\,$
$\varphi$
Due to the statement above, the kernel is a linear subspace of $V$.

## Remark

For an $m\times n$-matrixMDLD/matrix $M$, the kernelMDLD/kernel (vs) of the linear mappingMDLD/linear mapping
    $K^{n}\longrightarrow K^{m},x\longmapsto Mx,$
given by $M$ is just the solution space of the homogeneous linear systemMDLD/linear system
    $Mx=0\,.$
The following criterion for injectivity is important.

## LemmaCreate referencenumber

Let $K$ denote a field,MDLD/field let $V$ and $W$ denote $K$-vector spaces,MDLD/vector spaces and let
    $\varphi \colon V\longrightarrow W$
denote a $K$-linear mapping.MDLD/linear mapping Then $\varphi$ is injectiveMDLD/injective if and only if $\operatorname {kern} \varphi =0$

### Proof

If the mapping is injective, then there can exist, apart from $0\in V$, no other vector $v\in V$ with $\varphi (v)=0$. Hence, $\varphi ^{-1}(0)=\{0\}$. So suppose that $\operatorname {kern} \varphi =0$, and let $v_{1},v_{2}\in V$ be given with $\varphi (v_{1})=\varphi (v_{2})$. Then, due to linearity,
    $\varphi (v_{1}-v_{2})=\varphi (v_{1})-\varphi (v_{2})=0\,.$
Therefore, $v_{1}-v_{2}\in \operatorname {kern} \varphi$, and so $v_{1}=v_{2}$.
$\Box$
    The dimension formula
The following statement is called dimension formula.

## TheoremCreate referencenumber

Let $K$ denote a field,MDLD/field let $V$ and $W$ denote $K$-vector spaces,MDLD/vector spaces and let
    $\varphi \colon V\longrightarrow W$
denote a $K$-linear mapping.MDLD/linear mapping Suppose that $V$ has finite dimension. Then
    $\dim _{K}{\left(V\right)}=\dim _{K}{\left(\operatorname {kern} \varphi \right)}+\dim _{K}{\left(\operatorname {Im} \varphi \right)}\,$

### Proof

Set $n=\dim _{K}{\left(V\right)}$. Let $U=\operatorname {kern} \varphi \subseteq V$ denote the kernelMDLD/kernel (vs) of the mapping and let $k=\dim _{K}{\left(U\right)}$ denote its dimensionMDLD/dimension (fgvs) ($k\leq n$). Let
    $u_{1},\ldots ,u_{k}$
be a basisMDLD/basis (vs) 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 combinationMDLD/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,MDLD/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,MDLD/field let $V$ and $W$ denote $K$-vector spaces,MDLD/vector spaces and let
    $\varphi \colon V\longrightarrow W$
denote a $K$-linear mapping.MDLD/linear mapping Suppose that $V$ has finite dimension. Then we call
    $\operatorname {rk} \,\varphi :=\dim _{K}{\left(\operatorname {Im} \varphi \right)}\,$
$\varphi$
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,MDLD/linear mapping where $V$ has finite dimension.MDLD/finite dimension (vs) 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,MDLD/injective then $\operatorname {kern} \varphi =0$ and
    $\operatorname {rk} \,\varphi =\dim _{K}{\left(V\right)}\,.$
The rankMDLD/rank (linear mapping) is always between $0$ and the dimension of the source space $V$. If $\varphi$ is surjective,MDLD/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 Create referencenumber

We consider the linear mappingMDLD/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,MDLD/kernel (linear mapping) we have to solve the homogeneous linear systemMDLD/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.

## CorollaryCreate referencenumber

Let $K$ denote a field,MDLD/field let $V$ and $W$ denote $K$-vector spacesMDLD/vector spaces with the same dimensionMDLD/dimension (fgvs) $n$. Let
    $\varphi \colon V\longrightarrow W$
denote a linear mapping.MDLD/linear mapping Then $\varphi$ is injectiveMDLD/injective if and only if $\varphi$ is

### Proof

This follows from the dimension formula and fact.
$\Box$
    Composition of linear mappings and matrices
In the last lecture we have discussed the correspondence between linear mappings and matrices under the condition that bases are fixed. This correspondence respects also compositions of mappings and matrix multiplication, as the following lemma shows.

## LemmaCreate referencenumber

In the correspondenceMDLD/correspondence between linear mappingsMDLD/linear mappings and matrices,MDLD/matrices the compositionMDLD/composition of linear mappings corresponds to the matrix multiplication.MDLD/matrix multiplication More precisely: let $U,V,W$ denote vector spacesMDLD/vector spaces over a fieldMDLD/field $K$ with basesMDLD/bases (vs)
    ${\mathfrak {u}}=u_{1},\ldots ,u_{p},\,{\mathfrak {v}}=v_{1},\ldots ,v_{n}{\text{ and }}{\mathfrak {w}}=w_{1},\ldots ,w_{m}.$
Let
    $\psi :U\longrightarrow V{\text{ and }}\varphi :V\longrightarrow W$
denote linear mappings. Then, for the describing matrix of $\psi ,\,\varphi$, and of the composition $\varphi \circ \psi$, the relation
    $M_{\mathfrak {w}}^{\mathfrak {u}}(\varphi \circ \psi )=(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))\circ (M_{\mathfrak {v}}^{\mathfrak {u}}(\psi ))\,$

