## help with #3 |
wllmskn |
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Can someone explain the polynomial notation of the linear transformation? I always get confused by it. How to I write it as a matrix? |

Jackhe: June 2, 2015, 10:17 p.m. |
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you simply put the coefficients in the matrix which gives you: \begin{bmatrix} 1 & -3 & 3\\ 3 & -5 & 3\\ 6 & -6 & 4 \end{bmatrix} |

Jackhe: June 2, 2015, 10:24 p.m. |
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Reading the other comments now makes me think that my approach is wrong, not sure anymore |

Jackhe: June 2, 2015, 10:47 p.m. |
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Ok, now I'm sure that my answer is correct. So, we start with the standard basis 1, x, x^2 which would be (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively (vertical). T(1) is 1+3x+6x^2, T(x) is -3-5x-6x^2 and T(x^2) is 3+3+4x^2. These are respectively the vertical vectors (1, 3 ,6), (-3, -5, -6) and (3, 3, 4) and thus are the columns of the transformation matrix |

alanvce: June 2, 2015, 10:54 p.m. |
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T(a +bx +cX^2) how i think about it is as follow, Xo X X^2 (a 3a 6a) (-3b -5b -6b) (3c 3c 4c) which translates to just ( 1 3 6) ( -3 -5 -6) ( 3 3 4 ) Those are matrices btw, idk how to write matrices here |

Jackhe: June 3, 2015, 1:20 a.m. |
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I use this website and then copy and paste it here: http://www.codecogs.com/latex/eqneditor.php |

wllmskn: June 3, 2015, 1:34 p.m. |
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What I mostly don't get is how a sum of things (the polynomial) becomes something that isn't a sum (the matrix). How can we do that? It seems to me like we're changing the nature of the expression. I just can't quite wrap my head around it |