# Choose $\mathbf{B}$ such that eigenvalues are un/controllable

I have the state space system $\dot{x} = Ax + Bu$

with $A = \begin{bmatrix} 1 & -5 \\ -5 & 1\end{bmatrix}$.

I have to find a $B$ vector such that the system has $\lambda = 6$ as controllable eigenvalue and $\lambda = -4$ as uncontrollable eigenvalue.

I think I can just choose either $B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ or $B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, depending on where you would put the eigenvalues of $A$ on the diagonal. I think however that I am making a error here...

edit

I understand the question now some what better. But $\lambda_i$ is said to be a controllable eigenvalue if $\text{rank}\begin{bmatrix}A - \lambda_i I & B\end{bmatrix} = n$.

Thus I have to find a $B$ for which $\text{rank}\begin{bmatrix}A - \lambda_i I & B\end{bmatrix} \neq n$ for $\lambda_i = -4$ and $\text{rank}\begin{bmatrix}A - \lambda_i I & B\end{bmatrix} = n$ for $\lambda_i = 6$

• What is the meaning of controllable/uncontrollable? – Christopher A. Wong Jan 28 '13 at 20:34
• When you say you can choose either...Is $B$ a matrix or a vector? It looks like a matrix in your state space system. What is $u$, a fixed vector? – Alex R. Jan 28 '13 at 21:07
• @Alex $\mathbf{B}$ is indeed a vector. Sorry for the confusion. $u$ is a arbitrary input vector. You need not to know the input vector. – WG- Jan 28 '13 at 21:39
• @ChristopherA.Wong See def 2.1 goo.gl/dw6bi – WG- Jan 28 '13 at 21:40
• @WG-: I'm sorry, I still don't follow. If $\mathbf{B}$ is a vector, a column vector at that by your reasoning, then what is $\mathbf{B}u$? From your equation $\mathbf{B}u$ must be a vector. I think $\mathbf{B}$ is a $n\times n$ matrix where $n$ is the dimension of your system. – Alex R. Jan 28 '13 at 21:53

$\lambda_i$ is said to be a controllable eigenvalue if $\text{rank}\begin{bmatrix}A - \lambda_i I & B\end{bmatrix} = n$.

$\begin{bmatrix} A - \lambda_i & B\end{bmatrix} = \begin{bmatrix} 1-\lambda_i & -5 & b_1 \\ -5 & 1 - \lambda_i & b_2\end{bmatrix}$

For $\lambda_i = 6$

$\begin{bmatrix} A - \lambda_i & B\end{bmatrix} = \begin{bmatrix} -5 & -5 & b_1 \\ -5 & -5 & b_2\end{bmatrix} \sim \begin{bmatrix} -5 & -5 & b_1 \\ 0 & 0 & b_2 - b_1\end{bmatrix}$, this has full rank for $b_2 - b_1 \neq 0$.

For $\lambda_i = -4$

$\begin{bmatrix} A - \lambda_i & B\end{bmatrix} = \begin{bmatrix} 5 & -5 & b_1 \\ -5 & 5 & b_2\end{bmatrix} \sim \begin{bmatrix} 5 & -5 & b_1 \\ 0 & 0 & b_2 + b_1\end{bmatrix}$, this has not full rank for $b_2 + b_1 = 0$.

Hence take $b_1 = -1$ and $b_2 = 1$ thus $B = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$