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Basis for Range of Linear transformation

  1. Oct 29, 2012 #1
    The problem is attached. The problem is "find a basis for the range of the linear transformation T."



    p(x) are polynomials of at most degree 3. R(T)={p''+p'+p(0) of atmost degree 2}

    This is pretty much as far as I got. I'm not sure how to do the rest.


    I'm thinking of picking a random function, let's say:
    ax^3+bx^2+cx+d

    p'(x)=3ax^2+2bx+c
    p''(x)=6ax+2b

    p"+p'+p(0)=(6ax+2b)+(3ax^2+2bx+c)+(d)

    The book says the answer is {1, x, x^2}. Is it because in the line above, you can see that x^2, x are in there and 1 is there in the form of d?
     

    Attached Files:

  2. jcsd
  3. Oct 29, 2012 #2

    Bacle2

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    Science Advisor

    Have you tried seeing the effect of R(T) on a basis for P3?
     
  4. Oct 29, 2012 #3
    What does the transformation due to each elmeent in the basis?
     
  5. Oct 29, 2012 #4
    Yes, I did, but I am not getting there answer.

    Here is my modified procedure (ignore original post, i didn' tknow what i was doing).


    A basis for P3 is {1, x, x^2, x^3}

    T(1)=0+0+1=1
    T(x)=0+1+0
    T(x^2)=2+2x+0
    T(x^3)=6x+3x^2+0



    This is basically
    T(1)=1+0x+0x^2+0x^3
    T(x)=1+0x+0x^2+0x^3
    T(x^2)=2+2x+0x^2+0x^3
    T(x^3)=0+6x+3x^2+0x^3
    Where these coefficients represent the column for the transformation matrix:
    1 1 2 0
    0 0 2 6
    0 0 0 3
    0 0 0 0

    rref of this matrix is
    1 1 0 0
    0 0 1 0
    0 0 0 1
    0 0 0 0

    There's a pivot in the 1st, 3rd, and 4th column which would correspond to: 1, x^2, x^3 as the basis.


    How did they get 1, x, x^2?
     
  6. Oct 29, 2012 #5

    Mark44

    Staff: Mentor

    You have:
    T(x3) = 3x2 + 6x
    T(x2) = 2x + 2
    T(x) = 1
    T(1) = 1

    Every output "vector" is a linear combination of 1, x, and x2.
     
  7. Oct 29, 2012 #6
    Yes, but why does my rref form of the matrix say it's 1, x, x^3?

    The pivot columns correspond to {1, x,x^3}.


    Wait....I think I got this problem confused with another problem. I put the coefficients in the columns instead of rows.

    So the matrix should look like this
    1 1 2 0
    0 0 2 6
    0 0 0 3
    0 0 0 0 excepted it should be transposed.


    Is that right?
     
    Last edited: Oct 29, 2012
  8. Oct 29, 2012 #7

    Bacle2

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    Science Advisor

    Notice there is no multiple of x^3 in the image of T, so your matrix should

    not correspond to {1,x,x^3}..
     
  9. Oct 29, 2012 #8
    Yeah, I did my matrix wrong (did it in columns instead of rows.
     
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