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anonymous

  • one year ago

hello can anyone help me solve this? the questions in english are : a) show that F is linear b) determine the associated matrix relative to the basis A and B . Is this matrix invertible? yes, give the reverse c) determine by hand: the image of the vector the polynomial Px d)determine the kernel ker(f) and image Im(f)

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  1. anonymous
    • one year ago
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  2. Loser66
    • one year ago
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    To show F is linear, we need show F(aP + Q)(x) = aF(P(x) +F(Q(x) YOu can break it out by doing 2 parts : F(P + Q)(x) = F(P)+F(Q) and F(aP(x))= aF(P(x)) I do the first one Let P, Q in R^2[x], a in R then aP (x) + Q(x) = (aP+Q)(x) Now, apply F on \(F(aP + Q)(x) = ((aP+Q)(0), (aP+Q)'(1),\int_0^1 (aP+Q)(x))\)

  3. Loser66
    • one year ago
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    All you need is "translate" the right hand side to \(F(aP(x)) +F(Q(x)\)

  4. Loser66
    • one year ago
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    for the first term: \((aP + Q)(x)\) , the property of function give us \((aP+Q)(x) = aP(x) +Q(x)\) It likes (f+g)(x) = f(x) +g(x)

  5. Loser66
    • one year ago
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    the second term is derivative, and we know that derivative of the sum = sum of derivative, right? like (f+g)' = f'+g' hence we have \((aP+Q)'(1) = (aP)'(1) +Q'(1)\)

  6. Loser66
    • one year ago
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    the last term is integral, exactly the same, we still have \[\int_0^1 (aP+Q)(x)dx=\int_0^1aP(x) dx +\int_0^1 Q(x) dx\]

  7. Loser66
    • one year ago
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    Now, combine all

  8. Loser66
    • one year ago
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    You know that if I have vectors \[\left[\begin{matrix}a +b \\c +d\end{matrix}\right] =\left[\begin{matrix}a \\c \end{matrix}\right]+\left[\begin{matrix}b \\c \end{matrix}\right]\], right?

  9. Loser66
    • one year ago
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    hence, your result can be broken down like that

  10. Loser66
    • one year ago
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    \[\left[\begin{matrix}aP(x) + Q(x)\\aP'(1) +Q'(1)\\\int_0^1aP(x)dx +\int_0^1Q(x)\end{matrix}\right]=\left[\begin{matrix}aP(x) \\aP'(1) \\\int_0^1aP(x)dx \end{matrix}\right]+\left[\begin{matrix} Q(x)\\Q'(1)\\\int_0^1Q(x)\end{matrix}\right]\]

  11. Loser66
    • one year ago
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    From the right, the first matrix is F(aP(x), and the second one is F(Q(x) Done, right?

  12. anonymous
    • one year ago
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    ok that seems quite logical , i really got confused by the writing

  13. Loser66
    • one year ago
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    Of course, you must jot down in neat. I inserted the explanation when solving it. It makes the proof has many irrelevant terms. hehehe...

  14. anonymous
    • one year ago
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    okay thank you , and do you know b, c and d? ( I'm very grateful that you help me)

  15. Loser66
    • one year ago
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    For b) matrix A is invertible since its determinant =-12 \(\neq 0\)

  16. Loser66
    • one year ago
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    So does B, det(B) = 2

  17. anonymous
    • one year ago
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    oh but the question is to combine those two bases to get the associated matrix

  18. Loser66
    • one year ago
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    And you know how to find their inverse, right?

  19. anonymous
    • one year ago
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    yes the inverse and determinant aren't the problem , that's just math but i need the matrix that is formed by those two bases given F

  20. Loser66
    • one year ago
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    I don't get the question.

  21. anonymous
    • one year ago
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    it's called the associated matrix in dutch. But i can't find a decent translation for it

  22. Loser66
    • one year ago
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    You meant the transformation that changes base A to base B??

  23. anonymous
    • one year ago
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    yes i think so

  24. Loser66
    • one year ago
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    so, just put [B|A] and try to change to [Identity| required matrix] by rref. Dat sit

  25. anonymous
    • one year ago
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    this is an example of what we did on another exercise

  26. Loser66
    • one year ago
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    |dw:1437695240946:dw|

  27. Loser66
    • one year ago
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    wait, my master here, let me tag him @oldrin.bataku

  28. anonymous
    • one year ago
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    with base of starting vector space [(1,0,0),(1,1,0),(1,1,1) ] to the ending vector space [(0,1,0,0),(0,0,1,0),(0,0,0,1),(1,0,0,0)]

  29. anonymous
    • one year ago
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    b) determine the associated matrix relative to the basis A and B . Is this matrix invertible? yes, give the reverse what are your bases?

