Let $G(z)$ be the transfer function matrix with minimal realization given by Eqs. (11.14) and (11.15). If the system input vector process $d(k)$ is of the form $d(k)={\lambda }^{k}d(0)$ for each $k⩾0$ and the system initial conditions satisfy $x(0)={(\lambda I-A)}^{-1}Bd(0)$, in which λ is a constant scalar not equal to any eigenvalue of the matrix A, and $d(0)$ is a constant vector with compatible dimension. Then for an arbitrary integer $k⩾0$, the system output vector process $y(k)$ can be expressed as

$y(k)={\lambda }^{k}G(\lambda )d(0).$

When λ is not an eigenvalue of the matrix A, the matrix $\lambda I-A$ is invertible, which means that the condition $x(0)={(\lambda I-A)}^{-1}Bd(0)$ is well defined. Note that when a system has a minimal realization of Eqs. (11.14) and (11.15), the $\mathcal{Z}$-transformation of its output vector process $y(k)$ can be expressed as

$y(z)=C{(zI-A)}^{-1}[x(0)+Bd(z)]+Dd(z).$

Hence, when $d(k)={\lambda }^{k}d(0)$ for each $k⩾0$, we have that

$\begin{array}{ccc}\hfill y(z)& \hfill =\hfill & C{(zI-A)}^{-1}\{x(0)+B[{(z-\lambda )}^{-1}d(0)]\}+D[{(z-\lambda )}^{-1}d(0)]\hfill \\ \hfill & \hfill =\hfill & [C{(\lambda I-A)}^{-1}B+D][{(z-\lambda )}^{-1}d(0)]\hfill \\ \hfill & \hfill \hfill & +C{(zI-A)}^{-1}\{x(0)+B[{(z-\lambda )}^{-1}d(0)]\}\hfill \\ \hfill & \hfill \hfill & -C{(\lambda I-A)}^{-1}B[{(z-\lambda )}^{-1}d(0)]\hfill \\ \hfill & \hfill =\hfill & G(\lambda )[{(z-\lambda )}^{-1}d(0)]\hfill \\ \hfill & \hfill \hfill & +C{(zI-A)}^{-1}\{x(0)+[I-(zI-A){(\lambda I-A)}^{-1}]B[{(z-\lambda )}^{-1}d(0)]\}\hfill \\ \hfill & \hfill =\hfill & G(\lambda )[{(z-\lambda )}^{-1}d(0)]+C{(zI-A)}^{-1}\{x(0)-{(\lambda I-A)}^{-1}Bd(0)\}.\hfill \end{array}$

Therefore, the condition $x(0)={(\lambda I-A)}^{-1}Bd(0)$ leads to the following equality:

$y(z)=G(\lambda )[{(z-\lambda )}^{-1}d(0)]$

Recall that λ is a constant scalar. The proof can now be completed through taking the inverse $\mathcal{Z}$-transform of the vector-valued function $y(z)$. □

A vector $x(0)\in {\mathcal{R}}^n$ is called weakly unobservable if there exists an input vector sequence $d(k){|}_{k=0}^{\infty }$, such that the corresponding output vector sequence $y(k)$ is constantly equal to zero, that is, $y(k,x(0),d(j){|}_{j=0}^{k-1})=0$ for each $k=0,1,2,\cdots$.

When the initial state vector of the system described by Eqs. (11.14) and (11.15) is restricted to belong to its weakly unobservable subspace $\mathcal{V}$, its associated input–output relation is called its zero dynamics.

Concerning the system described by Eqs. (11.14) and (11.15), assume that the matrix $\left[\begin{array}{c}\hfill B\hfill \\ \hfill D\hfill \end{array}\right]$ is of full column rank. Define its system matrix

$P(z)=\left[\begin{array}{cc}\hfill zI-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right].$

Suppose that there exist a vector $x(0)\in {\mathcal{R}}^n$ and a vector $d(0)\in {\mathcal{R}}^q$ such that at least one of these two vectors is not equal to zero, and at a real value of the complex variable z, denoted λ, the following equality is satisfied:

$P(\lambda )\left[\begin{array}{c}\hfill x(0)\hfill \\ \hfill d(0)\hfill \end{array}\right]=0.$

Then for each $k=1,2,\cdots$, the input vector sequence $d(k)={\lambda }^{k}d(0)$ satisfies both $y(k,x(0),d(i){|}_{i=0}^k)=0$ and $d(k)\in {\mathcal{R}}^q$.

