This question is from a System Theory test without answers or solutions:

Consider the system $\dot{x}=Ax$, where

$A=\begin{bmatrix}-3&1&2&-1&1&0\\2&-2&0&2&-2&0\\0&0&0&0&0&0\\1&0&-1&0&-1&0\\-2&1&1&-1&0&0\\0&-1&1&1&0&-1 \end{bmatrix}$

with the eigenvalues given by $\lambda_{1,2}=-2,\ \lambda_{3,4}=0,\ \lambda_{5,6}=-1$ and the corresponding (generalized) eigenvectors given by


Compute the state response $x(t)$ for $x_0=\begin{bmatrix}0&0&0&0&0&1\end{bmatrix}^\top$.

$\textbf{hint:}$ Do not compute $M^{-1}$

My approach:

The state response follows from $x(t)=e^{At}x_0$

Normally I would compute this using $e^{At}=MDM^{-1}$ in which

$D=\begin{bmatrix}e^{-2t}&0&0&0&0&0\\0&e^{-2t}&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&e^{-t}&0\\0&0&0&0&0&e^{-t} \end{bmatrix}$

Using this method the state response is $\begin{bmatrix}0&0&0&0&0&e^{-t}\end{bmatrix}^{\top}$

An answer is not provided but I assume this to be correct.

But the hint bothers me.

I do not know how to achieve this result without computing/using $M^{-1}$


You could answer this even without knowing the eigenvalues and $M$. Namely $\dot{x}(0)=A\,x_0=-x_0$ also lies in the span of $x_0$, so $x(dt)$ should also remain in this span as well. From this it can be generalized that $x(t)$ for all $t>0$ will also remain in that span. From this it can be concluded that your proposed solution is indeed correct.

I do have to note that your expression for $D$ might not be correct, because due to the repeated eigenvalues you might be dealing with Jordan block of size bigger than one by one. To check this you could see if it holds that $A\,\vec{m} = \lambda\,\vec{m}$ for all eigenvalue (generalized)eigenvector pairs.

  • 1
    $\begingroup$ I confirm that $A$ is not diagonizable. $\endgroup$ – Digitalis Sep 11 '18 at 13:11

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