Thread: Jordan Normal Form quick help!!!

1. Jordan Normal Form quick help!!!

I have two matricies A and B:

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

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

(Not sure if I got this right)

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

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

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

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

(Not sure if I got this right)

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

3. 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).

4. The Jordan form you have for A is wrong.

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

Characteristic polynomial :

$
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

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

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