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

Linear subspace under a linear mapping

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

Lemma

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

$\varphi \colon V\longrightarrow W$

be a

${}K$-linear mapping. Then the following hold.

1. For a linear subspace ${}S\subseteq V$, the image ${}\varphi (S)$ is a linear subspace of ${}W$.
2. In particular, the image ${}\operatorname {Im} \varphi =\varphi (V)$ of the mapping is a linear subspace of ${}W$.
3. For a linear subspace ${}T\subseteq W$, the preimage ${}\varphi ^{-1}(T)$ is a linear subspace of ${}V$.
4. In particular, ${}\varphi ^{-1}(0)$ is a linear subspace of ${}V$.

Proof

See Exercise 11.2 .

$\Box$

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. Then

${}\operatorname {kern} \varphi :=\varphi ^{-1}(0)={\left\{v\in V\mid \varphi (v)=0\right\}}\,$

is called the kernel of ${}\varphi$.

Due to the statement above, the kernel is a linear subspace of ${}V$.

Remark

For an ${}m\times n$-matrix ${}M$, the kernel of the linear mapping

$K^{n}\longrightarrow K^{m},x\longmapsto Mx,$

given by ${}M$ is just the solution space of the homogeneous linear system

${}Mx=0\,.$

The following criterion for injectivity is important.

Lemma

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. Then ${}\varphi$ is injective if and only if ${}\operatorname {kern} \varphi =0$

holds.

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.

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 Theorem 8.10 , 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 Lemma 11.4 .

$\Box$

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.

Lemma

In the correspondence between linear mappings and matrices, the composition of linear mappings corresponds to the matrix multiplication. More precisely: let ${}U,V,W$ denote vector spaces over a field ${}K$ with bases

${\mathfrak {u}}=u_{1},\ldots ,u_{p},\,{\mathfrak {v}}=v_{1},\ldots ,v_{n}{\text{ and }}{\mathfrak {w}}=w_{1},\ldots ,w_{m}.$

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

${}M_{\mathfrak {w}}^{\mathfrak {u}}(\varphi \circ \psi )=(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))\circ (M_{\mathfrak {v}}^{\mathfrak {u}}(\psi ))\,$

holds.

Proof

We consider the 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 Lemma 10.14 . The (inverse) 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 10.20 , 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

Lemma

Let ${}K$ denote a field, and let ${}V$ and ${}W$ denote finite-dimensional ${}K$-vector spaces. Let ${}{\mathfrak {v}}$ and ${}{\mathfrak {u}}$ be bases of ${}V$ and ${}{\mathfrak {w}}$ and ${}{\mathfrak {z}}$ bases of ${}W$. Let

$\varphi \colon V\longrightarrow W$

denote a linear mapping, which is described by the 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, which describe the change of basis from ${}{\mathfrak {v}}$ to ${}{\mathfrak {u}}$ and from

${}{\mathfrak {w}}$ to ${}{\mathfrak {z}}$.

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 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 Lemma 9.1 and Lemma 10.14 . In this situation, we have altogether

${}{\begin{aligned}M_{\mathfrak {z}}^{\mathfrak {u}}(\varphi )&=\Psi _{\mathfrak {z}}^{-1}\circ \varphi \circ \Psi _{\mathfrak {u}}\\&=\Psi _{\mathfrak {z}}^{-1}\circ (\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ \Psi _{\mathfrak {v}}^{-1})\circ \Psi _{\mathfrak {u}}\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {v}}^{-1}\circ \Psi _{\mathfrak {u}})\\&=(\Psi _{\mathfrak {z}}^{-1}\circ \Psi _{\mathfrak {w}})\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (\Psi _{\mathfrak {u}}^{-1}\circ \Psi _{\mathfrak {v}})^{-1}\\&=M_{\mathfrak {z}}^{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\circ (M_{\mathfrak {u}}^{\mathfrak {v}})^{-1}.\end{aligned}}$

$\Box$

Corollary

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping. Let ${}{\mathfrak {u}}$ and ${}{\mathfrak {v}}$ denote bases of ${}V$. Then the matrices that 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 Lemma 11.11 .

$\Box$

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“.

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