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Let $T:\mathbb R^3\to\mathbb R^3$ be the operator given by $$T(v)=\left[\begin{matrix}-3 & 1 & 2\\-4 & 1& 4\\0&0 &-1\end{matrix}\right]v$$ Determine whether $T$ is decomposable or indecomposable.

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What is a decomposable operator? Google did not help me... –  1015 Feb 26 '13 at 19:36
By this definition your matrix is decomposable, if we take $I=\{ 3\}$... Does it make sense? –  Ludolila Feb 26 '13 at 19:41
I would take this definition here. It's decomposable if its similar to lower block triangular matrix, where the similarity transform must be a permutation. –  Elmar Zander Feb 26 '13 at 19:49
You live and you learn. By the way, I think that @Elmar Zander 's definition coincides with the one I found (according to this mathworld.wolfram.com/ReducibleMatrix.html ). Nice... =) –  Ludolila Feb 26 '13 at 19:54
@ElmarZander That's equivalent to the definition Ludilla linked to in the Cartan Matrices wikipedia. –  Thomas Andrews Feb 26 '13 at 19:54
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1 Answer

After a long and extremely educating discussion (above), we will use the following definition:

"A $n \times n$ matrix $A$ is decomposable if there exists a nonempty proper subset $I \subseteq \{1,2,...,n\}$ such that $a_{ij}=0$ whenever $i\in I$ and $j \notin I$ ".

According to this definition, the matrix in question is decomposable, since we can take $I=\{3\}$, and indeed $a_{31}=a_{32}=0$ (for $2,3 \notin I$).

Note, though, that we answered the question of decomposability of the matrix, and not the operator...

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Why did you decide to take I={3}? –  AdamFox Feb 27 '13 at 2:23
Because $a_{31}=a_{32}=0$. So $i$ has to be $3$. Is that what you meant? –  Ludolila Feb 27 '13 at 9:09
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