Note that the system defined by Eqs. (11.14) and (11.15) is well defined. This means that for a fixed initial state vector $x(0)$ and each input vector sequence $d(k){|}_{k=0}^{\infty }$, there are only one state vector sequence $x(k){|}_{k=0}^{\infty }$ and only one output vector sequence $y(k){|}_{k=0}^{\infty }$ that satisfy these two equations simultaneously.

When condition (11.23) is satisfied, define $x(k)$ as $x(k)={\lambda }^{k}x(0)$ for every $k=0,1,2,\cdots$. Then, we can straightforwardly prove that

$x(1)=\lambda x(0)=Ax(0)+Bd(0).$

Assume now that, for an integer i with $i⩾0$, Eq. (11.14) is satisfied by $x(i)={\lambda }^{i}x(0)$ and $d(i)={\lambda }^{i}d(0)$. Then, we can straightforwardly prove that

$\begin{array}{cc}\hfill x(i+2)& ={\lambda }^{i+2}x(0)=\lambda x(i+1)=\lambda [Ax(i)+Bd(i)]=A[\lambda x(i)]+B[\lambda d(i)]\hfill \\ \hfill & =Ax(i+1)+Bd(i+1),\hfill \end{array}$

that is, when $k=i+1$, the aforementioned state vector process and input vector process also satisfy Eq. (11.14). Therefore, $x(k)={\lambda }^{k}x(0)$ and $d(k)={\lambda }^{k}d(0)$ satisfy this equation for every $k⩾0$. Hence, $x(k)={\lambda }^{k}x(0)$ is the solution to this difference equation with its initial state vector $x(0)$ and input vector process $d(k)={\lambda }^{k}d(0)$.

On the other hand, note that, for each $k=0,1,2,\dots$,

$Cx(k)+Dd(k)=C({\lambda }^{k}x(0))+D({\lambda }^{k}d(0))={\lambda }^k[Cx(0)+Dd(0)]=0,$

provided that the vectors $x(0)$ and $d(0)$ satisfy conditions (11.23). Therefore, $y(0)=0$, and for each $k=1,2,\cdots$,

$y(k,x(0),d(i){|}_{i=0}^{k-1})=0.$

Now, assume that λ takes a complex value. Denote its real and imaginary parts respectively by ${\lambda }_r$ and ${\lambda }_j$. Recall that the matrices A, B, C, and D and the vectors $x(0)$ and $d(0)$ are real valued. Satisfaction of Eq. (11.23) implies that

$\left[\begin{array}{c}\hfill {\lambda }_{r}I-A\hfill \\ \hfill C\hfill \end{array}\right]x(0)+\left[\begin{array}{c}\hfill -B\hfill \\ \hfill D\hfill \end{array}\right]d(0)=0,{\lambda }_{j}x(0)=0.$

In addition, note that

$\left[\begin{array}{c}\hfill -B\hfill \\ \hfill D\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -I\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill I\hfill \end{array}\right]\left[\begin{array}{c}\hfill B\hfill \\ \hfill D\hfill \end{array}\right],$

which means that

$\mathrm{rank}\left(\left[\begin{array}{c}\hfill -B\hfill \\ \hfill D\hfill \end{array}\right]\right)=\mathrm{rank}\left(\left[\begin{array}{c}\hfill B\hfill \\ \hfill D\hfill \end{array}\right]\right).$

If ${\lambda }_j\ne 0$, then it is necessary that $x(0)=0$. As the matrix $\left[\begin{array}{c}\hfill B\hfill \\ \hfill D\hfill \end{array}\right]$ is assumed to be of full column rank, $x(0)=0$ and the first subequation in the above equation imply that the vector $d(0)$ is also equal to zero. This is a contradiction with the requirement that at least one of the vectors $x(0)$ and $d(0)$ is not equal to zero. Hence, ${\lambda }_j=0$, that is, λ takes a real value.

These arguments show that if there is λ satisfying Eq. (11.23), then it is necessary that this λ is real valued. Hence, ${\lambda }^{k}d(0)$ is real valued for each $k=0,1,2,\cdots$. This completes the proof. □

If for all permissible input vector sequences $u_1(k){|}_{k=0}^{\infty }$ and $u_2(k){|}_{k=0}^{\infty }$, the satisfaction by the system output vector sequence $y(k){|}_{k=0}^{\infty }$ of the condition

$y(k,0,u_1(j){|}_{j=0}^k)=y(k,0,u_2(j){|}_{j=0}^k)$

for all $k=0,1,2,\dots$ implies that

$u_1(k)=u_2(k)$

for all $k=0,1,2,\dots$, then this system is called left invertible.

