# How to prove the observability condition for the following state sytem?

A linear state system is given with the following matrices $\boldsymbol{A}$ and $\boldsymbol{C}$. $$\boldsymbol{A} = \begin{bmatrix} μ & 1 & 0 & \dots & 0 \\ 0 & μ & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 &\ldots & 1 \\ 0 & 0 & 0 &\ldots & μ \end{bmatrix}$$$$\boldsymbol{C} = \begin{bmatrix} c_1 & c_2 & \dots & c_n \\ \end{bmatrix}$$ And the observability matrix $\boldsymbol{\mathcal{O}}$ is equal to

\begin{bmatrix} C\\ CA\\ CA^2\\ \vdots\\ CA^{n-1} \end{bmatrix} How can we show that the rank of the observability matrix is $n$ (the state system is observable) if and only if $c_1$ is nonzero? Remark: empirically, I found that the determinant of $\boldsymbol{\mathcal{O}}$ is equal to ${c_1}^n$, however, I need analytical proof.

• I edited your $\boldsymbol{A}$ matrix because it was not square. Please check if it is correct now. – MrYouMath Mar 12 '18 at 14:15
• Thanks, A needs to be square. – Saeed Mar 12 '18 at 17:07

Checking the rank of the observability matrix $\mathcal{O}$ is equivalent to the Hautus test, which for observability can be formulated as

$$\text{rank} \begin{bmatrix} C \\ A - \lambda\,I \end{bmatrix}=n \quad \forall\,\lambda \in \mathbb{C}$$

with $A\in\mathbb{R}^{n \times n}$.

In this case it is easier to proof observability using the Hautus test. Since the given $A$ is just one Jordan block with eigenvalues $\mu$, then the matrix, whose rank it tested, might only lose rank when $\lambda = \mu$. Performing this substitution gives

$$\begin{bmatrix} C \\ A - \mu\,I \end{bmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 &\cdots & c_n \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & 0 \\ 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & \cdots & 0 & 0 \end{bmatrix}$$

From here it is straightforward to show that the rank of that matrix is only equal to $n$ if $c_1 \neq 0$.

• Thanks a lot Kwin – Saeed Mar 14 '18 at 8:42

Kalman's observability criterion states that the observability matrix $\boldsymbol{\mathcal{O}}$ has to have full rank in order to guarantee observability for the linear time-invariant system. Your observability matrix $\boldsymbol{\mathcal{O}}$ is a square matrix. And a square matrix has a full rank if it is invertible, which is equivalent to the statement that the determinant is not equal to zero.

You have found out that the determinant is equal to (in order to prove this you can use mathematical induction)

$$\det \boldsymbol{\mathcal{O}} = c_1^n.$$

And $c_1\neq 0$ guarantees that it is not equal to $0$. Hence, $c_1\neq 0$ guarantees observability.

Note, that for practical purposes a very small value of $c_1$ is not desirable as the determinant comes closer to the value of zero. Which might mean that practically your system is unobservable.

• Thanks, however, I am still looking for analytical proof for the determinant of \boldsymbol ${\mathcal{O}}$ (or equivalently how to prove that it is nonsingular if ${c_1\neq 0}$ – Saeed Mar 12 '18 at 17:22
• Thanks. I tried induction and I failed to reach the desired result though. – Saeed Mar 13 '18 at 7:16