Pole placement in uncontrollable systems with state-feedback. I am somewhat confused on how the standard form for an uncontrollable system can be used to implement state-feedback on the controllable subspace.


*

*I understand how to use feedback to achieve pole-placement in completely controllable systems.

*I don't understand how this can then be applied to controllable subspaces of uncontrollable systems.
For example, suppose I have an LTI system with eigenvalues {-1, ,1, 2} where eigenvalues {1, 2} are part of the controllable submatrix while -1 is uncontrollable. How do change the pole-locations for the controllable eigenvalues?  
Thank you. 
 A: Since you only mentioned state feedback I assume that you are dealing with a system of the following form
$$
\dot{x} = A\,x + B\,u
$$
with $x \in \mathbb{R}^n$, $u \in \mathbb{R}^m$ and you want to find a state feedback $u = -K\,x$ such that the dynamics of the controllable parts of the system can be chosen. The entire state $x$ is considered to be known and thus observability does not have to be taken into consideration.
By looking at the controllable and uncontrollable states it is possible to find a similarity transformation which decomposes these states, similar to the Kalman decomposition but without also considering the observability. In order to do this you need the controllability matrix $\mathcal{C}$ defined as
$$
\mathcal{C} = \begin{bmatrix}
B & A\,B & \cdots & A^{n-1}B
\end{bmatrix}.
$$
A controllable decomposition can be found using the similarity transformation $T = \begin{bmatrix}T_1 & T_2\end{bmatrix}$, with the columns of $T_1$ having the same span as the controllability matrix and $T_2$ chosen such that $T$ is invertible. In the new coordinates, denoted by $x = T\,\hat{x}$, the dynamics are as follows
$$
\dot{\hat{x}} = \underbrace{T^{-1} A\,T}_{\hat{A}}\,\hat{x} + \underbrace{T^{-1} B}_{\hat{B}}\,u
$$
such that
$$
\hat{A} = 
\begin{bmatrix}
A_{11} & A_{12} \\ 0 & A_{22}
\end{bmatrix}, \quad
\hat{B} = 
\begin{bmatrix}
B_1 \\ 0
\end{bmatrix}
$$
where $A_{11}$ has the same size as the rank of $\mathcal{C}$ and the pair $(A_{11}, B_1)$ is controllable (their associated controllability matrix is full rank). So for this pair you can do pole placement, such that the poles of $A_{11}-B_1\,K_1$ can be chosen arbitrarily. Therefore, $u = -\begin{bmatrix}K_1 & 0\end{bmatrix} \hat{x}$ will place the controllable poles of $\hat{A}$ arbitrarily as well.
By using the definition of $\hat{x}$ it is possible to express this state feedback in terms of $x$ instead, namely $\hat{x} = T^{-1} x$. Or in other words your feedback gain can be found with $K = \begin{bmatrix}K_1 & 0\end{bmatrix}\,T^{-1}$.

For example given the system
$$
A = 
\begin{bmatrix}
4 & 0 & 0 \\
-4 & 0 & -2 \\
-2 & -2 & 0
\end{bmatrix}, \quad
B = 
\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix},
$$
which has the eigenvalues $2$, $-2$ and $4$. This pair has the following controllability matrix
$$
\mathcal{C} = 
\begin{bmatrix}
1 & 4 & 16 \\
0 & -2 & -12 \\
-1 & -2 & -4
\end{bmatrix},
$$
with has rank 2. It can be shown that its first two columns are linearly independent of each other, so an option for the similarity transformation would be
$$
T = 
\begin{bmatrix}
1 & 2 & 0 \\
0 & -1 & 0 \\
-1 & -1 & 1
\end{bmatrix}.
$$
For the added last column it can be shown that it makes $T$ invertible (full rank). Performing the similarity transformation gives
$$
\hat{A} = 
\begin{bmatrix}
0 & -4 & -4 \\
2 & 6 & 2 \\
0 & 0 & -2
\end{bmatrix}, \quad
\hat{B} = 
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix},
$$
such that
$$
A_{11} = 
\begin{bmatrix}
0 & -4 \\
2 & 6
\end{bmatrix}, \quad
B_1 = 
\begin{bmatrix}
1 \\
0
\end{bmatrix}.
$$
Finding a feedback gain for this pair, which places the poles at $-1$ and $-2$ gives $K_1 = \begin{bmatrix}9 & 24\end{bmatrix}$. So the actual state feedback gain can be calculated with
$$
K = \begin{bmatrix}K_1 & 0\end{bmatrix}\,T^{-1} = 
\begin{bmatrix}9 & -6 & 0\end{bmatrix},
$$
which also places the two controllable poles of $A$, which are initially located at $2$ and $4$, at $-1$ and $-2$ respectively.
