[proofplan] The numerator has degree at most $n-1$ while the denominator is monic of degree $n$, so strict properness follows from the degree comparison. To compute the transfer function of the proposed realization, solve the linear system obtained from the companion matrix resolvent. The companion structure forces the solution vector to be a scalar multiple of $(1,s,\dots,s^{n-1})^\top$, and the last row identifies that scalar as the reciprocal of the denominator polynomial. Multiplying this vector by $C_c$ recovers exactly the prescribed numerator divided by the prescribed denominator. [/proofplan]

[step:Verify strict properness from the polynomial degrees] Let $p: \mathbb{C} \to \mathbb{C}$ denote the denominator polynomial \begin{align*} p(s)=s^n+a_{n-1}s^{n-1}+\cdots+a_1s+a_0. \end{align*} Let $q: \mathbb{C} \to \mathbb{C}$ denote the numerator polynomial \begin{align*} q(s)=b_{n-1}s^{n-1}+\cdots+b_1s+b_0. \end{align*} The polynomial $p$ is monic of degree $n$, and $q$ has degree at most $n-1$, with the convention that the zero polynomial has degree less than every nonzero polynomial degree for this purpose. Hence $\deg q < \deg p$, so the rational function $G=q/p$ is strictly proper, meaning that the degree of its numerator is strictly smaller than the degree of its denominator. [/step]

[step:Solve the companion resolvent equation]Fix $s \in \mathbb{C}$ with $p(s)\neq 0$. Since $(A_c,B_c)$ is the controllable companion pair, the relevant entries are $(A_c)_{i,i+1}=1$ for $1 \le i \le n-1$, $(A_c)_{n,j}=-a_{j-1}$ for $1 \le j \le n$, all remaining entries of $A_c$ are zero, and $B_c=(0,\dots,0,1)^\top$. First define the [linear map](/page/Linear%20Map) $L_s: \mathbb{C}^n \to \mathbb{C}^n$ by sending $z \in \mathbb{C}^n$ to $(sI_n-A_c)z$. We show that $L_s$ is invertible. Let $w=(w_1,\dots,w_n)^\top \in \mathbb{C}^n$ satisfy $L_s w=0$. The first $n-1$ rows give $sw_i-w_{i+1}=0$ for $1 \le i \le n-1$, so induction gives $w_j=s^{j-1}w_1$ for $1 \le j \le n$. The last row gives $a_0w_1+a_1w_2+\cdots+a_{n-1}w_n+sw_n=0$. Substituting $w_j=s^{j-1}w_1$ yields \begin{align*} p(s)w_1=0. \end{align*} Since $p(s)\neq 0$, we have $w_1=0$, hence $w=0$. Therefore $\ker L_s=\{0\}$. Since $L_s$ is a linear map from the finite-dimensional [vector space](/page/Vector%20Space) $\mathbb{C}^n$ to itself, the [Injective iff Surjective in Finite Dimensions](/theorems/386) theorem implies that $L_s$ is invertible. Define $v \in \mathbb{C}^n$ by $v=(sI_n-A_c)^{-1}B_c$. Equivalently, $v$ is the unique solution of $(sI_n-A_c)v=B_c$. Writing $v=(v_1,\dots,v_n)^\top$, the first $n-1$ rows of this system are $sv_i-v_{i+1}=0$ for $1 \le i \le n-1$. Therefore $v_{i+1}=sv_i$ for $1 \le i \le n-1$, and induction gives $v_j=s^{j-1}v_1$ for $1 \le j \le n$. The last row of $(sI_n-A_c)v=B_c$ is $a_0v_1+a_1v_2+\cdots+a_{n-1}v_n+sv_n=1$. Substituting $v_j=s^{j-1}v_1$ gives \begin{align*} p(s)v_1=1. \end{align*} Since $p(s)\neq 0$, this yields \begin{align*} v_1=\frac{1}{p(s)}. \end{align*} Consequently \begin{align*} (sI_n-A_c)^{-1}B_c=\frac{1}{p(s)}(1,s,\dots,s^{n-1})^\top. \end{align*}[/step]

