Consider an uncontrolled linear dynamical system observed with an output $Y(t)$ in state space form: $$\left|\begin{array}{lll}\dot{X}=AX\\Y=CX\end{array}\right.$$ $X(t)$ is the state of the system in the state space (size $n\times1$), $Y(t)$ is the output vector (size $p\times 1$ for $p$ outputs), $A$ is the state matrix (size $n\times n)$, and $C$ is the output matrix (size $p\times n$).
We know that the unobservable subset of the state space is the null space of the observability matrix $$\mathfrak{O}=\begin{bmatrix}C\\CA\\\vdots\\CA^{n-1}\end{bmatrix}$$ And in particular, the system is fully observable iff $\text{rank}(\mathfrak{O})=n$.
Now, I'd like to derive this result in a simple way. I've seen a very simple derivation of the controlability matrix, and the author hinted that this approach can also work similarly to derive the observability matrix ; but I can't seem to work it out. Here it goes :
A state $X_0$ is unobservable iff the output $Y(t)$ is always $0$ when the system evolves freely from the said state $X_0$.
Now given an arbitrary initial state $X(0)=X_0$, by integrating the state equation $\dot{X}=AX$ we get that for $t\geq0$, $$X(t)=e^{At}X_0$$ And by the output equation $Y=CX$, the output is then $$Y=Ce^{At}X_0$$ Now, using the Cayley-Hamilton theorem, we can prove that the matrix $e^{At}$ can always be expressed as a finite polynomial of $A$ up to the power $n-1$, whose coefficient depends on $t$: $$e^{At}=\phi_0(t)I_n+\phi_1(t)A+\phi_2(t)A^2+..+\phi_{n-1}(t)A^{n-1}$$ I won't prove this result here because it's beyond my point, but it is not that complicated to show.
We can now use this expression of $e^{At}$ in the expression of $Y(t)$ above: \begin{align} Y(t)&=Ce^{At}X_0\nonumber\\ &=C\left(\phi_0(t)I_n+\phi_1(t)A+..+\phi_{n-1}(t)A^{n-1}\right)X_0\nonumber\\ &=C\cdot\phi_0(t)X_0+CA\cdot\phi_1(t)X_0+..+CA^{n-1}\cdot\phi_{n-1}(t)X_0\nonumber \end{align}
This last equality can be expressed as a matrix product: $$Y(t)=\begin{bmatrix}C&CA&CA^2&..&CA^{n-1}\end{bmatrix}\begin{bmatrix}\phi_0(t)X_0\\\phi_1(t)X_0\\\vdots\\\phi_{n-1}(t)X_0\end{bmatrix}$$ And the state $X_0$ is in the unobservable space iff $Y(t)=0$ for all $t\geq0$ ; and I feel the last equality would enable us to conclude that the unobervable space of the system is the null space of the matrix $\begin{bmatrix}C&CA&CA^2&..&CA^{n-1}\end{bmatrix}$, which would mean that this is the observability matrix.
But it's not: it's constructed horizontaly from the blocs of $A$ and $C$'s, not verticaly. I can't seem to figure any way of expressing the last equations in terms of the observability matrix in it's official, vertical form. The thing is, the observability matrix i'm trying to get to is a $np\times n$ matrix, so there is no way we can obtain $Y(t)$ (which is a $p\times1$ matrix) by multiplicating the observability matrix by some other matrix.
And I am utterly lost there. I have no idea where the vertical construction of the observability matrix comes from, and I don't understand why the horizontal construction I got to wouldn't work. Plus, this approach i'm taking is supposed to lead me to the real observability matrix.
Any idea ?
You would want to factor out as much as possible, so also $X_0$, and use the fact that each $\phi_i(t)$ is scalar and thus commutes. Therefore, it is also possible to write your expression for $Y(t)$ as
$$ Y(t) = \begin{bmatrix} \phi_0(t)\,I & \phi_1(t)\,I & \cdots & \phi_{n-1}(t)\,I \end{bmatrix} \begin{bmatrix} C \\ C\,A \\ \vdots \\ C\,A^{n-1} \end{bmatrix} X_0, $$
with $I$ the identity matrix of size $p \times p$.