# Given a system $\dot{x}(t)=Ax(t) + Bu(t)$ find $S$ such that $s^{-1}AS=\bar{A}$, $S^{-1}B=B$ is in controller form.

I am given the system $\dot{x}(t)=Ax(t) + Bu(t)$ where $$A = \left( \begin{matrix} -1 & 0& 2\\ 0 & -3 & 0 \\ 1&0&0 \end{matrix} \right), \quad B = \left( \begin{matrix} 1\\1\\0 \end{matrix} \right)$$ Now this system is not stable as $A$ has eigenvalues $-3,-2$ and $1$, the characteristic polynomial of $A$ is $\lambda ^3 + 4 \lambda^2 +\lambda -6$. It is stabilizable though. Now I need to put it into controller form, that is find an invertible $S$ such that $\bar{A} =S^{-1}AS$ and $\bar{B} = S^{-1}B$ where $$\bar{A} = \left( \begin{matrix} 0 & 1& 0\\ 0 & 0 & 1 \\ 6&-1&-4 \end{matrix} \right),\quad \bar{B} = \left( \begin{matrix} 0\\0\\1 \end{matrix} \right)$$ Where the third row of $\bar{A}$ is given by the coefficients of the characteristic polynomial of $A$.

Now I have tried for hours now to find $S$ using the algorithm I am given but I can't get the right matrix $S$. I keep checking $AS = S\bar{A}$ but every time this doesn't work out. If anyone can show me their workings and method of how to find $S$ I would be very happy and thankful! Thank you.

Also, it is possible of course I made a mistake somewhere along the road, if so please point it out and I will see if I can finally figure this problem out. Thanks :)

• – copper.hat Dec 9 '13 at 17:52
• Hey thanks for your comment! That is actually my question :) No matter how I try using the method described there I cant seem to get the right answer for $S$. So im looking for someone to give a brief outline of their workings – Slugger Dec 9 '13 at 18:13
• Let me check, I may have made a mistake. – copper.hat Dec 9 '13 at 19:16
• Nope, seems OK. Check your calculations. – copper.hat Dec 9 '13 at 20:09
• As an aside, the matrix is real diagonalizable, so stabilizing via the diagonal route seems more straightforward... – copper.hat Dec 9 '13 at 20:12

Following the procedure in https://math.stackexchange.com/a/596574/27978:

Let $P = \begin{bmatrix} b & A b &A^2 b\end{bmatrix} = \begin{bmatrix} 1 & -1 & 3 \\ 1 & -3 & 9 \\ 0 & 1 & -1 \end{bmatrix}$. Let $e_3 = (0,0, 1)^T$.

Form $W = \begin{bmatrix} P^{-T} e_3 & A^TP^{-T} e_3 & (A^2)^TP^{-T} e_3 \end{bmatrix} = {1 \over 4} \begin{bmatrix} -1 & 3 & -5 \\ 1 & -3 & 9 \\ 2 & -2 & 6 \end{bmatrix}$.

Then $W^T A W^{-T} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -1 & -4 \end{bmatrix}$, and $W^T b = e_3$, as required (with $S= W^{-T}$).

Just to be sure:

a = [-1 0 2
0 -3 0
1 0 0]
b = [1 1 0]'

e3 = [0 0 1]'
p = [b a*b a*a*b]
pi = inv(p)
wt = [e3'*pi ; e3'*pi*a ; e3'*pi*a*a]
w = wt'


This gives

octave:8> w'*a*inv(w)'
ans =

0   1   0
0   0   1
6  -1  -4

octave:9> w'*b
ans =

0
0
1