Some Categorical Properties of Linear Systems

Abstract

Linear control systems are studied by means of a state-space approach. Feedback morphisms are presented as natural generalization of feedback equivalences. The set of feedback morphisms between two linear systems is a vector space. Kernels, cokernels, as well as monomorphisms, epimorphisms, sections, and retracts of feedback morphisms are studied in the category of linear systems over finite dimensional vector spaces. Finally, a classical Kalman’s decomposition of linear systems over vector spaces is presented as a split short exact sequence in the category.

1. Introduction

Morphisms of linear (control) systems have been recently introduced [1] in order to study the structure of linear systems up to feedback actions. New feedback invariants were found [2] and thus classical results [3] were extended to a complete classification of regular linear systems over commutative rings R.

The purpose of this paper is to continue developing the theory of the feedback morphisms of linear systems. The kernels and cokernels of feedback morphisms are computed when they exist. The category $S_R$ of linear systems over a commutative ring R, together with feedback morphisms, is not abelian (note that $S_R$ contains R-Mod as a subcategory). The use of categories in the study of linear systems goes back at least to [4], where the authors proved the wildness of the feedback classification problem over arbitrary commutative rings.

In this paper, we focus on linear systems over finite dimensional vector spaces, that is to say, on the category $S_{\mathbb{K}}$, where $\mathbb{K}$ denotes the field of scalars. Monomorphisms and epimorphisms are identified in the category $S_{\mathbb{K}}$ of linear systems, as well as sections and retracts. The classical Kalman’s decomposition Theorem [5] is stated in terms of a kernel–cokernel split exact sequence of linear systems.

2. Some Preliminaries

Let $\mathbb{K}$ be a field and V a finite dimensional $\mathbb{K}$-vector space. A linear system over $\mathbb{K}$ is a triple $\mathsf{\Sigma }=(V,f,B)$, where $f:V\to V$ is an endomorphism, and $B\subseteq V$ is a subspace (the subspace of controls).

Let $\mathsf{\Sigma }=(V,f,B)$ and ${\mathsf{\Sigma }}^{\prime }=(V^{\prime },f^{\prime },B^{\prime })$ be linear systems over $\mathbb{K}$. A morphism of linear systems, or feedback morphism, between $\mathsf{\Sigma }$ and ${\mathsf{\Sigma }}^{\prime }$ is a linear map $\left(\mathfrak{a}:V\to V^{\prime }\right)\in {hom}_{\mathbb{K}}(V,V^{\prime })=hom(V,V^{\prime })$, such that [6]

$$\begin{array}{cc}1.\hfill & \mathfrak{a}\left(B\right)\subset B^{\prime }\hfill \\ 2.\hfill & Im(f^{\prime }\circ \mathfrak{a}-\mathfrak{a}\circ f)\subset B^{\prime }\hfill \end{array}$$

(1)

Denote by $\mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })={\mathfrak{hom}}_{S_{\mathbb{K}}}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })$ the $\mathbb{K}$-vector space [1] of morphisms of linear systems between $\mathsf{\Sigma }$ and ${\mathsf{\Sigma }}^{\prime }$. Note that $\mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })$ is a subspace of ${hom}_{\mathbb{K}}(V,V^{\prime })$.

The composition of feedback morphisms is defined as the composition of linear maps. It is routine to check that the composition of feedback morphisms is a feedback morphism. It is also straightforward to check that identities are morphisms of systems, as well as zero maps. Associative laws follow from the associative laws of composition of linear maps. Thus, $S_{\mathbb{K}}$ denotes the category whose objects are linear systems over $\mathbb{K}$ and whose morphisms are feedback morphisms.

Since all arrows in category $S_{\mathbb{K}}$ are obtained from linear maps, and since identities in $S_{\mathbb{K}}$ are obtained from identity linear maps, it follows that $S_{\mathbb{K}}$ is an enriched category over the category $\mathbb{K}$-Vect of finite dimensional $\mathbb{K}$-vector spaces.

Note 1.

Functor $F:S_{\mathbb{K}}\to \mathbb{K}$-Vect sending $F\left(V,f,B\right)=V$ on objects and $F\left(\mathfrak{a}\right)=\mathfrak{a}$ on morphisms is obviously injective on morphisms. Hence, F is a faithful functor. Functor F is also dense because every vector space V is of the form $F(V,0,0)$.

Isomorphisms in the category $S_{\mathbb{K}}$ are exactly the classical feedback equivalences (see [6] for details).

Some other categories of linear systems might be considered. Here are some examples.

Example 1

(Zero-controlled linear systems). Let $S_{\mathbb{K}}^0$ be the full subcategory of $S_{\mathbb{K}}$ gathering all linear systems on the form $(V,f,0)$. Then, feedback morphisms a are linear maps commuting with f.

