Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 13

Projections

For a direct sum decomposition ${}V=U_{1}\oplus U_{2}$ of a ${}K$-vector space ${}V$, the linear mapping

$p_{1}\colon V\longrightarrow U_{1},v_{1}+v_{2}\longmapsto v_{1},$

is called the first projection (or projection onto ${}U_{1}$ with respect to the given decomposition or projection onto ${}U_{1}$ along ${}U_{2}$) and, accordingly,

$p_{2}\colon V\longrightarrow U_{2},v_{1}+v_{2}\longmapsto v_{2},$

the second projection of this decomposition. Since ${}U_{1}$ and ${}U_{2}$ are linear subspaces of ${}V$, it is reasonable to call the composed mapping

$V{\stackrel {p_{1}}{\longrightarrow }}U_{1}\longrightarrow V$

a projection as well. Then we have a projection in the sense of the following definition.

Definition

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and ${}U\subseteq V$ a linear subspace. A linear mapping

$\varphi \colon V\longrightarrow V$

is called a projection of ${}V$ onto ${}U$, if ${}U=\operatorname {Im} \varphi$ and ${}\varphi {|}_{U}=\operatorname {Id} _{U}$

holds.

Example

Let ${}V$ be a finite-dimensional ${}K$-vector space and ${}v_{i}$, ${}i\in I$, a basis of ${}V$. For a subset ${}J\subseteq I$, set

${}V_{J}=\langle v_{i},\,i\in J\rangle \,,$

the linear subspace corresponding to ${}J$. Moreover, let

$p_{J}\colon V\longrightarrow V,\sum _{i\in I}a_{i}v_{i}\longmapsto \sum _{i\in J}a_{i}v_{i},$

the corresponding projection. The image of this projection is ${}V_{J}$. On ${}V_{J}$, this mapping is the identity, one can also consider this mapping as

$p_{J}\colon V\longrightarrow V_{J}.$

The kernel of this mapping is

${}\operatorname {kern} p_{J}=\langle v_{i},\,i\notin J\rangle \,.$

Example

For ${}\mathbb {R} ^{3}$, equipped with the standard basis, we consider subsets ${}J\subset \{1,2,3\}$ with two elements and the corresponding projections (in the sense of Example 13.2 ) onto the coordinate planes. The projections are

$p_{\{1,2\}}\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3},(a,b,c)\longmapsto (a,b,0),$

$p_{\{1,3\}}\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3},(a,b,c)\longmapsto (a,0,c),$

$p_{\{2,3\}}\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3},(a,b,c)\longmapsto (0,b,c).$

The images of some object in ${}\mathbb {R} ^{3}$ under these projections carry names like ground view etc.

For a subset ${}\{j\}\subseteq \{1,2,3\}$ with one element, the projections map to an axis.

A more abstract definition is the following that does not refer to a linear subspace.

Definition

Let ${}K$ be a field and ${}V$ a ${}K$-vector space. A linear mapping

$\varphi \colon V\longrightarrow V$

is called a projection, if

${}\varphi ^{2}=\varphi \,$

holds.

The identity and the zero mapping are projections.

Lemma

Let ${}K$ be a field and ${}V$ a ${}K$-vector space. For a direct sum decomposition

${}V=U\oplus U'\,,$

the projection onto ${}U$ is a projection in the sense of Definition. Such a projection

$\varphi \colon V\longrightarrow V$

gives a decomposition

${}V=\operatorname {kern} \varphi \oplus \operatorname {Im} \varphi \,,$

and ${}\varphi$ is the projection onto ${}\operatorname {Im} \varphi$.

Proof

Let ${}\pi _{U}$ be the projection onto ${}U$. Write ${}v=u+u'$ with ${}u\in U,\,u'\in U'$. Then we have

${}\pi _{U}(\pi _{U}(u+u'))=\pi _{U}(u)=u=\pi _{U}(u+u')\,,$

and hence

${}\pi _{U}^{2}=\pi _{U}\,.$

Suppose now that

$\varphi \colon V\longrightarrow V$

is an endomorphism with

${}\varphi ^{2}=\varphi \,.$

Let ${}x\in \operatorname {kern} \varphi \cap \operatorname {Im} \varphi$. Then there exists some ${}y\in V$ such that

${}\varphi (y)=x\,.$

Then

${}x=\varphi (y)=\varphi (\varphi (y))=\varphi (x)=0\,.$

This means that the intersection of these linear subspaces is the zero space. For an arbitrary ${}v\in V$, we write

${}v=\varphi (v)+(v-\varphi (v))\,.$

Here, the first summand belongs to the image and, because of

${}\varphi (v-\varphi (v))=\varphi (v)-\varphi (\varphi (v))=\varphi (v)-\varphi (v)=0\,,$

the second summand belongs to the kernel. Therefore, we have a direct sum decomposition.