If for every permissible initial state vector $x(0)$ and every feasible input vector sequence $u(k){|}_{k=0}^{\infty }$, the system output vector sequence $y(k){|}_{k=0}^{\infty }$ satisfies the equality

$y(k,0,u(j){|}_{j=0}^k)=0$

for all $k=0,1,2,\cdots$ only when

$x(0)=0,$

then this system is called strongly observable.

A system with its state space model described by Eqs. (11.14) and (11.15) is strongly observable, if and only if at every complex value of the complex variable z,

$\mathrm{rank}\left(\left[\begin{array}{cc}\hfill zI-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]\right)=n+\mathrm{rank}\left(\left[\begin{array}{c}\hfill B\hfill \\ \hfill D\hfill \end{array}\right]\right).$

Assume that the system described by Eqs. (11.14) and (11.15) is left invertible. Then there exists at least one nonzero initial state vector $x(0)$ such that there is an input vector sequence $d(k){|}_{k=0}^{\infty }$ that makes the system output vector $y(k,x(0),d(j){|}_{j=0}^k)$ constantly equal zero if and only if it has a transmission zero.

Assume that the system has a real transmission zero. Then by Lemma 11.2 there certainly exists a nonzero initial state vector satisfying the requirements.

Now, assume that all the transmission zeros of the system are complex. Let λ be any one of them. Then there exist vectors $x_{⁎}(0)$ and $d_{⁎}(0)$ such that at least one of them is not equal to zero and the following equality is satisfied:

$\left[\begin{array}{cc}\hfill \lambda I-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_{⁎}(0)\hfill \\ \hfill d_{⁎}(0)\hfill \end{array}\right]=0.$

Recall that all the system parameter matrices A, B, C, and D are real. Taking the conjugates of both sides of this equation, we obtain the following equality:

$\overline{\left[\begin{array}{cc}\hfill \lambda I-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_{⁎}(0)\hfill \\ \hfill d_{⁎}(0)\hfill \end{array}\right]}=\left[\begin{array}{cc}\hfill \overline{\lambda }I-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]\left[\begin{array}{c}\hfill {\overline{x}}_{⁎}(0)\hfill \\ \hfill {\overline{d}}_{⁎}(0)\hfill \end{array}\right]=0.$

As $\mathbf{col}\{x_{⁎}(0),d_{⁎}(0)\}$ does not equal zero, it is obvious that the vector $\mathbf{col}\{{\overline{x}}_{⁎}(0),{\overline{d}}_{⁎}(0)\}$ is not a zero vector either. These imply that ${\overline{\lambda }}_0$ is also a transmission zero of the system.

Define $d_{⁎}(k)$ as $d_{⁎}(k)={\lambda }^{k}d(0)$, $k=0,1,2,\cdots$. Let $d_{⁎}(z)$ and $d_{\diamond }(z)$ denote respectively the $\mathcal{Z}$-transformations of the sequences $d_{⁎}(k){|}_{k=0}^{\infty })$ and ${\overline{d}}_{⁎}(k){|}_{k=0}^{\infty })$. Moreover, let $y_{⁎}(z)$ and $y_{\diamond }(z)$ denote respectively the $\mathcal{Z}$-transformations of the output vector sequences $y(k,x_{⁎}(0),d_{⁎}(j){|}_{j=0}^k)$ and $y(k,{\overline{x}}_{⁎}(0),{\overline{d}}_{⁎}(j){|}_{j=0}^k)$. According to the proof of Lemma 11.2, we have that

$y(k,x_{⁎}(0),d_{⁎}(j){|}_{j=0}^{k-1})=0,y(k,{\overline{x}}_{⁎}(0),{\overline{d}}_{⁎}(j){|}_{j=0}^{k-1})=0\text{for all}k=0,1,2,\cdots .$

Define $x(0)$ and $d(k)$ further as

$x(0)=x_{⁎}(0)+{\overline{x}}_{⁎}(0),d(k)=d_{⁎}(k)+{\overline{d}}_{⁎}(k)\text{for all}k=0,1,2,\cdots .$

Then, both $x(0)$ and $d(k)$ with $k\in \{0,1,2,\cdots \}$ are real valued and therefore can be realized in principle.