Example 2

(Full-controlled linear systems). Let $S_{\mathbb{K}}^{\mathrm{Full}}$ be the subcategory of $S_{\mathbb{K}}$ gathering all linear systems of the form $(V,f,V)$. In this subcategory of $S_{\mathbb{K}}$, every linear map arises as a feedback morphism, that is to say ${\mathfrak{hom}}_{S_{\mathbb{K}}^{\mathrm{Full}}}={hom}_{\mathbb{K}}$. Functor $F:S_{\mathbb{K}}^{\mathrm{Full}}\to \mathbb{K}-\mathrm{Vect}$ sends $F(V,f,V)=V$ and $F\left(\mathfrak{a}\right)=\mathfrak{a}$ is faithful and dense. F is also full because the map $\mathfrak{hom}\left(\right(V,f,V),(W,g,W\left)\right)\to hom(V,W)$ is surjective, and in fact bijective. Therefore, F is, in this case, an equivalence of categories $S_{\mathbb{K}}^{\mathrm{Full}}$ and $\mathbb{K}$-Vect.

Example 3

(Reachable linear systems). Let $\mathsf{\Sigma }=(V,f,B)$ be a linear system. Consider the chain of vector subspaces of V given by $N_0=0$, $N_1=B$, $N_2=B+f\left(B\right)$, and in general $N_i=B+f\left(N_{i-1}\right)$. It is clear that $N_0\subseteq N_1\subseteq N_2\subseteq \cdots \subseteq N_i\subseteq N_{i+1}\subseteq \cdots$, and, on the other side, if $N_i=N_{i+1}$, then $N_{i+1}=N_{i+2}$ and the chain stabilizes forever. Since $dimV<+\infty$, there exists an s such that $N_s=N_{s+1}$. Define the degree of system Σ, and write $s=deg\left(\mathsf{\Sigma }\right)$ as the least integer s such that $N_s=N_{s+1}$.

Linear system $\mathsf{\Sigma }=(V,f,B)$ is called reachable if $N_{deg\left(\mathsf{\Sigma }\right)}=V$. Consider $A_{\mathbb{K}}$ the full subcategory of $S_{\mathbb{K}}$, whose objects are the reachable linear systems. Brunovsky’s theorem [3] implies in particular that every reachable system is feedback isomorphic to a Brunovsky canonical form. Hence, the skeleton of $A_{\mathbb{K}}$ is the full subcategory of $A_{\mathbb{K}}$ gathering all Brunovsky’s canonical forms.

3. Kernels and Cokernels of Feedback Morphisms

Let $\mathsf{\Sigma }=(V,f,B),{\mathsf{\Sigma }}^{\prime }=(V^{\prime },f^{\prime },B^{\prime })$ be linear systems. Then, the set $\mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })$ is a vector subspace of $hom(V,V^{\prime })$ and hence an abelian group. Moreover, compositions are bilinear. That is to say, for objects $\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime },{\mathsf{\Sigma }}^{\prime \prime }$ and $\mathfrak{f}\in \mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })$ and $\mathfrak{g}\in \mathfrak{hom}({\mathsf{\Sigma }}^{\prime },{\mathsf{\Sigma }}^{\prime \prime })$ the composition maps

$$\begin{array}{ccc}\begin{array}{cccc}{\mathfrak{f}}_{*}:& \mathfrak{hom}({\mathsf{\Sigma }}^{\prime },{\mathsf{\Sigma }}^{\prime \prime })& ⟶& \mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime \prime })\\ & \mathfrak{b}& \mapsto & \mathfrak{b}\circ \mathfrak{f}\end{array}& ,& \begin{array}{cccc}{\mathfrak{g}}^{*}:& \mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })& ⟶& \mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime \prime })\\ & \mathfrak{a}& \mapsto & \mathfrak{g}\circ \mathfrak{a}\end{array}\end{array}$$

(2)

are group homomorphisms.

On the other hand, system $0=(0,0,0)$ is the zero object in the category $S_{\mathbb{K}}$. Finite direct products of linear systems do exist in $S_{\mathbb{K}}$: just take $\mathsf{\Sigma }\times {\mathsf{\Sigma }}^{\prime }=(X\oplus X^{\prime },f\oplus f^{\prime },B\oplus B^{\prime })$. Hence, category $S_{\mathbb{K}}$ (and also $A_{\mathbb{K}})$ are additive categories.

In an additive category, it makes sense to search for kernels and cokernels of morphisms as well as natural (canonical) decompositions of morphisms.

Kernels and Cokernels

The next result is a classical characterization of kernels and cokernels.

Lemma 1.