$\Box$

Spaces of homomorphisms

Definition

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

${}\operatorname {Hom} _{K}{\left(V,W\right)}={\left\{f:V\rightarrow W\mid f{\text{ linear mapping}}\right\}}\,$

is called the space of homomorphisms. It is endowed with the addition defined by

${}(f+g)(v):=f(v)+g(v)\,,$

and the scalar multiplication defined by

${}(\lambda f)(v):=\lambda \cdot f(v)\,.$

Due to Exercise 13.8 , this is indeed a ${}K$-vector space.

Example

Let ${}W$ be a ${}K$-vector space over the field ${}K$. Then the mapping

$\operatorname {Hom} _{K}{\left(K,W\right)}\longrightarrow W,\varphi \longmapsto \varphi (1),$

is an isomorphism of vector spaces, see Exercise 13.10 .

The space of homomorphisms ${}\operatorname {Hom} _{K}{\left(V,K\right)}$ plays an important role. It is called the dual space of ${}V$. In the next lectures, we will discuss it in more details.

Lemma

Let ${}K$ be a field, and let ${}V$ and ${}W$ be

${}K$-vector spaces. Then the following hold.

1. A linear mapping

   $\varphi \colon U\longrightarrow V$

   from another vector space ${}U$ induces a linear mapping

   $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Hom} _{K}{\left(U,W\right)},f\longmapsto f\circ \varphi .$

2. A linear mapping

   $\psi \colon W\longrightarrow T$

   to another vector space ${}T$ induces a linear mapping

   $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Hom} _{K}{\left(V,T\right)},f\longmapsto \psi \circ f.$

Proof

See Exercise 13.15 .

$\Box$

Lemma

Let ${}V$ be a ${}K$-vector space together with a direct sum decomposition

${}V=U_{1}\oplus U_{2}\,.$

Let ${}W$ be another ${}K$-vector space and let

$\varphi _{1}\colon U_{1}\longrightarrow W$

and

$\varphi _{2}\colon U_{2}\longrightarrow W$

denote linear mappings. Then we get, by setting

${}\varphi (v):=\varphi _{1}(v_{1})+\varphi _{2}(v_{2})\,,$

where ${}v=v_{1}+v_{2}$ is the direct decomposition, a linear mapping

$\varphi \colon V\longrightarrow W.$

Proof

The mapping is well-defined, since the representation ${}v=v_{1}+v_{2}$ with ${}v_{1}\in U_{1}$ and ${}v_{2}\in U_{2}$ is unique. The linearity follows from

${}{\begin{aligned}\varphi (av+bv')&=\varphi _{1}{\left((av+bv')_{1}\right)}+\varphi _{2}{\left((av+bv')_{2}\right)}\\&=\varphi _{1}{\left(av_{1}+bv'_{1}\right)}+\varphi _{2}{\left(av_{2}+bv'_{2}\right)}\\&=a\varphi _{1}(v_{1})+b\varphi _{1}(v'_{1})+a\varphi _{2}(v_{2})+b\varphi _{2}(v'_{2})\\&=a\varphi _{1}(v_{1})+a\varphi _{2}(v_{2})+b\varphi _{1}(v'_{1})+b\varphi _{2}(v'_{2})\\&=a\varphi (v)+b\varphi (v')\end{aligned}}$

$\Box$

Lemma

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

${}V=V_{1}\oplus \cdots \oplus V_{n}\,$

and

${}W=W_{1}\oplus \cdots \oplus W_{m}\,$

de direct sum decompositions and let

$p_{j}\colon W\longrightarrow W_{j}$

denote the canonical projections. Then the mapping

$\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \prod _{1\leq i\leq n,\,1\leq j\leq m}\operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)},f\longmapsto p_{j}\circ {\left(f{|}_{V_{i}}\right)},$

is an isomorphism. If we consider ${}\operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)}$ as linear subspaces of ${}\operatorname {Hom} _{K}{\left(V,W\right)}$, then we have the direct sum decomposition

${}\operatorname {Hom} _{K}{\left(V,W\right)}\cong \bigoplus _{1\leq i\leq n,\,1\leq j\leq m}\operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)}\,.$

Proof

It follows directly from Lemma 13.8 that the given mapping is linear. In order to prove injectivity, let ${}f\in \operatorname {Hom} _{K}{\left(V,W\right)}$ with ${}f\neq 0$ be given. Then there exists some ${}v\in V$ such that

