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 Post subject: Jordan Matrix
PostPosted: Sat, 12 Jun 2010 09:45:58 UTC 
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Let A=\left[\begin{array}{ccc}2&10&5\\-2&-4&-4\\3&5&6\end{array}\right].
Find matrix T \ni TJT^{-1}=A.

Spoiler:
I have found \lambda_1=\lambda_2=1,\lambda_3=2
J_A=\left[\begin{array}{ccc}1&1&0\\0&1&0\\0&0&2\end{array}\right]
Matrix T consists of eigen vectors of A, hence
T=\left[\begin{array}{ccc}5&2&-5\\5&2&-5\\1&1&-2\end{array}\right]
but it means |T|=0 implies T has no invers. So..... ???? Please help..

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 Post subject:
PostPosted: Sat, 12 Jun 2010 09:51:44 UTC 
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Sorry... my matrix T doesn't display correctly...
I don't know. I think I have writen the latex correctly

This is the right one (sorry not using latex, it doesn't work)

T =

5 2 -5
5 2 -5
1 1 -2

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 Post subject:
PostPosted: Mon, 14 Jun 2010 19:09:53 UTC 
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Is that supposed to be A or T? And what do you mean by TJT^{-1}\in T?

Oh, I see, you are using this strange convention of \ni meaning "such that" rather than "contains the element".

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Tue, 22 Jun 2010 09:04:17 UTC 
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I'm confused as to the necessity of your response outermeasure...especially the remark about "strange convention"...

in regards to the OP:

\left[
\begin{matrix}
1 & 3.75 & 2.5\\
 0 & -2& -1\\
 1 & 5&  3
\end{matrix}
\right]
 \cdot A \cdot 
\left[
\begin{matrix}
-4   &  5 &    5 \\
    -4   &  2    & 4 \\
     8   & -5   & -8
\end{matrix}
\right]
= J


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