- #1
ysebastien
- 6
- 0
Homework Statement
[tex]
A=\left[\begin{array}{cccc}
5 & 4 & 2 & 1\\
0 & 1 & -1 & -1\\
-1 & -1 & 3 & 0\\
1 & 1 & -1 & 2
\end{array}\right]
[/tex]
- Find the Jordan form of A
- Find a matrix P such that A = PJP-1
Homework Equations
The Attempt at a Solution
I've already found, and verified the Jordan form it is,
[tex]
J=\left[\begin{array}{cccc}
4 & 1 & 0 & 0\\
0 & 4 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 1
\end{array}\right]
[/tex]
from the characteristic polynomial [itex] q(x) = (x-4)^{2}(x-2)(x-1) [/itex].
To find P, I found a basis vector for each of null spaces of each eigenvalue.
[tex] \ker{(A-4I)} [/tex]
Basis: [itex] (1,0,-1,1) [/itex]
[tex] \ker{(A-4I)^{2}} [/tex]
Basis: [itex] (0,0,-1,1), (1,0,0,0) [/itex]
[tex] \ker{(A-2I)} [/tex]
Basis: [itex] (1,-1,0,1) [/itex]
[tex] \ker{A-I)} [/tex]
Basis: [itex] (-1,1,0,0) [/itex]
(these have all been verified with Mathematica)
Now I'm supposed to construct the columns of P by picking a set of 4 linearly independent vectors, which they all are. Here is my struggle. In choosing between the two basis vectors for [itex] \ker{(A-4I)^{2}} [/itex], choosing [itex] (1,0,0,0) [/itex] works fine i.e.
[tex]
P=\left[\begin{array}{cccc}
1 & 1 & 1 & -1\\
0 & 0 & -1 & 1\\
-1 & 0 & 0 & 0\\
1 & 0 & 1 & 0
\end{array}\right]
[/tex]
but if I choose the other vector,
[tex]
P=\left[\begin{array}{cccc}
1 & 0 & 1 & -1\\
0 & 0 & -1 & 1\\
-1 & -1 & 0 & 0\\
1 & 1 & 1 & 0
\end{array}\right]
[/tex]
and P doesn't satisfy [itex] A=PJP^{-1} [/itex]. Oddly, if I take [itex] (0,0,1,-1) [/itex] it does work. Does anyone see if my logic is going awry somewhere? (0,0,1,-1) is just the negative of (0,0,-1,1) so shouldn't they in fact both work, satisfying A=PJP-1?
Sorry for the long post.