${}f(v)\neq 0\,.$

Let ${}v=v_{1}+\cdots +v_{n}$ with ${}v_{i}\in V_{i}$. Then also ${}f(v_{i})\neq 0$ for some ${}i$. Therefore, ${}(f(v_{i}))_{j}\neq 0$ for some ${}j$. Hence

${}f_{ij}=p_{j}\circ {\left(f{|}_{V_{i}}\right)}\neq 0\,.$

In order to prove surjectivity, let a family of homomorphisms ${}f_{ij}\in \operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)},\,1\leq i\leq n,\,1\leq j\leq m$, be given, which we consider as mappings to ${}W$. Then the

${}f_{i}=\sum _{j=1}^{m}f_{ij}\,$

are linear mappings from ${}V_{i}$ to ${}W$. This yields via Lemma 13.9 a linear mapping ${}\sum _{i=1}^{n}f_{i}$ from ${}V$ to ${}W$, which restricts to the given mappings.

$\Box$

Theorem

Let ${}K$ denote a field and let ${}V$ and ${}W$ denote finite-dimensional ${}K$-vector spaces. Let ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ be a basis of ${}V$ and ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$ be a basis of ${}W$. Then the assignment

$\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Mat} _{m\times n}(K),f\longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(f),$

is an isomorphism

of ${}K$-vector spaces.

Proof

The bijectivity was shown in Theorem 10.15 . The additivity follows from

${}{\left(M_{\mathfrak {w}}^{\mathfrak {v}}(f+g)\right)}_{ij}=((f+g)(v_{j}))_{i}=(f(v_{j})+g(v_{j}))_{i}=f(v_{j})_{i}+g(v_{j})_{i}={\left(M_{\mathfrak {w}}^{\mathfrak {v}}(f)\right)}_{ij}+{\left(M_{\mathfrak {w}}^{\mathfrak {v}}(g)\right)}_{ij}\,,$

where the index ${}i$ denotes the ${}i$-th component with respect to the basis ${}{\mathfrak {w}}$.

$\Box$

This means that we can consider the direct sum decomposition for the one-dimensional linear subspaces ${}Kv_{i}$ bzw. ${}Kw_{j}$ corresponding to bases and apply Lemma 13.10 .

Corollary

Let ${}K$ denote a field, and let ${}V$ and ${}W$ denote finite-dimensional ${}K$-vector spaces with dimensions ${}n$ and ${}m$. Then

${}\dim _{K}{\left(\operatorname {Hom} _{K}{\left(V,W\right)}\right)}=nm\,.$

Proof

This follows immediately from Theorem 13.11 .

$\Box$

Linear subspaces of homomorphism spaces

Lemma

Let ${}V$ and ${}W$ be vector spaces over a field ${}K$. Then the following subsets are linear subspaces

of ${}\operatorname {Hom} _{K}{\left(V,W\right)}$.

1. For a linear subspace ${}U\subseteq V$,

   ${}S={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi {|}_{U}=0\right\}}\,$

   is a linear subspace of ${}\operatorname {Hom} _{K}{\left(V,W\right)}$. If ${}V$ and ${}W$ are finite-dimensional, then

   ${}\dim _{K}{\left(S\right)}={\left(\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\right)}\dim _{K}{\left(W\right)}\,.$

2. For a linear subspace ${}U\subseteq W$,

   ${}T={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \operatorname {Im} \varphi \subseteq U\right\}}\,$

   is a linear subspace of ${}\operatorname {Hom} _{K}{\left(V,W\right)}$, which is isomorphic to ${}\operatorname {Hom} _{K}{\left(V,U\right)}$. If ${}V$ and ${}W$ are finite-dimensional, then

   ${}\dim _{K}{\left(T\right)}=\dim _{K}{\left(V\right)}\cdot \dim _{K}{\left(U\right)}\,.$

3. For linear subspaces ${}V_{1}\subseteq V$ and ${}W_{1}\subseteq W$,

   ${}H={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{1})\subseteq W_{1}\right\}}\,$

   is a linear subspace of ${}\operatorname {Hom} _{K}{\left(V,W\right)}$. If ${}V$ and ${}W$ finite-dimensional, then

   $\dim _{K}{\left(H\right)}=\dim _{K}{\left(V_{1}\right)}\dim _{K}{\left(W_{1}\right)}+{\left(\dim _{K}{\left(V\right)}-\dim _{K}{\left(V_{1}\right)}\right)}\dim _{K}{\left(W\right)}\,.$

4. For linear subspaces ${}V_{1},\ldots ,V_{n}\subseteq V$ and ${}W_{1},\ldots ,W_{n}\subseteq W$,

   ${}L={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{1})\subseteq W_{1}{\text{ and }}\varphi (V_{2})\subseteq W_{2}{\text{ and }}\ldots {\text{ and }}\varphi (V_{n})\subseteq W_{n}\right\}}\,$

   is a linear subspace of ${}\operatorname {Hom} _{K}{\left(V,W\right)}$.