Let $\mathfrak{a}:\mathsf{\Sigma }\to {\mathsf{\Sigma }}^{\prime }$ be a feedback morphism.

kernels.Morphism $\iota :K\to \mathsf{\Sigma }$ defines the kernel of $\mathfrak{a}$ if and only if, for every $\mathsf{\Gamma }\in Obj\left(S_{\mathbb{K}}\right)$, the following sequence of abelian groups is exact.

$$\begin{array}{ccccccc}0& \to & \mathfrak{hom}(\mathsf{\Gamma },K)& \stackrel{{\iota }_{*}}{\to }& \mathfrak{hom}(\mathsf{\Gamma },\mathsf{\Sigma })& \stackrel{{\mathfrak{a}}_{*}}{\to }& \mathfrak{hom}(\mathsf{\Gamma },{\mathsf{\Sigma }}^{\prime })\end{array}$$

(3)

kernel diagram.Equivalently, morphism $\iota :K\to \mathsf{\Sigma }$ defines the kernel of $\mathfrak{a}$ if there exists a system K and a feedback morphism ι making the top triangle commutative, and for each linear system Γ and each morphism j making the bottom triangle commutative, there exists a unique feedback morphism $\mathfrak{f}$ making the left triangle commutative.

cokernels.Morphism $\mathfrak{c}:{\mathsf{\Sigma }}^{\prime }\to C$ defines the cokernel of $\mathfrak{a}$ if and only if, for every $\mathsf{\Gamma }\in Obj\left(S_{\mathbb{K}}\right)$, the following sequence of abelian groups is exact.

$$\begin{array}{ccccccc}\mathfrak{hom}(C,\mathsf{\Gamma })& \stackrel{{\mathfrak{c}}^{*}}{\to }& \mathfrak{hom}({\mathsf{\Sigma }}^{\prime },\mathsf{\Gamma })& \stackrel{{\mathfrak{a}}^{*}}{\to }& \mathfrak{hom}(\mathsf{\Sigma },\mathsf{\Gamma })& \to & 0\end{array}$$

(4)

cokernel diagram.Equivalently, morphism $\mathfrak{c}:{\mathsf{\Sigma }}^{\prime }\to C$ defines the cokernel of $\mathfrak{a}$ if there exists a system C and a feedback morphism 𝔠 making the top triangle commutative, and for each linear system Γ and each morphism j making the bottom triangle commutative, there exists a unique feedback morphism $\mathfrak{f}$ making the right triangle commutative

In the following, the kernel (respectively, cokernel) of a feedback map $\mathfrak{a}$ in $S_{\mathbb{K}}$ is denoted by $\mathfrak{Ker}\left(\mathfrak{a}\right)$ (respectively, $\mathfrak{Coker}\left(\mathfrak{a}\right)$) while $ker\left(\mathfrak{a}\right)$ (respectively, $coker\left(\mathfrak{a}\right)$) are often used, by abuse of notation, for the kernel (respectively, cokernel) of linear map $F\left(\mathfrak{a}\right)$ in $\mathbb{K}$-Vect, where $F:S_{\mathbb{K}}\to \mathbb{K}-\mathrm{Vect}$ is the functor defined in the above Note 1.

Lemma 2.

Let $\mathfrak{a}:(V,f,B)\to (V^{\prime },f^{\prime },B^{\prime })$ be a feedback morphism. Assume that its kernel $\mathfrak{Ker}\mathfrak{a}$ does exist in $S_{\mathbb{K}}$. Then, $F\left(\mathfrak{Ker}\mathfrak{a}\right)$ injects into $kerF\left(\mathfrak{a}\right)$ (we often write $\mathfrak{Ker}\mathfrak{a}\subseteq ker\mathfrak{a}$).

On the other hand, $coker\left(F\right(\mathfrak{a}\left)\right)$ maps onto $F\left(\mathfrak{Coker}\mathfrak{a}\right)$ (we often write $coker\mathfrak{a}$ maps onto $\mathfrak{Coker}\mathfrak{a}$).

Proof. 

It is immediate because $kerF\left(\mathfrak{a}\right)$ is the kernel of $F\left(\mathfrak{a}\right)$ in $\mathbb{K}$-Vect and $F\left(\mathfrak{a}\right)\left(F\right(\mathfrak{Ker}\mathfrak{a}\left)\right)=0$. The dual argument works for cokernels. □

Corollary 1.

Let $\mathfrak{a}:(V,f,B)\to (V^{\prime },f^{\prime },B^{\prime })$ be a feedback morphism.

(i) If $\mathfrak{a}:V\to V^{\prime }$ is an injective map, then $\mathfrak{Ker}\mathfrak{a}=(0,0,0)$.

(ii) If $\mathfrak{a}:V\to V^{\prime }$ is a surjective map, then $\mathfrak{Coker}\mathfrak{a}=(0,0,0)$.

Theorem 1.

Category $S_{\mathbb{K}}$ has all cokernels.

Proof. 

Let $\mathfrak{a}:(V,f,B)\to (V^{\prime },f^{\prime },B^{\prime })$ be a feedback morphism. Assume $n=dimV$ and $n^{\prime }=dim{V}^{\prime }$. Set bases in V and $V^{\prime }$ so that the matrix of $\mathfrak{a}$ (as a linear map between vector spaces) is of the Hermite form. Obtain the block matrices associated with f and $f^{\prime }$, as well as the basis of subspaces B and $B^{\prime }$. Then, the above systems and morphism are transformed into:

$$\left({\mathbb{K}}^r\oplus {\mathbb{K}}^{n-r},\left(\begin{array}{cc}{F}_{11}\hfill & F_{12}\hfill \\ F_{21}\hfill & F_{22}\hfill \end{array}\right),\left(\begin{array}{c}{B}_1\\ B_2\end{array}\right)\right)\frac{\left(\begin{array}{cc}{\mathbf{1}}_r& 0\\ 0& 0\end{array}\right)}{\mathfrak{a}}\left({\mathbb{K}}^r\oplus {\mathbb{K}}^{n^{\prime }-r},\left(\begin{array}{cc}{F}_{11}^{\prime }& F_{12}^{\prime }\\ F_{21}^{\prime }& F_{22}^{\prime }\end{array}\right),\left(\begin{array}{c}{B}_1^{\prime }\hfill \\ B_2^{\prime }\hfill \end{array}\right)\right)$$