### Proof

We consider the commutative diagramMDLD/commutative diagram
    ${\begin{matrix}K^{p}&{\stackrel {M_{\mathfrak {v}}^{\mathfrak {u}}(\psi )}{\longrightarrow }}&K^{n}&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )}{\longrightarrow }}&K^{m}&\\\!\!\!\!\!\Psi _{\mathfrak {u}}\downarrow &&\!\!\!\!\!\Psi _{\mathfrak {v}}\downarrow &&\downarrow \Psi _{\mathfrak {w}}\!\!\!\!\!&&\\U&{\stackrel {\psi }{\longrightarrow }}&V&{\stackrel {\varphi }{\longrightarrow }}&W&\!\!\!\!\!,\\\end{matrix}}$
where the commutativity rests on the identities
    $\varphi \circ \Psi _{\mathfrak {v}}=\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\,$
from fact. The (inverse) coordinate mappingsMDLD/coordinate mappings $\psi _{\mathfrak {v}}$ are bijective. Therefore, we have
    $M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )=\Psi _{\mathfrak {w}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {v}}\,.$
Hence, we get altogether
    ${\begin{aligned}M_{\mathfrak {w}}^{\mathfrak {u}}(\varphi \circ \psi )&=\Psi _{\mathfrak {w}}^{-1}\circ (\varphi \circ \psi )\circ \Psi _{\mathfrak {u}}\\&={\left(\Psi _{\mathfrak {w}}^{-1}\circ \varphi \right)}\circ {\left(\Psi _{\mathfrak {v}}\circ \Psi _{\mathfrak {v}}^{-1}\right)}\circ {\left(\psi \circ \Psi _{\mathfrak {u}}\right)}\\&={\left(\Psi _{\mathfrak {w}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {v}}\right)}\circ {\left(\Psi _{\mathfrak {v}}^{-1}\circ \psi \circ \Psi _{\mathfrak {u}}\right)}\\&=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ M_{\mathfrak {v}}^{\mathfrak {u}}(\psi ),\end{aligned}}$
where we have everywhere compositions of mappings. Due to exercise, the composition of mappings corresponds to the matrix multiplication.
$\Box$
This implies immediately that the multiplication of matrices is associative.
    Linear mappings and base change

## LemmaCreate referencenumber

Let $K$ denote a field,MDLD/field and let $V$ and $W$ denote finite-dimensional $K$-vector spaces.MDLD/vector spaces Let ${\mathfrak {v}}$ and ${\mathfrak {u}}$ be basesMDLD/bases (vs) of $V$ and ${\mathfrak {w}}$ and ${\mathfrak {z}}$ bases of $W$. Let
    $\varphi \colon V\longrightarrow W$
denote a linear mapping,MDLD/linear mapping which is described by the matrixMDLD/matrix $M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$ with respect to the bases ${\mathfrak {v}}$ and ${\mathfrak {w}}$. Then $\varphi$ is described with respect to the bases ${\mathfrak {u}}$ and ${\mathfrak {z}}$ by the matrix
    $M_{\mathfrak {z}}^{\mathfrak {w}}\circ (M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1},$
where $M_{\mathfrak {u}}^{\mathfrak {v}}$ and $M_{\mathfrak {z}}^{\mathfrak {w}}$ are the transformation matrices,MDLD/transformation matrices which describe the change of basis from ${\mathfrak {v}}$ to ${\mathfrak {u}}$ and from
${\mathfrak {w}}$
${\mathfrak {z}}$

### Proof

The linear standard mappings $K^{n}\rightarrow V$ and $K^{m}\rightarrow W$ for the various bases are denoted by $\Psi _{\mathfrak {v}},\,\Psi _{\mathfrak {u}},\,\Psi _{\mathfrak {w}},\,\Psi _{\mathfrak {z}}$. We consider the commutative diagramMDLD/commutative diagram
    ${\begin{matrix}K^{n}&&&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )}{\longrightarrow }}&&&K^{m}\\&\searrow \Psi _{\mathfrak {v}}\!\!\!\!\!&&&&\Psi _{\mathfrak {w}}\swarrow \!\!\!\!\!&\\\!\!\!\!\!M_{\mathfrak {u}}^{\mathfrak {v}}\downarrow &&V&{\stackrel {\varphi }{\longrightarrow }}&W&&\,\,\,\,\downarrow M_{\mathfrak {z}}^{\mathfrak {w}}\\&\nearrow \Psi _{\mathfrak {u}}\!\!\!\!\!&&&&\Psi _{\mathfrak {z}}\nwarrow \!\!\!\!\!&\\K^{n}&&&{\stackrel {M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )}{\longrightarrow }}&&&K^{m},\!\!\!\!\!\end{matrix}}$
where the commutativity rests on fact and fact. In this situation, we have altogether
    ${\begin{aligned}M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )&=\Psi _{\mathfrak {z}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {u}}\\&=\Psi _{\mathfrak {z}}^{-1}\circ (\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ \Psi _{\mathfrak {v}}^{-1})\circ \Psi _{\mathfrak {u}}\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {v}}^{-1}\circ \Psi _{\mathfrak {u}})\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {u}}^{-1}\circ \Psi _{\mathfrak {v}})^{-1}\\&=M_{\mathfrak {z}}^{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}.\end{aligned}}$
$\Box$

## CorollaryCreate referencenumber

Let $K$ denote a field,MDLD/field and let $V$ denote a $K$-vector spaceMDLD/vector space of finite dimension. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping.MDLD/linear mapping Let ${\mathfrak {u}}$ and ${\mathfrak {v}}$ denote basesMDLD/bases (vs) of $V$. Then the matrices which describe the linear mapping with respect to ${\mathfrak {u}}$ and ${\mathfrak {v}}$ respectively (on both sides), fulfil the relation
    $M_{\mathfrak {u}}^{\mathfrak {u}}(\varphi )=M_{\mathfrak {u}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}\,.$

### Proof

This follows directly from fact.
$\Box$
It is an important goal of linear algebra to find, for a given linear mapping $\varphi \colon V\rightarrow V$, a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ such that the describing matrix $M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )$ becomes "quite simple“.