# Jordan Normal Form quick help!!!

• Jun 30th 2009, 06:24 PM
realpart1/2
Jordan Normal Form quick help!!!
I have two matricies A and B:

$\displaystyle A=\left[\begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & -3\\ 0 & 1 & 3\end{array}\right]$ $\displaystyle B=\left[\begin{array}{ccccc} 3 & 1 & 0 & 1 & -2\\ 1 & 3 & -1 & 0 & 1\\ -1 & -1 & 4 & 3 & -3\\ 1 & 1 & -1 & 2 & 1\\ -2 & -2 & 2 & 2 & 1\end{array}\right]$

which I need to put in Jordan Normal Form:

$\displaystyle J_{A}=\left[\begin{array}{ccc} -1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{array}\right]$ $\displaystyle J_{B}=\left[\begin{array}{ccccc} 2 & 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0 & 0\\ 0 & 0 & 3 & 1 & 0\\ 0 & 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 0 & 3\end{array}\right]$

(Not sure if I got this right)

and then (my biggest problem) construct matrices S an T so that
$\displaystyle S^{-1}AS=J_{A}$ and $\displaystyle T^{-1}BT=J_{B}$ (I know th significance, but I'm not sure how to construct these matricies.)
• Jun 30th 2009, 06:33 PM
Bruno J.
Quote:

Originally Posted by realpart1/2
I have two matricies A and B:

$\displaystyle A=\left[\begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & -3\\ 0 & 1 & 3\end{array}\right]$ $\displaystyle B=\left[\begin{array}{ccccc} 3 & 1 & 0 & 1 & -2\\ 1 & 3 & -1 & 0 & 1\\ -1 & -1 & 4 & 3 & -3\\ 1 & 1 & -1 & 2 & 1\\ -2 & -2 & 2 & 2 & 1\end{array}\right]$

which I need to put in Jordan Normal Form:

$\displaystyle J_{A}=\left[\begin{array}{ccc} -1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{array}\right]$ $\displaystyle J_{B}=\left[\begin{array}{ccccc} 2 & 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0 & 0\\ 0 & 0 & 3 & 1 & 0\\ 0 & 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 0 & 3\end{array}\right]$

(Not sure if I got this right)

and then (my biggest problem) construct matrices S an T so that
$\displaystyle S^{-1}AS=J_{A}$ and $\displaystyle T^{-1}BT=J_{B}$ (I know th significance, but I'm not sure how to construct these matricies.)

Can you do $\displaystyle A$? From what I can understand, you manage to get to the Jordan form but you can't find the base change matrices?

What are you able to do:

Can you find the eigenvalues?
Do you know the definition of generalized eigenspaces?
From the definition you should be able to find a basis for the generalized eigenspaces corresponding to the eigenvalues -1 and 1.

You have to say where you are stuck precisely.
• Jun 30th 2009, 06:50 PM
realpart1/2
I got the Eigenvalues by using the respective characteristic polynomials of A and B, and as far as I understood, the Jordan form is made by putting the eigenvalues on the diagonal, keeping the identical values together and adding 1's in-between. The rest was never fully explained. I know what the inverse matrices around A and B are supposed to do but not how to construct them (If I'm right, their existence somehow defines the Jordan form).
• Jul 1st 2009, 10:06 AM
Bruno J.
The Jordan form you have for A is wrong.

$\displaystyle A=\left[\begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & -3\\ 0 & 1 & 3\end{array}\right]$

Characteristic polynomial :

$\displaystyle f(t)=\mbox{det }\left[\begin{array}{ccc} -t& 0 & 1\\ 1 & -t & -3\\ 0 & 1 & 3-t\end{array}\right] = t^2(3-t)+1-3t = -(t-1)^3$

So 1 is the only eigenvalue. You will have

$\displaystyle K_1 = \mbox{Kernel }(A-I)^3 = \mbox{Kernel }\left[\begin{array}{ccc} -1& 0 & 1\\ 1 & -1 & -3\\ 0 & 1 & 2\end{array}\right]^3=\mbox{Kernel }\left[ \begin{array}{lll} -1 & 0 & 1 \\ 1 & -1 & -27 \\ 0 & 1 & -1 \end{array} \right]$

Now find a vector $\displaystyle v$ in the kernel such that $\displaystyle (A-I)^2v, (A-I)v$ are nonzero. Your basis for $\displaystyle K_1$ will be given by $\displaystyle \{(A-I)^2v, (A-I)v, v\}$.