(5)

where $r=\mathrm{rank}\left(F\right(\mathfrak{a}\left)\right)$.

It is important to note here that, since $\mathfrak{a}$ is a feedback morphism, it follows that $\mathrm{Im}(\mathfrak{a}\circ f-f^{\prime }\circ \mathfrak{a})\subseteq B^{\prime }$. Hence, one has that the image of block $F_{21}^{\prime }$ is contained in the image of block $B_2^{\prime }$.

Consider the cokernel diagram:

A cokernel $\mathfrak{c}:(V^{\prime },f^{\prime },B^{\prime })\to (C,\gamma ,L)$ makes the upper triangle commutative and for all systems $(Y,g,D)$ and morphism $\mathfrak{d}$ making the lower triangle commutative, there exists a unique feedback morphism $\mathfrak{f}$ making the right triangle commutative.

Define the linear system $(C,\gamma ,L)=\left({\mathbb{K}}^{n^{\prime }-r},F_{22}^{\prime },B_2^{\prime }\right)$ and linear map $\mathfrak{c}=\left(\begin{array}{cc}0& {\mathbf{1}}_{n^{\prime }-r}\end{array}\right)$.

(i)

$\mathfrak{c}\left(B^{\prime }\right)=\left(\begin{array}{cc}0& {\mathbf{1}}_{n^{\prime }-r}\end{array}\right)\left(\begin{array}{c}{B}_1^{\prime }\\ B_2^{\prime }\end{array}\right)=B_2^{\prime }=L$

(ii)

$\mathfrak{c}\circ f^{\prime }-\gamma \circ c=\left(\begin{array}{cc}0& {\mathbf{1}}_{n^{\prime }-r}\end{array}\right)\left(\begin{array}{cc}{F}_{11}^{\prime }& F_{12}^{\prime }\\ F_{21}^{\prime }& F_{22}^{\prime }\end{array}\right)-F_{22}^{\prime }\left(\begin{array}{cc}0& {\mathbf{1}}_{n^{\prime }-r}\end{array}\right)=\left(\begin{array}{cc}{F}_{21}^{\prime }& 0\end{array}\right)\subseteq B_2^{\prime }=L$

Moreover, $\mathfrak{c}$ is a feedback morphism. On the other hand, it is clear that

(iii)

$\mathfrak{c}\circ \mathfrak{a}=0$

Now, set any linear system $({\mathbb{K}}^p,g,D)$ and note that:

(iv)

A morphism $\mathfrak{d}:({\mathbb{K}}^{n^{\prime }},f^{\prime },B^{\prime })\to ({\mathbb{K}}^p,g,D)$ must be on the block form $\left(\begin{array}{cc}0& d\end{array}\right)$ in order to assure $\mathfrak{d}\circ \mathfrak{a}=0$

(v)

$d{B}_2^{\prime }\subseteq D$ and $\left(\begin{array}{cc}d{F}_{21}^{\prime }& d{F}_{22}^{\prime }-gd\end{array}\right)\subseteq D$ because $\mathfrak{d}$ is a feedback morphism

We claim that $\mathfrak{f}=\left(d\right):({\mathbb{K}}^{n^{\prime }-r},F_{22}^{\prime },B_2^{\prime })\to ({\mathbb{K}}^p,g,L)$ is the unique feedback morphism making the right triangle commutative.

(vi)

$b{B}_2^{\prime }\subseteq D$ (by iv) yields $\mathfrak{f}\left(L\right)\subset D$

(vii)

$\mathrm{Im}(\mathfrak{f}\circ \gamma -g\circ \mathfrak{f})=⟨\left(\begin{array}{cc}0& d{F}_{22}^{\prime }-gd\end{array}\right)⟩\subseteq D$ by (v).

Hence, $\mathfrak{f}$ is a feedback morphism. It is quite clear that $\mathfrak{f}=\left(d\right)$ is the unique solution to equation $\mathfrak{f}\circ \mathfrak{c}=\mathfrak{d}$, which is, in matrix form, $\mathfrak{f}\left(\begin{array}{cc}0& {\mathbf{1}}_{n^{\prime }-r}\end{array}\right)=\left(\begin{array}{cc}0& d\end{array}\right)$ □

Next, some sufficient conditions for the existence of kernels in the category $S_{\mathbb{K}}$ of linear systems are given.

