How do I find the transition matrix of this dynamic system?

In summary, the matrix that is given to me is the Jordan form of a matrix and I am supposed to find the transition matrix. I have found the eigenvalues and eigenvectors, but I am having problems with the middle one. I think it is supposed to be the following: ## e^{At} = Me^{\Lambda t}M^{-1} ## but no matter how I multiply with e and the eigenvalue,regardless of the "combination" I cannot get to the solution.
  • #1
arhzz
268
52
Homework Statement
Find the transition matrix
Relevant Equations
-
Hello! I have the following matrix (picture 1.)and I am susposed to find the transition matrix ($$ \phi $$) now for that I need the eigenvalue and vectors of this matrix A. The eigenvalues are 1,1 and 2. The eigenvectors I have found to be (1 0 0) (1 1 0) (5 3 1). Now to find the transition matrix we are given this formula(picture 2) where V is a matrix comprised of my eigenvectors and V^-1 simply the inverse. And what the middle one is the part I am having problems with. This phi wave lets call it, is susposed to be the following. In a case of where you have more than one eigenvalue that is the same ( in my case 1 is double) and where the geometrical value of those eigenvalues is 1. Phi wave is calculated like this(picture) where S is the Jordan form of the matrix and beta (this weird looking b) are the eigenvectors of the matrix. Now I dont know how to construct this,even though it might seem straight forward. What is confusing me is I know that lambda is the eigenvalue but,I have 3 where am I susposed to put the third one,what I mean by that is the lambda in the picture is I am assuming reffering to the 1 (the double eigenvalue) but how what do I do with the 2? When you multiply the Jordan form with t and proceed to multiply with V and V-1 you dont get the solution. Also I have tried reading this schematic and I am pretty sure that the Matrix we need to multiply with e is simply the first 3 rows and coloumns. So basically its going to be exactly what is in this picture but the under the t should be a 1. But no matter how I multiply with e and the eigenvalue,regardless of the "combination" I cannot get to the solution.(picture 4). Thanks for the help in advance!

Picture 1. The matrix that is given to me
ssss.png


Picture 2. The formula to calculate Phi
eeeee.png
Picture 3.How to construct Phi wave
ffffff.png
Picture 4. Solution
ssssrr.png
 
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  • #2
I may be completely off base here. For a transition matrix, I would have just added some columns to ##A## for the derivatives of ##\vec x## and some rows with the integrator operator, ##1/s##, to integrate the derivatives and obtain ##\vec x##.
 
  • #3
FactChecker said:
I may be completely off base here. For a transition matrix, I would have just added some columns to ##A## for the derivatives of ##\vec x## and some rows with the integrator operator, ##1/s##, to integrate the derivatives and obtain ##\vec x##.
Okay,but I've never seen that in class,tbh. We have only been shown to do it via matrices.
 
  • #4
arhzz said:
Okay,but I've never seen that in class,tbh. We have only been shown to do it via matrices.
It is for matrices, but it must be completely different from the approach in your class.
 
  • #5
FactChecker said:
It is for matrices, but it must be completely different from the approach in your class.
Oh, okay. Yea its a different approach, I didnt even realise it involved matrices.
 
  • #6
arhzz said:
The eigenvectors I have found to be (1 0 0) (1 1 0) (5 3 1).
Recheck your eigenvectors:

##
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & 0 & 2
\end{bmatrix}

\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}
=
\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}
##

PS: Wait, I see. It's not an eigenvector but it is part of the "Jordan Chain". So you are very close.

You have already found ##
M =
\begin{bmatrix}
1 & 1 & 5 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{bmatrix}
##

Then ## M^{-1}AM =
\begin{bmatrix}
1 & -1 & -2 \\
0 & 1 & -3 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 5 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 0\\
0 & 1 & 0 \\
0 & 0 & 2
\end{bmatrix}
\equiv \Lambda
##

Look at the link below to find ##e^{At} = Me^{\Lambda t}M^{-1}##
 
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  • #8
DaveE said:
These links look helpful, although it's been so long since I did this stuff I might have missed something.
https://en.wikipedia.org/wiki/Jordan_normal_form
https://faculty.washington.edu/chx/teaching/me547/1-7_ss_sol_slides.pdf
Okay I think I am slowly getting there. What you posted in your first post (#6),the very last matrix should be the Jordan Matrix (in my case that is the S). I have been able to get to find that matrix (S or as you put it that weird A) but take a look at my formula in the original post. The Phi wave should be ## e^{S*t} ## which equals the part that is confusing me. After I get my S matrix what do I need to do next?
Regarding the link you shared,I found something very interesting. Take a look at this picture;

ttttt.png

Taking a look from the taylor expansion we can see how they derive the matrix (this is the "schematic" of how I am susposed to construct my Phi wave) and look at how their matrix looks like. Basically its the same as in my formula but the ##e^{\lambda}t## is inside the matrix itself.Now the only part that I need to figure out is what is lambda refering to? I know its the eigenvalues but which one? I have two that are the same (1) and a 2,how do I know where to put them,in what order?
 
  • #9
arhzz said:
as you put it that weird A
Capital Lambda, a greek letter often used for the eigenvalue matrix. But everyone seems to have their own notation here.

arhzz said:
Basically its the same as in my formula but the eλt is inside the matrix itself.Now the only part that I need to figure out is what is lambda refering to? I know its the eigenvalues but which one? I have two that are the same (1) and a 2,how do I know where to put them,in what order?
That example only has a single eigenvalue. The derivation is on page ten of that link, read the whole thing:

##
\Lambda =
\begin{bmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3
\end{bmatrix}
\Rightarrow
e^{\Lambda t} =
\begin{bmatrix}
e^{\lambda_1 t} & 0 & 0 \\
0 & e^{\lambda_2 t} & 0 \\
0 & 0 & e^{\lambda_3 t}
\end{bmatrix}
##

This only works for a diagonal matrix, that's why they split it into two parts.
 
