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Homework Help: Linear map

  1. Mar 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the linear map [tex]f:R^2 \rightarrow R^3[/tex], with f(1,2) = (2,1,0) and f(2,1)=(0,1,2)

    2. Relevant equations



    3. The attempt at a solution

    I actually don't understand this task. PLease help! Thank you...
     
  2. jcsd
  3. Mar 24, 2008 #2
    Hi
    If you don't know where to begin, use that a linear map can be written:
    [tex]\[\left(\begin{array}{c}f_{1} & f_{2} & f_{3}\end{array}\right)=
    F\left(\begin{array}{c}x_{1} & x_{2}\end{array}\right)\][/tex]
    Where F is a two-by-three matrix.
     
  4. Mar 25, 2008 #3

    HallsofIvy

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    (1, 2) and (2, 1) are independent vectors and so form a basis for R2. Knowing what a linear map does to a basis tells you what it does to any vector.

    You need to show what f(x,y) is for any pair (x, y). To do that write (x, y) as a linear combination of (1, 2) and (2, 1): That is, find a, b so that a(1, 2)+ b(2, 1)= (x, y). Then f(x,y)= af(1,2)+ bf(2,1).
     
  5. Mar 25, 2008 #4
    2(0,1)+(1,0)=(1,2)
    (1,0)+2(0,1)=(2,1)

    like this?
     
  6. Mar 25, 2008 #5

    HallsofIvy

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    Does that look anything like "a(1, 2)+ b(2,1)= (x,y)"?

    a(1, 2)+ b(2, 1)= (a, 2a)+ (2b, b)= (a+ 2b, 2a+ b)= (x, y) so a+ 2b= x, 2a+ b= y. Solve that for a and b in terms of x and y.
     
  7. Mar 25, 2008 #6
    a=1
    b=1
    x=3
    y=3
    like this?
     
  8. Mar 26, 2008 #7

    HallsofIvy

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    NO! Find a and b for any x and y!!!
     
    Last edited by a moderator: Mar 26, 2008
  9. Mar 26, 2008 #8
    x=1
    y=1

    a+2b=1

    2a+b=1

    should I make system out of those equations?

    Actually, I don't understand the point of this task... I don't know what should I find...
     
  10. Mar 26, 2008 #9

    HallsofIvy

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    Well, I've told you twice already: Solve the equations a+ 2b= x, 2a+ b= y for a and b. you should get a= an expression with x and y in it, b= an expression with x and y in it. Once you have done that you will be able to say that (x, y)= a(1, 2)+ b(2, 1) so that
    f(x,y)= af(1,2)+ bf(2,1) and you know what f(1,2) and f(2,1) are.

    You want to be able to find f(x,y) for any x and y. Do not assume x= y= 1!
     
  11. Mar 26, 2008 #10
    [tex]a=-\frac{x-2y}{3}[/tex]

    [tex]b=\frac{2x-y}{3}[/tex]
    ,
    now what to do next?
     
  12. Mar 26, 2008 #11

    dynamicsolo

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    We've dropped a dimension somewhere, haven't we? How does this map get the third coordinate?

    The approach eys_physics suggested might be more straightforward. Write your argument vector and your resulting vector as either row or column vectors. If you choose row vectors, you have

    [1 2] ยท [first row: a b c , second row: d e f] = [2 1 0] ,

    and similarly for the other set. You will have a 2 x 1 matrix (2-d row vector) times a 2 x 3 mapping matrix giving you a 3 x 1 matrix (3-d row vector). Following the rules of matrix multiplication will give you a set of three equations in two distinct variables (the coefficients of the mapping matrix) for the first transformation shown, and a second set of three equations for the transformation of [2 1] to [0 1 2].

    You now have a pair of equations relating a and d, another pair for b and e, and a third pair for c and f. All of these sets are simple to solve.
     
  13. Mar 27, 2008 #12
    Actually, I don't know why you multiply by [1 2] and why it is equal to [2 1 0]
     
  14. Mar 27, 2008 #13

    HallsofIvy

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    Excellent!

    Haven't we already been through what a "basis" is? R2 has dimension 2 and, since (1, 2) and (2, 1) are two independent vectors (one is not a multiple of the other) they form a basis. Every vector in can be written as a linear combination of those two vectors. You have now determined how they can be written: for any x, y,
    [tex](x,y)= \frac{2y-x}{3}(1, 2)+ \frac{2x-y}{3}(2, 1)[/tex]
    Since f is a linear map, by definition of "linear",
    [tex]f(x,y)= \frac{2y-x}{3}f(1, 2)+ \frac{2x-y}{3}f(2, 1)[/tex]
    Now, you are told in the original problem that f(1,2)= (2, 1, 0) and that f(2, 1)= (0, 1, 2).

    [tex]f(x,y)= \frac{2y-x}{3}(2, 1, 0)+ \frac{2x-y}{3}(0, 1, 2)[/tex]
    [tex]= \left(\frac{4y- 2x}{3},\frac{2y-x}{3}+ \frac{2x-y}{3},\frac{4x- 2y}{3}\right)[/tex]
    [tex]= \left(\frac{4y-2x}{3},\frac{x+y}{3},\frac{4x-2y}{3}\right)[/tex]
     
  15. Mar 27, 2008 #14
    Ok, I understand now how you find it. But I can't understand what we do actually to find it.
    f(x,y)= af(1,2)+ bf(2,1)

    f(x,y)= a(2,1,0)+ b(0,1,2)

    we actually make linear combination of the 2 vectors in the basis, why?
     
  16. Mar 27, 2008 #15

    HallsofIvy

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    Because they were the only ones ones for which we knew what f does!

    I hope that, by this time, you understand what a basis is.
     
  17. Mar 27, 2008 #16
    Is it (x,y)= a(2,1,0)+ b(0,1,2) or (x,y)= a(1,2)+ b(2,1)
     
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