Theorem 2

(Sufficient condition of kernels). Let $\mathfrak{a}:(V,f,B)\to (V^{\prime },f^{\prime },B^{\prime })$ be a feedback morphism. If the kernel of $\mathfrak{a}$ as linear map $ker(\mathfrak{a}:V\to V^{\prime })$ is f-invariant (i.e., $f(ker\mathfrak{a})\subset ker\mathfrak{a}$), then the natural inclusion $\iota :ker\mathfrak{a}\hookrightarrow V$ defines a feedback morphism and

$$\mathfrak{Ker}(\mathfrak{a}:(V,f,B)\to (V^{\prime },f^{\prime },B^{\prime }))=(\iota :(ker\mathfrak{a},f{|}_{ker\mathfrak{a}},B\cap ker\mathfrak{a})\to (V,f,B))$$

(7)

Proof. 

Consider the kernel diagram Then, $\mathfrak{f}\left(y\right)=j\left(y\right)$ is a well defined feedback morphism and it is unique, making the left triangle commutative □

Next, we give an example of the above condition of existence of kernels in $S_{\mathbb{K}}$:

Example 4.

Consider the linear systems over ${\mathbb{K}}^1$ and ${\mathbb{K}}^2$ given by

$$\begin{array}{ccc}{\mathsf{\Sigma }}^{\left[1\right]}=(\mathbb{K},\left(0\right),⟨1⟩)& ,& {\mathsf{\Sigma }}^{\left[2\right]}=\left({\mathbb{K}}^2,\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),⟨\left(\begin{array}{c}1\\ 0\end{array}\right)⟩\right)\end{array}$$

(8)

The linear map ${\mathbb{K}}^2\underset{\left(\begin{array}{cc}1& 0\end{array}\right)}{\overset{a}{\to }}\mathbb{K}$ verifies the following properties:

$\mathfrak{a}\in \mathfrak{hom}\left({\mathsf{\Sigma }}^{\left[2\right]},{\mathsf{\Sigma }}^{\left[1\right]}\right)$ is a feedback morphism because:

(i) 

$\mathfrak{a}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(1\right)\subset ⟨1⟩$

(ii) 

$\mathrm{Im}\left(\mathfrak{a}\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)-\left(0\right)a\right)=\mathrm{Im}\left(\left(\begin{array}{cc}1& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)-\left(0\right)\left(\begin{array}{cc}1& 0\end{array}\right)\right)=\mathrm{Im}\left(\left(\begin{array}{cc}0& 0\end{array}\right)\right)=0$

Thus, $\mathfrak{a}\in \mathfrak{hom}\left({\mathsf{\Sigma }}^{\left[2\right]},{\mathsf{\Sigma }}^{\left[1\right]}\right)$ is a feedback morphism. $ker\left(\mathfrak{a}\right)=⟨\left(\begin{array}{c}0\\ 1\end{array}\right)⟩$ and $f(ker\mathfrak{a})=0\subseteq B$. Then, the kernel is

$$\mathfrak{Ker}\mathfrak{a}=\left((\mathbb{K},0,0),\iota =\left(\begin{array}{c}0\\ 1\end{array}\right):(\mathbb{K},0,0)\to {\mathsf{\Sigma }}^{\left[2\right]}\right)$$

(9)

Note 2.

The above sufficient condition to the existence of kernels in $S_{\mathbb{K}}$ is, in general, not necessary. The feedback map $(V,f,B)\stackrel{\mathfrak{a}}{\to }(V^{\prime },f^{\prime },B^{\prime })$ given by

$$\left({\mathbb{K}}^2,\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),⟨\left(\begin{array}{c}1\\ 0\end{array}\right)⟩\right)\underset{\mathfrak{a}}{\overset{\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)}{\to }}\left({\mathbb{K}}^2,\left(\begin{array}{cc}0& 0\\ 1& 1\end{array}\right),⟨\left(\begin{array}{c}0\\ 1\end{array}\right)⟩\right)$$

(10)

verifies:

(i) 

$\mathfrak{a}\in \mathfrak{hom}(\mathsf{\Sigma },{\mathsf{\Sigma }}^{\prime })$ (trivial).

(ii) 

$ker\mathfrak{a}={\displaystyle ⟨\left(\begin{array}{c}1\\ 0\end{array}\right)⟩}$ is not f-invariant because $f(ker\mathfrak{a})=⟨\left(\begin{array}{c}0\\ 1\end{array}\right)⟩\not\subset ker\mathfrak{a}$

(iii) 

However, the morphism has a kernel in $S_{\mathbb{K}}$. In fact, $\mathfrak{Ker}\mathfrak{a}=(0,0,0)$ in $S_{\mathbb{K}}$ because of any feedback morphism j in the below diagramneeds to be $j=0$.

Note also that $\mathfrak{Ker}\mathfrak{a}=(0,0,0)$, while $ker\mathfrak{a}\cong \mathbb{K}$. That is to say, $F\left(\mathfrak{Ker}\mathfrak{a}\right)$ is in general not isomorphic to $ker\left(F\right(\mathfrak{a}\left)\right)$.