Proof

(1). It is clear that we have a linear subspace. In order to prove the statement about the dimension, let ${}U'$ be an direct complement of ${}U$ in ${}V$, so that

${}V=U\oplus U'\,.$

Let ${}u_{1},\ldots ,u_{r}$ be a basis of ${}U'$. Every linear mapping from ${}S$ maps ${}U$ to ${}0$, and on ${}U'$ (or on a basis thereof) we have free choice. Therefore

${}S\cong \operatorname {Hom} _{K}{\left(U',W\right)}\,$

and the statement follows from Corollary 13.12 .

(2). It is clear that we have a linear subspace. The natural mapping

$\operatorname {Hom} _{K}{\left(V,U\right)}\longrightarrow \operatorname {Hom} _{K}{\left(V,W\right)}$

of Lemma 13.8   (2) is injective in this case. Therefore,

${}T\cong \operatorname {Hom} _{K}{\left(V,U\right)}\,.$

(3). It is clear that we have a linear subspace. In the finite-dimensional case, let

${}V=V_{1}\oplus V_{2}\,$

be a direct sum decomposition. Due to Lemma 13.10 , we have

${}\operatorname {Hom} _{K}{\left(V,W\right)}=\operatorname {Hom} _{K}{\left(V_{1},W\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,$

and

${}H=\operatorname {Hom} _{K}{\left(V_{1},W_{1}\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,.$

Therefore, the dimension equals

$\dim _{K}{\left(V_{1}\right)}\cdot \dim _{K}{\left(W_{1}\right)}+\dim _{K}{\left(V_{2}\right)}\cdot \dim _{K}{\left(W\right)}=\dim _{K}{\left(V_{1}\right)}\dim _{K}{\left(W_{1}\right)}+{\left(\dim _{K}{\left(V\right)}-\dim _{K}{\left(V_{1}\right)}\right)}\dim _{K}{\left(W\right)}\,.$

(4). Setting

${}L_{i}={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{i})\subseteq W_{i}\right\}}\,,$

we have ${}L=\bigcap _{i=1}^{k}L_{i}$. Hence, (4) follows from (3).

$\Box$

Remark

Let ${}K$ denote a field and let ${}V$ and ${}W$ denote finite-dimensional ${}K$-vector spaces. We consider the natural mapping

$\Psi \colon V\times \operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow W,(v,\varphi )\longmapsto \varphi (v),$

where we have the product space on the left. This mapping is not linear in general. We have, on one hand,

${}\varphi (v_{1}+v_{2})=\varphi (v_{1})+\varphi (v_{2})\,$

and, on the other hand,

${}{\left(\varphi _{1}+\varphi _{2}\right)}(v)=\varphi _{1}(v)+\varphi _{2}(v)\,,$

so, if we fix one component, we have additivity (and also compatibility with the scalar multiplication) in the other component. In the product space, we have

${}(v_{1},\varphi _{1})+(v_{2},\varphi _{2})=(v_{1}+v_{2},\varphi _{1}+\varphi _{2})\,,$

and, therefore,

${}{\begin{aligned}\Psi ((v_{1},\varphi _{1})+(v_{2},\varphi _{2}))&=\Psi ((v_{1}+v_{2},\varphi _{1}+\varphi _{2}))\\&={\left(\varphi _{1}+\varphi _{2}\right)}{\left(v_{1}+v_{2}\right)}\\&=\varphi _{1}{\left(v_{1}+v_{2}\right)}+\varphi _{2}{\left(v_{1}+v_{2}\right)}\\&=\varphi _{1}(v_{1})+\varphi _{1}(v_{2})+\varphi _{2}(v_{1})+\varphi _{2}(v_{2})\\&\neq \varphi _{1}(v_{1})+\varphi _{2}(v_{2})\\&=\Psi ((v_{1},\varphi _{1}))+\Psi ((v_{2},\varphi _{2}))\end{aligned}}$

(only in exceptional cases we have ${}\varphi _{1}(v_{2})+\varphi _{2}(v_{1})=0$).

Footnotes

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