Help with Modal canonical form linear systems.

Question

I'm trying to find the modal canonical form of a linear system, the book I'm using is Linear systems theory and design by Chen, but has no clear explanation about this topic, and I'm having a hard time calculating the eigenvectors for matrix $Q$. The problem goes like this:

Find the modal equivalent system of:

$\dot{x}=\begin{bmatrix} -2&0&0\\ 1&0&1\\ 0&-2&-2 \end{bmatrix}x + \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}u(t)$

$y=\begin{bmatrix} 1&-1&0 \end{bmatrix}x$

So, as a first step I calculated the eigenvalues of system matrix "A", in order to obtain the the equivalent form, I need to calculate matrix $P$, wich is defined as $P^{-1}:=Q$, where Q is an eigenvectors matrix. So, the rest of the parameters are defined as:

$\bar{x}=Px$, $\bar{A}=PAP$, $\bar{B}=PB$, $\bar{C}=CP^{-1}$

The eigenvalues are $\lambda_{1}=-2$, $\lambda_{2}=-1+i$, $\lambda_{3}=-1+i$. Thus, according to the theory, in orther to obtain the eigenvectors, the following equality should hold with each eigenvalue:

$(\lambda_{i}I-A)v_{i}=0$

So that,

$P^{-1}=\begin{bmatrix} v_{1}&v_{2}&v_{3} \end{bmatrix}$

The thing is that when I try to calculate the eigenvectors, some of them make no sense to me. The idea is to find a non-zero vector that makes $\lambda_{i}I-A$ equals to zero. I've seen that in some approaches for complex eigenvalues, eigenvectors have complex components but in my case I'm obtaining real numbers, which doesn't match with the solution's manual. I think there might be different solutions for eigenvectors but I would appreciate if someone could guide me in to the proper way to obtain the eigenvectors for $P$. For example, I calculated

For $\lambda_{1}=-2$:

$(\lambda_{1}I-A)v_{1}=0$

$\begin{bmatrix} 0&0&0\\ -1&-2&-1\\ 0&2&0 \end{bmatrix}\begin{bmatrix} v_{1x}\\ v_{1y}\\ v_{1z} \end{bmatrix}=0$

$v_{1y}=0$, $v_{1z}=k---->k=1$ (free variable), $v_{1x}=-v_{1z}=-1$

Thus,

$v_{1}=\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$

Answer 1

Hey the equation you want to be using for this is $(A-\lambda_{i} I)v_{i} = 0$. The final matrix you presented will have different entries.

Answer 2

You have to keep in mind that the eigenvector can be complex. However, such eigenvector can also be multiplied by any constant, including a complex constant. So if you parameterize each eigenvector as

$$ v_i = \begin{bmatrix} v_{i,x1} + v_{i,x2} \, i \\ v_{i,y1} + v_{i,y2} \, i \\ v_{i,z1} + v_{i,z2} \, i \end{bmatrix} $$

you essentially have six instead of three degrees of freedom, but because you are allowed to multiply this vector with any complex number also means that two instead of one of those degrees of freedom can remain free when solving for $v_i$.

Often the resulting eigenvectors are normalized in length, however this still doesn't give a unique solution, since multiplying such vector by any number of the complex unit circle ($\cos(\theta) + i\sin(\theta)$) is still a vector of unit length. Therefore, it could be that your solution for each $v_i$ differs by such factor on the complex unit circle compared to the solutions manual.