4. Feedback Morphisms, Monomorphisms, and Epimorphisms

A morphism $\mathfrak{a}:\mathsf{\Sigma }\to {\mathsf{\Sigma }}^{\prime }$ is a monomorphism if for each object $\mathsf{\Gamma }$ in $S_{\mathbb{K}}$ and each $\mathfrak{b},\mathfrak{c}\in \mathfrak{hom}(\mathsf{\Gamma },\mathsf{\Sigma })$, one has:

$$\left(\mathfrak{a}\circ \mathfrak{b}=\mathfrak{a}\circ \mathfrak{c}\right)\Rightarrow \left(\mathfrak{b}=\mathfrak{c}\right)$$

(11)

A morphism $\mathfrak{a}:\mathsf{\Sigma }\to {\mathsf{\Sigma }}^{\prime }$ is an epimorphism if for each object $\mathsf{\Gamma }$ in $S_{\mathbb{K}}$ and each $\mathfrak{b},\mathfrak{c}\in \mathfrak{hom}({\mathsf{\Sigma }}^{\prime },\mathsf{\Gamma })$, one has:

$$\left(\mathfrak{b}\circ \mathfrak{a}=c\circ \mathfrak{a}\right)\Rightarrow \left(\mathfrak{b}=\mathfrak{c}\right)$$

(12)

Since $F:S_{\mathbb{K}}\to \mathbb{K}-\mathrm{Vect}$ is a faithful functor, then F reflects monomorphisms and epimorphisms ([7] 7.44, 7.46). That is to say:

(i)

If $F\left(\mathfrak{a}\right)$ is a monomorphism in $\mathbb{K}$-Vect (i.e., injective), then $\mathfrak{a}$ is a monomorphism in $S_{\mathbb{K}}$.

(ii)

If $F\left(\mathfrak{a}\right)$ is an epimorphism in $\mathbb{K}$-Vect (i.e., surjective), then $\mathfrak{a}$ is an epimorphism in $S_{\mathbb{K}}$.

Moreover, a stronger result holds in the case of epimorphisms.

Theorem 3.

Feedback morphism $\mathfrak{a}:\mathsf{\Sigma }=(V,f,B)\to {\mathsf{\Sigma }}^{\prime }=(V^{\prime },f^{\prime },B^{\prime })$ is an epimorphism in $S_{\mathbb{K}}$ if and only if $F\left(\mathfrak{a}\right):V\to V^{\prime }$ is surjective.

Proof. 

We claim that for every pair of linear maps $f,g:V^{\prime }\to W$ such that $f\circ F\left(\mathfrak{a}\right)=g\circ F\left(\mathfrak{a}\right)$, it follows that $f=g$.

However, this is clear from the fact that every arrow $f:V^{\prime }\to W$ is in fact $f=F\left(\mathfrak{f}\right)$ for the feedback morphism $\mathfrak{f}:(V^{\prime },f^{\prime },B^{\prime })\to (W,0,W)$, because functor $F:S_{\mathbb{K}}^{\mathrm{full}}\to \mathbb{K}-\mathrm{Vect}$ is dense. Therefore,

$$\left(f\circ F\left(\mathfrak{a}\right)=g\circ F\left(\mathfrak{a}\right)\right)\Rightarrow \left(F\left(\mathfrak{f}\right)\circ F\left(\mathfrak{a}\right)=F\left(\mathfrak{g}\right)\circ F\left(\mathfrak{a}\right)\right)\Rightarrow \left(F(\mathfrak{f}\circ \mathfrak{a})=F(\mathfrak{g}\circ \mathfrak{a})\right)\Rightarrow$$

$$\Rightarrow \left(\mathfrak{f}\circ \mathfrak{a}=\mathfrak{g}\circ \mathfrak{a}\right)\Rightarrow \left(\mathfrak{f}=g\right)\Rightarrow \left(f=g\right)$$

□

Note 3.

Not every monomorphism in $S_{\mathbb{K}}$ is injective: the above Note 2 gives an example of a feedback monomorphism $\mathfrak{a}$ that is not injective as linear map $F\left(\mathfrak{a}\right)$, but for every feedback morphism $\mathfrak{f}$ the condition $\mathfrak{a}\mathfrak{f}=0$ implies $\mathfrak{f}=0$. Thence, condition $\mathfrak{a}\circ \mathfrak{b}=\mathfrak{a}\circ \mathfrak{c}$ implies $\mathfrak{a}\circ (\mathfrak{b}-\mathfrak{c})=0$ and $\mathfrak{b}=\mathfrak{c}$. Therefore, a is a feedback monomorphism, while $F\left(\mathfrak{a}\right)$ is not monic.

Note 4.