Let $y(z)$ denote the $\mathcal{Z}$-transformation of the output vector sequence associated with the system initial state vector $x(0)$ and the input vector sequence $d(k){|}_{k=0}^{\infty }$. Then by Eq. (11.19) we have that

$\begin{array}{ccc}\hfill y(z)& \hfill =\hfill & C{(zI-A)}^{-1}[x(0)+Bd(z)]+Dd(z)\hfill \\ \hfill & \hfill =\hfill & C{(zI-A)}^{-1}\{[x_{⁎}(0)+{\overline{x}}_{⁎}(0)]+B[d_{⁎}(z)+d_{\diamond }(z)]\}+D[d_{⁎}(z)+d_{\diamond }(z)]\hfill \\ \hfill & \hfill =\hfill & \{C{(zI-A)}^{-1}[x_{⁎}(0)+B{d}_{⁎}(z)]+D{d}_{⁎}(z)\}\hfill \\ \hfill & \hfill \hfill & +\{C{(zI-A)}^{-1}[x_{⁎}(0)+B{d}_{\diamond }(z)]+D{d}_{\diamond }(z)\}\hfill \\ \hfill & \hfill =\hfill & y_{⁎}(z)+y_{\diamond }(z).\hfill \end{array}$

We can therefore declare from Eq. (11.26) that, for each $k=0,1,2,\cdots$,

$y(k,x(0),d(j){|}_{j=0}^k)=0.$

On the contrary, assume that the system described by Eqs. (11.14) and (11.15) does not have a transmission zero but there is a nonzero initial state vector $x(0)$ such that there exists an input sequence $d(k){|}_{k=0}^{\infty }$ that makes $y(k,x(0),d(j){|}_{j=0}^k)=0$ for every $k=0,1,2,\cdots$. Then, for arbitrary $\lambda \in \mathcal{C}$, the matrix

$\left[\begin{array}{cc}\hfill \lambda I-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]$

is of full column rank. Hence

$\mathrm{rank}\left(\left[\begin{array}{cc}\hfill \lambda I-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]\right)=n+q=n+\mathrm{rank}\left(\left[\begin{array}{c}\hfill B\hfill \\ \hfill D\hfill \end{array}\right]\right)$

By Lemma 11.3 this system is strongly observable. Hence, if there is an input sequence $d(k){|}_{k=0}^{\infty }$ that makes $y(k,x(0),d(j){|}_{j=0}^k)=0$ for every $k=0,1,2,\cdots$, then $x(0)=0$. This contradicts the assumption $x(0)\ne 0$. Therefore, the system must have some transmission zeros.

This completes the proof. □

If an LTI system is not left invertible, then it is certainly not attack detectable/identifiable.

Assume that a transfer function matrix $G(z)$ is not of FNCR. Then there exists a nonzero vector α such that, for each $\lambda \in \mathcal{C}$, the following equality is satisfied:

$G(z)\alpha =0.$

Let $(A,B,C,D)$ be a real matrix quadruplex that satisfies $C{(zI-A)}^{-1}B+D=G(z)$ and associates with a state space model having a minimal realization. Take an arbitrary real scalar λ that is not an eigenvalue of the matrix A and denote the associated real and imaginary parts of the vector α by ${\alpha }_r$ and ${\alpha }_j$, respectively. Then, the Eq. (11.29) means that the following two equalities are satisfied:

$[C{(\lambda I-A)}^{-1}B+D]{\alpha }_r=0,[C{(\lambda I-A)}^{-1}B+D]{\alpha }_j=0.$

Note that $\alpha \ne 0$ is equivalent to $\left[\begin{array}{c}\hfill {\alpha }_r\hfill \\ \hfill {\alpha }_j\hfill \end{array}\right]\ne 0$, that is, the assumption $\alpha \ne 0$ implies that at least one of ${\alpha }_r$ and ${\alpha }_j$ is not equal to zero.

Assume that ${\alpha }_r\ne 0$. Then, the first equality of Eq. (11.30) can be equivalently rewritten as

$\left[\begin{array}{cc}\hfill \lambda I-A\hfill & \hfill -B\hfill \\ \hfill C\hfill & \hfill D\hfill \end{array}\right]\left[\begin{array}{c}\hfill {(\lambda I-A)}^{-1}B{\alpha }_r\hfill \\ \hfill {\alpha }_r\hfill \end{array}\right]=0.$

Note that both vectors ${(\lambda I-A)}^{-1}B{\alpha }_r$ and ${\alpha }_r$ are real. Moreover, the vector $\left[\begin{array}{c}\hfill {(\lambda I-A)}^{-1}B{\alpha }_r\hfill \\ \hfill {\alpha }_r\hfill \end{array}\right]$ is not a zero vector. By Lemma 11.2 there always exist a nonzero $x(0)\in {\mathcal{R}}^n$ and an input vector sequence $d(k)\in {\mathcal{R}}^q$, not constantly equal to zero, such that the output vector process of the LTI system, which can be expressed as $y(k,x(0),d(i){|}_{i=0}^k)$, is constantly equal to zero for each $k=0,1,\cdots$. Hence the system is not attack detectable/identifiable.

When ${\alpha }_j\ne 0$, the same arguments show that the system is not attack detectable/identifiable.

This completes proof. □