Decomposing a Matrix into Jordan Normal Form

[PirateFan308]

Homework Statement

Let $A = \begin{pmatrix}4 & 1 & 0\\0 & 3 & 1\\-1 & -1 & 2\end{pmatrix}$, which has characteristic polynomial $(3-x)^{3}$. Find a matrix X such that:

$X^{-1}AX = \begin{pmatrix}3 & 1 & 0\\0 & 3 & 1\\0 & 0 & 3\end{pmatrix}$

The Attempt at a Solution

If $X^{-1}AX = \begin{pmatrix}3 & 1 & 0\\0 & 3 & 1\\0 & 0 & 3\end{pmatrix} =: B$

AX = XB

Suppose that the columns of X are $v_{1}, v_{2}, v_{3}$
The first column of B has the coordinates of $Av_{1}$

$Av_{1}=3v_{1}+0v_{2}+0v_{3}=3v_{3}~~~~(A-3I)v_{1}=0$

$v_{1} = \begin{pmatrix}a\\d\\g\end{pmatrix}$

$(A-3I) = \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\-1 & -1 & -1\end{pmatrix} → \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & -1\end{pmatrix} → \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}$

let d=1, a=-1, g=0

$v_{1} = \begin{pmatrix}-1\\1\\0\end{pmatrix}$

$Av_{2} = 1v_{1}+3v_{2}+0v_{3} = v_{1} + 3v_{2}$

$(A-3I)v_{2} = v_{1}$

$v_{2} = \begin{pmatrix}b\\e\\h\end{pmatrix}$

$\begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\-1 & -1 & -1\end{pmatrix} \begin{pmatrix}b\\e\\h\end{pmatrix} = \begin{pmatrix}-1\\1\\0\end{pmatrix}$

$\begin{pmatrix}1 & 1 & 0 & | ~~-1\\0 & 0 & 1 & |~~ 1\\-1 & -1 & -1 & |~~ 0\end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & | ~~-1\\0 & 0 & 1 & | ~~1\\0 & 0 & -1 & | ~~-1\end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & | ~~-1\\0 & 0 & 1 & |~~ 1\\0 & 0 & 0 & |~~ 0\end{pmatrix}$

$v_{2} = \begin{pmatrix}0\\-1\\0\end{pmatrix}$



$Av_{3} = 0v_{1}+1v_{2}+3v_{3} = v_{2} + 3v_{3}$

$(A-3I)v_{3} = v_{2}$

$v_{3} = \begin{pmatrix}c\\f\\i\end{pmatrix}$

$\begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\-1 & -1 & -1\end{pmatrix} \begin{pmatrix}c\\f\\i\end{pmatrix} = \begin{pmatrix}0\\-1\\0\end{pmatrix}$

$\begin{pmatrix}1 & 1 & 0 & | ~~0\\0 & 0 & 1 & | ~~-1\\-1 & -1 & -1 & |~~ 0\end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & |~~ 0\\0 & 0 & 1 & | ~~-1\\0 & 0 & -1 & |~~ \end{pmatrix} → \begin{pmatrix}1 & 1 & 0 & |~~ 0\\0 & 0 & 1 & | ~~-1\\0 & 0 & 0 & | ~~-1\end{pmatrix}$

Therefore, there does not exist an invertible matrix X because the last matrix can not be solved to get $v_{3}$ so that X can be created.

I don't think this is right as my teacher generally doesn't give us 'unsolvable' questions. I'm not sure what I've done wrong, and I've tried it a few different ways and I can never get it to work. Thanks!

 

[kru_]

I've only just finished my linear algebra course. This looks like you're being given the Jordan Normal form of the matrix, and asked to find the decomposition. Every square matrix has a Jordan decomposition.