Category $S_{\mathbb{K}}$ is additive but not balanced, that is to say, monomorphism plus epimorphism does not yield isomorphism. The following minimal example applies on every field $\mathbb{K}$:

$$\left(\mathbb{K},0,0\right)\underset{\mathfrak{f}}{\overset{\left(1\right)}{\to }}\left(\mathbb{K},0,\mathbb{K}\right)$$

(13)

The linear map $F\left(\mathfrak{f}\right)=F\left(1\right)={\mathbf{1}}_{\mathbb{K}}$ is the identity, thus a monomorphism and an epimorphism in $\mathbb{K}$-Vect. Hence, $\mathfrak{f}$ is a monomorphism and an epimorphism in $S_{\mathbb{K}}$. However, $\mathfrak{f}$ has no inverse in $S_{\mathbb{K}}$ because $\mathfrak{hom}\left(\left(\mathbb{K},0,\mathbb{K}\right),\left(\mathbb{K},0,0\right)\right)=\left\{0\right\}$. Therefore, $\mathfrak{f}$ is not an isomorphism.

Note 5.

A pointed category (a category with a zero object) is normal if every monomorphism is the kernel of some morphism. The category is said to be conormal if every epic is the cokernel of some morphism.

The example in Note 4 proves that $S_{\mathbb{K}}$ is neither normal nor conormal because $\mathfrak{f}$ is both monic and epic but neither a kernel nor a cokernel. To see this, just note that neither the identity morphism ${\mathbf{1}}_{\left(\mathbb{K},0,\mathbb{K}\right)}$ nor the identity ${\mathbf{1}}_{\left(\mathbb{K},0,0\right)}$ factors through $\mathfrak{f}$.

5. Sections, Retracts, and Feedback Decompositions

A morphism $\mathfrak{a}:\mathsf{\Sigma }\to {\mathsf{\Sigma }}^{\prime }$ is a retract (denoted by $\mathsf{\Sigma }\stackrel{\mathfrak{a}}{\twoheadrightarrow }{\mathsf{\Sigma }}^{\prime })$ if it has a right inverse, that is to say, if there exists a morphism ${\mathfrak{a}}^{\prime }:{\mathsf{\Sigma }}^{\prime }\to \mathsf{\Sigma }$ such that $\mathfrak{a}\circ {\mathfrak{a}}^{\prime }={\mathbf{1}}_{{\mathsf{\Sigma }}^{\prime }}$

A morphism $\mathfrak{a}:\mathsf{\Sigma }\to {\mathsf{\Sigma }}^{\prime }$ is a section (denoted by $\mathsf{\Sigma }\stackrel{\mathfrak{a}}{\hookrightarrow }{\mathsf{\Sigma }}^{\prime }$) if it has a left inverse, that is to say, if there exists a morphism ${\mathfrak{a}}^{\prime }:{\mathsf{\Sigma }}^{\prime }\to \mathsf{\Sigma }$ such that ${\mathfrak{a}}^{\prime }\circ \mathfrak{a}={\mathbf{1}}_{\mathsf{\Sigma }}$

Since every functor preserves sections and retracts ([7] 7.22, 7.28) the result follows.

Theorem 4.

If a feedback morphism $\mathfrak{a}$ is a section (respectively, retract) in $S_{\mathbb{K}}$, then $\mathfrak{a}$ is a section (respectively, retract) in $\mathbb{K}$-Vect.

Because one-side inverses of feedback morphisms need not be feedback morphisms, we have that the converse result of the above does not hold. That is to say, neither sections nor retracts are reflected by F.

Example 5.

A section in $\mathbb{K}$-Vect that is a feedback morphism but not a section in $S_{\mathbb{K}}$ is the following one

$$\left(\mathbb{K},0,0\right)\underset{a}{\overset{\left(\begin{array}{c}1\\ 0\end{array}\right)}{\to }}\left({\mathbb{K}}^2,\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\left(\begin{array}{c}0\\ 0\end{array}\right)\right)$$

Example 6.

An example of a retract in $\mathbb{K}$-Vect that is a feedback morphism but fails to be a retract in $S_{\mathbb{K}}$ is the following: consider the linear systems over ${\mathbb{K}}^1$ and ${\mathbb{K}}^2$ given by

$$\begin{array}{ccc}{\mathsf{\Sigma }}^{\left[1\right]}=(\mathbb{K},\left(0\right),⟨1⟩)& ,& {\mathsf{\Sigma }}^{\left[2\right]}=\left({\mathbb{K}}^2,\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),⟨\left(\begin{array}{c}1\\ 0\end{array}\right)⟩\right)\end{array}$$

(14)

The linear map ${\mathbb{K}}^2\underset{\left(\begin{array}{cc}1& 0\end{array}\right)}{\overset{a}{\to }}\mathbb{K}$ verifies the following properties:

$\mathfrak{a}\in \mathfrak{hom}\left({\mathsf{\Sigma }}^{\left[2\right]},{\mathsf{\Sigma }}^{\left[1\right]}\right)$is a feedback morphism because:

(i) 

$\mathfrak{a}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(1\right)\subset ⟨1⟩$

(ii) 

$\mathrm{Im}\left(\mathfrak{a}\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)-\left(0\right)\mathfrak{a}\right)=\mathrm{Im}\left(\left(\begin{array}{cc}1& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)-\left(0\right)\left(\begin{array}{cc}1& 0\end{array}\right)\right)=\mathrm{Im}\left(\left(\begin{array}{cc}0& 0\end{array}\right)\right)=0$

