Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
What is a decomposable operator? Google did not help me... –  1015 Feb 26 '13 at 19:36
1  
By this definition your matrix is decomposable, if we take $I=\{ 3\}$... Does it make sense? –  Ludolila Feb 26 '13 at 19:41
1  
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
1  
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
1  
@ElmarZander That's equivalent to the definition Ludilla linked to in the Cartan Matrices wikipedia. –  Thomas Andrews Feb 26 '13 at 19:54

1 Answer 1

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...

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.