  30. anonymous
    • one year ago
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    those are on the first picture

  31. anonymous
    • one year ago
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    okay, so the columns of the matrix representing a given linear map or transformation correspond to the way the matrix transforms the basis vectors (in order)

  32. anonymous
    • one year ago
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    so here our ordered basis for the first space is \((3x^2-2,4x,1)\) -- now look at how our map \(F\) transforms these

  33. anonymous
    • one year ago
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    consider that if \(P(x)=3x^2-2\) then $$P(0)=3(0)^2-2=-2\\P'(x)=6x\implies P'(0)=0\\\int_0^1 P\, dx=\left[\frac13 x^3-2x\right]_0^1=\frac13-2=-\frac53$$in other words, \(F\) maps \(3x^2-2\to (-2,0,-5/3)\)

  34. anonymous
    • one year ago
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    now take \(P(x)=4x\), so: $$P(0)=0\\P'(x)=4\implies P'(0)=4\\\int_0^1 P\, dx=\left[2x^2\right]_0^1=2$$ so \(4x\mapsto (0,4,2)\) and lastly \(P(x)=1\) gives $$P(0)=1\\P'(x)=0\implies P'(0)=0\\\int_0^1 P\, dx=1$$ so \(1\mapsto (1,0,1)\)

  35. anonymous
    • one year ago
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    now, we want to express these output vectors in terms of the given basis on \(\mathbb{R}^3\), which is \((1,0,0),(0,2,1),(1,0,1)\)

  36. anonymous
    • one year ago
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    so we want to figure out \(c_1,c_2,c_3\): $$(-2,0,-5/3)=c_1 (1,0,0)+c_2 (0,2,1)+c_3(1,0,1)$$

  37. anonymous
    • one year ago
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    so we can do this by considering \(c_1+c_3=-2,c_3=-5/3\) which gives: $$c_1=-1/3\\c_3=-5/3\\c_2=0$$

  38. anonymous
    • one year ago
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    so in terms of coefficients of our basis vectors we have that \(3x^2-2\mapsto (-2,0,-5/3)\) translates to \(\begin{bmatrix}1\\0\\0\end{bmatrix}\to\begin{bmatrix}-1/3\\0\\-5/3\end{bmatrix}\) so our first column is that

  39. anonymous
    • one year ago
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    |dw:1437696322479:dw| this is the solution

  40. Loser66
    • one year ago
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    Hold on, the second term is P'(1) , not P'(0)

  41. Loser66
    • one year ago
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    for 3x^2-2

  42. anonymous
    • one year ago
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    oops, then our results are actually: $$P(x)=3x^2-2\implies P'(x)=6x\implies P'(1)=6\\P(x)=4x\implies P'(x)=4\implies P'(1)=4\\P(x)=1\implies P'(x)=0\implies P'(1)=0$$

  43. anonymous
    • one year ago
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    so we have: $$3x^2-2\mapsto (-2,6,-5/3)$$

  44. anonymous
    • one year ago
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    which isn't the solution i'm afraid

  45. anonymous
    • one year ago
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    lol well obviously that's not the solution

  46. anonymous
    • one year ago
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    now express that in terms of the basis for \(\mathbb{R}^3\), and that's your first column

  47. anonymous
    • one year ago
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    then express the results for the next two in terms of those bases and those are your second nad third columns, and then you have your matrix

  48. anonymous
    • one year ago
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    i think you maybe understand my question wrong, and that's why you don't get to the solution

  49. anonymous
    • one year ago
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    and we solved it like this

  50. anonymous
    • one year ago
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    and i also think that px represents ax²+bx+c because its part of R2(x)

  51. Loser66
    • one year ago
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    I am with @oldrin.bataku He made mistake at \(F(3x^2 -2) \) \(P(x) = 3x^2 -2--> P(0) =-2\\P'(x) = 6x \rightarrow P'(1) =6\\\int_0^1 3x^2-2 dx= (\dfrac{3x^3}{3}-2x)|^1_0= -1\) Hence \(F(3x^2-2)= (-2,6,-1)~~or~~\left(\begin{matrix}-2\\6\\-1\end{matrix}\right)\). However, it is in standard base, not \(\beta\) base. We need transfer it to \(\beta\) base by doing: \[\left(\begin{matrix}-2\\6\\-1\end{matrix}\right)= a_1\left(\begin{matrix}-1\\0\\0\end{matrix}\right)+a_2\left(\begin{matrix}0\\2\\1\end{matrix}\right)+a_3\left(\begin{matrix}1\\0\\1\end{matrix}\right)\]

  52. Loser66
    • one year ago
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    sorry, the first entry of a_1 is 1, not -1 Solving it, we get \(a_1+a_3=-2\\2a_2=6\\a_2+a_3=-1\) Hence \(a_1= 2:a_2=3:a_3=-4\) .This is the first row of the matrix.

  53. Loser66
    • one year ago
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    Do the same with F(4x) and F(1), you will get other 2 rows. Combine all, that is the required matrix.

  54. anonymous
    • one year ago
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    okay i found the solutions for that and the inversible matrix but still got stuck for the vector 6x²-8x-1 . the solution should be (-1,4,-3) but i really can't find it . This exercise drives me nuts :(

  55. Loser66
    • one year ago
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    what is your matrix?

  56. anonymous
    • one year ago
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    the one we just calculated

  57. anonymous
    • one year ago
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    |dw:1437751753106:dw|

  58. Loser66
    • one year ago
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    (-1,4,-3) is the answer for standard base, not for \(\beta\) base Let's apply F to (6x^2-8x-1) P(0) =-1 P'(x) = 12x-8--> P'(1) = 4 \(\int_0^1 (6x^2-8x-1)dx = (2x^3-4x^2-x)|^1_0=-3\) hence, F(6x^2-8x-1)= (-1,4,-3)

  59. anonymous
    • one year ago
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    so u dont need the matrix to solve this? im so confused .Do you know any decent theory and exercises with the same level of difficulty for this? because in my text book it's just so badly explained

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