# Upper triangular matrix

1. Jan 17, 2012

### PirateFan308

1. The problem statement, all variables and given/known data
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:

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

3. 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!

2. Jan 18, 2012

### 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.