Moreover, $\mathfrak{a}\in hom({\mathbb{K}}^2,\mathbb{K})$ is an epimorphism in $S_{\mathbb{K}}$ because it is a feedback morphism and an epimorphism in $\mathbb{K}$-Vect, and in fact a retract in $\mathbb{K}$-Vect because

(iii) 

$\mathfrak{a}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(1\right):\mathbb{K}\to \mathbb{K}$

However, since

(iv) 

$\mathfrak{hom}\left({\mathsf{\Sigma }}^{\left[1\right]},{\mathsf{\Sigma }}^{\left[2\right]}\right)=\left\{0\right\}$ (see [1])

there is no right inverse of $\mathfrak{a}$ and hence $\mathfrak{a}\in \mathfrak{hom}({\mathsf{\Sigma }}^{\left[2\right]},{\mathsf{\Sigma }}^{\left[1\right]})$ is not a retract in the category of systems $S_{\mathbb{K}}$.

6. Split Exact Sequences of Linear Systems

Definition 1

(cf. [8]). A kernel–cokernel pair in $S_{\mathbb{K}}$ is a pair of feedback morphisms

$${\mathsf{\Sigma }}^{\prime }\stackrel{a}{\to }\mathsf{\Sigma }\stackrel{b}{\to }{\mathsf{\Sigma }}^{\prime \prime }$$

where $\mathfrak{a}$ is the kernel of $\mathfrak{b}$ and $\mathfrak{b}$ is the cokernel of $\mathfrak{a}$.

A kernel–cokernel pair where $\mathfrak{a}$ is a section and $\mathfrak{b}$ is a retract is called a split short exact sequence and denoted by

$${\mathsf{\Sigma }}^{\prime }\stackrel{\mathfrak{a}}{\hookrightarrow }\mathsf{\Sigma }\stackrel{\mathfrak{b}}{\twoheadrightarrow }{\mathsf{\Sigma }}^{\prime \prime }$$

Theorem 5

(Kalman’s Decomposition). Let $\mathsf{\Sigma }=(V,f,B)$ be a linear system in $S_{\mathbb{K}}$. Consider the sequence of vector subspaces of V given by $N_0=0$ and $N_i=B+f\left(N_{i-1}\right)$. Then, there exists an index $s\le dimV$ such that:

(i) 

$$N_0⊊N_1⊊\cdots ⊊N_{s-1}⊊N_s=N_{s+1}=\cdots$$

(ii) 

The following is a split short exact sequence of linear systems

$${\mathsf{\Sigma }}_{\mathrm{reach}}=\left(N_s{,f|}_{N_s},B\right)\stackrel{\iota }{\hookrightarrow }\mathsf{\Sigma }\stackrel{\pi }{\twoheadrightarrow }\left(V/N_s,\overline{f},0\right)={\mathsf{\Sigma }}_0$$

where ι is the standard inclusion, and π is quotient modulo subspace $N_s$.

Proof. 

Statement $\left(i\right)$ is given in Example 3.

In order to prove $\left(ii\right)$, note that $N_s$ is f-invariant, i.e., $f\left(N_s\right)\subseteq N_s$. Set a basis of $N_s$ and complete it to a basis of V. In this basis, system $\mathsf{\Sigma }$ is of the form

$$\mathsf{\Sigma }=\left(N_s\oplus W,\left(\begin{array}{cc}{f|}_{N_s}& f_{12}\\ 0& f_{22}\end{array}\right),B\right)$$

Then,

$$\left(N_s{,f|}_{N_s},B\right)\underset{\iota }{\overset{\left(\begin{array}{c}{\mathbf{1}}_{N_s}\\ 0\end{array}\right)}{\hookrightarrow }}\left(N_s\oplus W,\left(\begin{array}{cc}{f|}_{N_s}& f_{12}\\ 0& f_{22}\end{array}\right),B\right)\underset{\pi }{\overset{\left(\begin{array}{cc}0& {\mathbf{1}}_W\end{array}\right)}{\twoheadrightarrow }}\left(W,f_{22},0\right)$$

is a split short exact sequence of linear systems □

7. Conclusions

Some categorical properties of feedback morphisms of linear systems were introduced in order to study the additive category of linear systems. In particular, the classical Kalman’s decomposition theorem for linear systems over finite dimensional vector spaces was stated in these terms.

Some further developments of this approach should include the study of the subcategory $A_{\mathbb{K}}$ of reachable systems, where there exists a classification theorem and hence the objects of category $\mathrm{iso}\left(A_{\mathbb{K}}\right)$ are in bijective correspondence with the integer partitions of $dimV$ [2]. The objective would be setting up an exact structure on the additive category $S_{\mathbb{K}}$ such that Kalman’s decomposition is in fact an exact sequence. That exact structure should also reflect Brunovsky’s Theorem [3] as a factorization of reachable systems in the category $A_{\mathbb{K}}$.