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Here is the question.

I need to determine if there is linear transformation according to the given information. If there is I need to write it as $T(x,y,z)=(a,b,c)$. I don't know the algorithm for questions like that. I would be happy if someone could explain it to me. thanks for help. Here is the question.

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I've put an answer below, but I think your question is mis-aimed if what you're looking for is an algorithm. To become good at maths you need to develop the creativity to see an unfamiliar type of problem and find a solution to it. [In any case, I up-voted your question because I didn't think it deserved the downvote.] –  Clive Newstead Jan 2 '13 at 17:40
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  user53153 Jan 2 '13 at 17:55
    
thank you very much @PavelM –  user1816377 Jan 2 '13 at 17:56
    
It is better show us $T$ by written that. I couldn't see that pic. :-) –  B. S. Jan 2 '13 at 17:56
    
there are 6 vectors so I prefered to gave a link.. its not work? –  user1816377 Jan 2 '13 at 17:59

1 Answer 1

up vote 1 down vote accepted

If $T$ exists then it must be invertible since its image will have dimension three. You know what its inverse (if it exists) does to the standard basis vectors, so you can find the matrix of the inverse of $T$. Try and invert it. If you can, you end up with a matrix for $T$, and so $T$ is a linear map; else $T$ can't be a well-defined linear map.

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can you give me an exapmle for 1 vector? I didn't understand what to do.. –  user1816377 Jan 2 '13 at 17:40
    
@user1816377: Not right now, but you might find it useful to try to follow what my answer in one or two dimensions. –  Clive Newstead Jan 2 '13 at 17:43

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