Jordan Normal Form physical applications

Maybe_Memorie
Messages
346
Reaction score
0
Are there any applications of the Jordan Normal Form of a matrix in physics?
If so, please explain?
 
Physics news on Phys.org
The Jordan Normal Form is very often used to solve linear differential equations. Such an equations can often be represented in the form

y^\prime=Ay,~y(0)=y_0

with A a matrix. The solution to such an equation is y(t)=e^{At}y_0. The problem is to calculate the matrix exponential e^{At}, this can be very difficult. To ease the problem, we calculate the Jordan Normal Form of A. In this from, calculating the matrix exponential becomes very easy!
The fun part is of course to calculate the Jordan Normal Form, as this can be quite difficult...
 
A slight variation on what micromass said:
If you have the d.e. y'= Ay, and B is such that B^{-1}AB= J, the "Jordan form" for A, then we can multiply both sides of the equation by B^{-1}:
(B^{-1}y)'= B^{-1}Ay= B^{-1}A(BB^{-1})y= (B^{-1}AB)(B^{-1}y)
and, letting x= B^{-1}y write the equation as
x&#039;= Jx[/itex].<br /> <br /> Now, since J is a Jordan form matrix, that gives a system of &quot;almost uncoupled&quot; equations. If we write x as a column matrix<br /> \begin{bmatrix}x_1 \\ x_2 \\ ---\\ x_n\end{bmatrix}<br /> <br /> Then every equation is of the form x_k&amp;#039;= \lambda_kx_k or x_k&amp;#039;= \lambda_nx_k+ x_{k+1}. The very last equation is, of course, x_n&amp;#039;= \lambda_n x_n, because there is no &amp;amp;quot;x_{n+1}&amp;amp;quot; and is easy to solve. Then, if x_{n-1}&amp;amp;amp;#039;= \lambda_{n-1}x_{n-1}+ x_n, because we already have x_n, that is easy to solve, etc.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Finally, since x= B^{-1}y, y= Bx.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; If a matrix happens to be &amp;amp;quot;diagonalizable&amp;amp;quot; (all those &amp;amp;quot;1&amp;amp;quot; above the diagonal are unecessary), we can completely &amp;amp;quot;uncouple&amp;amp;quot; all those equations, whether differential equations or not, and have, basically, n separate, simpler, problems. If the matrix is NOT diagonalizable, we can still simplify as much as possible using the &amp;amp;quot;Jordan Normal Form&amp;amp;quot;.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

Similar threads

Replies
8
Views
2K
Replies
1
Views
2K
Replies
7
Views
2K
Replies
11
Views
2K
Replies
11
Views
5K
Replies
2
Views
2K
Replies
6
Views
7K
Replies
9
Views
4K
Replies
3
Views
1K
Back
Top