[guided]Fix $s \in \mathbb{C}$ with $p(s)\neq 0$. Before applying the resolvent $(sI_n-A_c)^{-1}$, we must prove that this inverse exists. Define the linear map $L_s: \mathbb{C}^n \to \mathbb{C}^n$ by sending $z \in \mathbb{C}^n$ to $(sI_n-A_c)z$. We prove that $L_s$ is injective. Let $w=(w_1,\dots,w_n)^\top \in \mathbb{C}^n$ satisfy $L_s w=0$. Since the only nonzero entry in row $i$ of $A_c$ for $1 \le i \le n-1$ is $(A_c)_{i,i+1}=1$, the first $n-1$ rows give $sw_i-w_{i+1}=0$ for $1 \le i \le n-1$. Thus $w_{i+1}=sw_i$, and induction gives $w_j=s^{j-1}w_1$ for $1 \le j \le n$. The last row of $L_s w=0$ is $a_0w_1+a_1w_2+\cdots+a_{n-1}w_n+sw_n=0$. Substituting $w_j=s^{j-1}w_1$ gives $(a_0+a_1s+\cdots+a_{n-1}s^{n-1}+s^n)w_1=0$. The scalar in parentheses is $p(s)$, so $p(s)w_1=0$. Because $p(s)\neq 0$, we get $w_1=0$, and then $w_j=s^{j-1}w_1$ gives $w_j=0$ for every $1 \le j \le n$. Hence $\ker L_s=\{0\}$. Since $L_s$ maps the finite-dimensional vector space $\mathbb{C}^n$ to itself, the [Injective iff Surjective in Finite Dimensions](/theorems/386) theorem implies that $L_s$ is surjective and hence bijective. Therefore $(sI_n-A_c)^{-1}$ exists. Now define $v \in \mathbb{C}^n$ by $v=(sI_n-A_c)^{-1}B_c$. Equivalently, $v$ is the unique solution of $(sI_n-A_c)v=B_c$. Write $v=(v_1,\dots,v_n)^\top$. The reason the companion form is useful is that the first $n-1$ equations contain no denominator coefficients; they only propagate powers of $s$. Indeed, since the only nonzero entry in row $i$ of $A_c$ for $1 \le i \le n-1$ is $(A_c)_{i,i+1}=1$, the $i$-th row of the equation is $sv_i-v_{i+1}=0$. Thus $v_{i+1}=sv_i$ for $1 \le i \le n-1$. Starting with $v_1$, this recursion gives $v_2=sv_1$. Applying the same recursion repeatedly gives $v_j=s^{j-1}v_1$ for $1 \le j \le n$. It remains to determine the scalar $v_1$. The last row is where the denominator coefficients enter. Since $(A_c)_{n,j}=-a_{j-1}$ for $1 \le j \le n$, and since the last component of $B_c$ is $1$, the last row of $(sI_n-A_c)v=B_c$ is $a_0v_1+a_1v_2+\cdots+a_{n-1}v_n+sv_n=1$. Substituting $v_j=s^{j-1}v_1$ into this equation yields $(a_0+a_1s+\cdots+a_{n-1}s^{n-1}+s^n)v_1=1$. The scalar multiplying $v_1$ is exactly $p(s)$, so \begin{align*} p(s)v_1=1. \end{align*} Because we fixed $s$ with $p(s)\neq 0$, division by $p(s)$ is valid, and therefore \begin{align*} v_1=\frac{1}{p(s)}. \end{align*} Combining this value with $v_j=s^{j-1}v_1$ gives \begin{align*} (sI_n-A_c)^{-1}B_c=\frac{1}{p(s)}(1,s,\dots,s^{n-1})^\top. \end{align*}[/guided]

[step:Multiply by the output map to recover the transfer function] For the state-space realization \begin{align*} \dot{x}=A_cx+B_cu, \qquad y=C_cx, \end{align*} by the transfer function definition for a state-space realization, its transfer function is the scalar rational function defined on the resolvent set of $A_c$ by $s \mapsto C_c(sI_n-A_c)^{-1}B_c$; there is no additional direct feedthrough term because the output equation contains no term proportional to $u$. Using the formula for the resolvent vector, \begin{align*} C_c(sI_n-A_c)^{-1}B_c=C_c\left(\frac{1}{p(s)}(1,s,\dots,s^{n-1})^\top\right). \end{align*} By the definition of the row map $C_c$, \begin{align*} C_c(sI_n-A_c)^{-1}B_c=\frac{1}{p(s)}\sum_{j=1}^{n} b_{j-1}s^{j-1}. \end{align*} Equivalently, \begin{align*} C_c(sI_n-A_c)^{-1}B_c=\frac{b_{n-1}s^{n-1}+\cdots+b_1s+b_0}{p(s)}. \end{align*} This is precisely $G(s)$ for every $s \in \mathbb{C}$ with $p(s)\neq 0$. Hence the system $\dot{x}=A_cx+B_cu$, $y=C_cx$ realizes the strictly proper transfer function $G$. [/step]