Question 12.8: Determine the controllability of the system, described by ......

(sI − A) = s \left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] -\left [ \begin{matrix} 0 & -1 \\ 2 & 3 \end{matrix} \right ] =\left [ \begin{matrix} s & 1 \\ -2 & s-3 \end{matrix} \right ]

The characteristic polynomial of the system, given by the determinant of this matrix, is

s^{2}-3s+2=(s-1)(s-2)

The roots of this equation are the two eigenvalues \{\lambda_{1}=1,\lambda_{2}=2\}. For finding the eigenvectors, we use the equation

(\lambda I-A)v=0

For λ = 1, we get

\left [ \begin{matrix} 1 & 1 \\ -2 & 1-3 \end{matrix} \right ] \left [ \begin{matrix} v(0) \\ v(1) \end{matrix} \right ] =0

We get

\begin{array}{c}{{v(0)+v(1)=0}}\\ {{-2v(0)-2v(1)=0}}\end{array}

As these equations are dependent, we take one of them and give any nontrivial solution. For example, {v(0) = 1, v(1) = –1}. For λ = 2, we get

\left [ \begin{matrix} 2 & 1 \\ -2 & 2-3 \end{matrix} \right ] \left [ \begin{matrix} v(0) \\ v(1) \end{matrix} \right ] =0

Similarly, {v(0) = 1, v(1) = –2}. Therefore,

P=\left\lbrack\begin{array}{l l}\ \ \ {{1}}\ \ \ \ \ \ {{1}}\\ {{-1 -2}}\end{array}\right\rbrack

The first and second columns are, respectively, the eigenvectors corresponding to eigenvalues 1 and 2.

P ^{−1}AP =\left [ \begin{matrix} 2 & 1 \\ -1 & -1 \end{matrix} \right ] \left [ \begin{matrix} 0 & -1 \\ 2 & 3 \end{matrix} \right ] \left [ \begin{matrix} 1 & 1 \\ -1 & -2 \end{matrix} \right ] =Λ=\left [ \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right ]

The rank of the matrix

[B\mid A B]=\left [ \begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix} \right ]

is 2. Therefore, the system is controllable.

P^{-1}B=\left[\!\!\begin{array}{c}{{1\!\!}}\\ {{-1}}\end{array}\right]

has no zero entries, and, therefore, the system is controllable.