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Let $T:V_3\to V_3 $ be a linear transformation such that

$$T(k)=2i+3j+5k,~~~~T(j+k)=i,~~~~T(i+j+k)=j-k$$

(both bases are usual unit coordinate vectors)

Now choose both bases to be $(e_1,e_2,e_3)$ where $e_1=(2,3,5)$, $e_2=(1,0,0)$, $e_3=(0,1,-1)$

Determine the matrix of T reative to the new bases.

My book says, that to determine the matrix of T relative to choice of bases, I need to transform each basis element of the first space and express it as a linear combination of the basis element of the second space. First of all, I express the transformation as follows:

$$T(i)=-i+j-k,~~~~T(j)=-i-3j-5k,~~~~T(k)=2i+3j+5k$$

Then for example $T(e_1)=-e_1+e_2-e_3=(-1,-4,-4)$ (right?)

Then I should express it a combination of $e_1,e_2,e_3$. If I do that, if I understand it correctly, I shuould get the first column of the matrix. However, the first column of the matrix has corresponding vaues of $2,1,2$ (I know this from the answers) and obviously something is wrong with my solution as these numbers does not correspond to the combination I am supposed to find.

I missunderstood something and looking forward to hints of how should I find the matrix to the new chosen bases. Thank you!

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1 Answer 1

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It seems to be you're mixing physics and mathematics notations: as I see it, you mean

$$i=(1,0,0)\;\;,\;\;j=(0,1,0)\;\;,\;\;k=(0,0,1)$$

so your transformation says:

$$T(0,0,1)=(2,3,5)\;\;,\;\;T(0,1,1)=(1,0,0)\;\;,\;\;T(1,1,1)=(0,1,-1)$$

As you can see, your transformation maps

$$\left\{\begin{pmatrix}0\\0\\1\end{pmatrix}\;,\;\;\begin{pmatrix}0\\1\\1\end{pmatrix}\;,\;\;\begin{pmatrix}1\\1\\1\end{pmatrix}\right\}\stackrel T\longrightarrow\left\{\begin{pmatrix}2\\3\\5\end{pmatrix}\;,\;\;\begin{pmatrix}1\\0\\0\end{pmatrix}\;,\;\;\begin{pmatrix}\;0\\\;1\\\!\!-1\end{pmatrix}\right\}$$

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It would be nice if the downvoter could explain what she/he sees wrong in my answer-hint. Maybe I can fix... –  DonAntonio Feb 16 '13 at 18:40
    
I think it does as it seems the OP missed the fact, if I translated correctly his notation, that the "new" basis vectors are the images of the old one... –  DonAntonio Feb 16 '13 at 18:47
    
Ok, yes, you're right, it is clearer like that and the OP probably missed this fact. –  1015 Feb 16 '13 at 18:51
    
In fact, it is my opinion that the hint is almost the whole answer, but of course: it may still be possible that my interpretation of $\,i,j,k\,$ is wrong, though it is the usual one in physics. Anyway, some people really rushes to downvote and even if someone's answer's wrong that is, I think, against the spirit of the site. Oh, well....:) –  DonAntonio Feb 16 '13 at 18:56
    
Almost, yes. Maybe you can add a few words to explain how to compute the matrix in the new basis. –  1015 Feb 16 '13 at 18:59

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