  • #10
DaveE said:
Capital Lambda, a greek letter often used for the eigenvalue matrix. But everyone seems to have their own notation here.That example only has a single eigenvalue. The derivation is on page ten of that link, read the whole thing:

##
\Lambda =
\begin{bmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3
\end{bmatrix}
\Rightarrow
e^{\Lambda t} =
\begin{bmatrix}
e^{\lambda_1 t} & 0 & 0 \\
0 & e^{\lambda_2 t} & 0 \\
0 & 0 & e^{\lambda_3 t}
\end{bmatrix}
##

This only works for a diagonal matrix, that's why they split it into two parts.
Okay that is the case when you have multiple eigenvalues,but it is required that all the eigenvalues are different from each other (we have covered that in class). But here I have 1 as an eigenvalue coming up twice. I have tried it with that approach but it didnt work for me.
 
  • #11
arhzz said:
Okay that is the case when you have multiple eigenvalues,but it is required that all the eigenvalues are different from each other (we have covered that in class). But here I have 1 as an eigenvalue coming up twice. I have tried it with that approach but it didnt work for me.
Try again. It doesn't matter if ##\lambda_1 = \lambda_2## the formula still works. Of course you have to do the off-diagonal matrix too.

Without seeing your work, I don't know where your problem is.
 
  • #12
Also, be careful to get the correct order for the modal matrix (eigenvector matrix) in the transformation back to the original state space.

## e^{At} = Me^{\Lambda t}M^{-1} \neq M^{-1}e^{\Lambda t}M ##

This is a common mistake.
 
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  • #13
DaveE said:
Try again. It doesn't matter if ##\lambda_1 = \lambda_2## the formula still works. Of course you have to do the off-diagonal matrix too.

Without seeing your work, I don't know where your problem is.
Okay I will try again tommorow,very carefully this time. If I am stuck I will post my work in detail.

Just to clarify, what do you mean with the "off diagonal matrix" ?
 
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  • #14
arhzz said:
Just to clarify, what do you mean with the "off diagonal matrix" ?
What they called "N" in that image you posted. What's left behind when you split your matrix into a diagonal form plus a non-diagonal form.
 
  • #15
DaveE said:
What they called "N" in that image you posted. What's left behind when you split your matrix into a diagonal form plus a non-diagonal form.
Ah so first I have to find the diagonal form, plug in the values I have for my eigenvalues that is my ## e^{\lambda t} ## Than what I have left is my ## e^{N} ## matrix. I multiply those two that gives me ## e^{At} ## and now that should be the Phi wave in my first post (I think it was S instead of an A). And the rest should be just multiplying the matrices right?
 
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  • #16
arhzz said:
Ah so first I have to find the diagonal form, plug in the values I have for my eigenvalues that is my ## e^{\lambda t} ## Than what I have left is my ## e^{N} ## matrix. I multiply those two that gives me ## e^{At} ## and now that should be the Phi wave in my first post (I think it was S instead of an A). And the rest should be just multiplying the matrices right?
Yes. It's all so that we can avoid actually calculating all of those Taylor expansion terms in ##e^{At}##. We have an easy way to do the diagonal matrix, then we hope that the non-diagonal matrix that's left behind won't have very many non-zero terms when we have to start calculating ##N, N^2, N^3,...##
 
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  • #17
Okay I know that everyone has probably forgot about this post,but I have been able to solve this actually. The method I described above worked. Thank you everyone for the help!
 
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  • #18
arhzz said:
Okay I know that everyone has probably forgot about this post,but I have been able to solve this actually. The method I described above worked. Thank you everyone for the help!
Congratulations, good work!
 
  • #19
For future readers... Here is my MathCad worksheet for this problem. Sorry, I'm just too lazy to describe it. The previous links show the method pretty well.

1688939769306.png
 

FAQ: How do I find the transition matrix of this dynamic system?

What is a transition matrix in a dynamic system?

A transition matrix in a dynamic system is a matrix that describes how the state of the system changes from one time step to the next. It encapsulates the probabilities or rules governing the transitions between different states in the system.

How do I derive the transition matrix from a system of linear equations?

To derive the transition matrix from a system of linear equations, you need to express the system in the form of a state-space representation. This involves identifying the state variables and writing the equations in matrix form, where the transition matrix is the matrix that multiplies the state vector to produce the next state vector.

Can I find the transition matrix for a non-linear dynamic system?

For non-linear dynamic systems, finding a transition matrix is more complex. One common approach is to linearize the system around an equilibrium point and then derive the transition matrix for the linearized system. This involves calculating the Jacobian matrix of the system's equations at the equilibrium point.

How do I use data to estimate the transition matrix for a Markov chain?

To estimate the transition matrix for a Markov chain using data, you can use the frequency of observed transitions between states. Count the number of times each transition occurs and divide by the total number of transitions from each state to get the probabilities. These probabilities form the entries of the transition matrix.

What role does the transition matrix play in the stability of a dynamic system?

The transition matrix plays a crucial role in determining the stability of a dynamic system. By analyzing the eigenvalues of the transition matrix, you can assess whether the system will converge to a steady state, oscillate, or diverge over time. If all eigenvalues have magnitudes less than one, the system is stable.

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