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Linear Transformation l:R3 to R2

  1. Jan 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that there exists only one linear transformation l: R3 to R2 such that:
    l(1,1,0) = (2,1)
    l(0,1,2) = (1,1)
    l(2,0,0) = (-1,-3)

    Find Ker(l), it's basis and dimension. Calculate l(1,2,-2)

    2. Relevant equations


    3. The attempt at a solution
    I still find linear transformations really confusing. Something about the notation and what is being asked, I don't know...

    The matrix associated to these linear transformations should be a 2x3 right?

    L=AX

    where L = linear transformation
    A = matrix associated to linear transformation
    X = vector

    so, (1,1,0) = A*(2,1)
    (0,1,2) = A* (1,1)
    (2,0,0) = A*(-1,-3)

    And there will be ONE matrix that is associated to all of these transformations? If this is correct, how can I find A?
     
  2. jcsd
  3. Jan 5, 2016 #2
    I've made a linear combination of
    l(1,1,0)
    l(0,1,2)
    l(2,0,0)

    and got the matrix:

    [ 1 0 2 ]
    [ 1 1 0 ]
    [ 0 2 0 ]

    so if I use the equation from my post above (L=AX) (I've formatted it below so L = X*A)

    [ 1 0 2 ]. .[ 2 -1 ]
    [ 1 1 0 ] = [ 1 1 ] * 2x3 MATRIX
    [ 0 2 0 ]. .[ 1 -3 ]

    I'm not sure how to find the matrix associated to this transformation.
    I know Ax = b, and if I find the inverse of A, I can find x, but I'm not sure how to solve the above...
     
  4. Jan 5, 2016 #3

    Ray Vickson

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    Personally, I would avoid using a matrix and just proceed directly.

    If e1=(1,0) and e2 = (0,1) are the standard basis in R2 and f1=(1,0,0), f2 = (0,1,0), f3=(0,0,1) are the standard basis in R3, then you set t1 = L(f1), t2=l(f2), t3 = l(f3) you are given that
    l(1,1,0) = l(f1+f2) = t1+t2 = (2,1) = 2e1+e2, so that t1+t2 = 2 e1 + e2.

    You can find two other similar equations by using the other two pieces of information about l. So now you will have three linear equations in the three unknowns t1, t2 and t3. Just figure out if they have a unique solution.
     
  5. Jan 5, 2016 #4
    We're been taught to use matrices. I know I can avoid it but I want to use them just so I can tie it all in with everything else.
     
  6. Jan 5, 2016 #5
    I tried your way, but I just followed the pattern and don't really understand what I've actually done.

    l(1,1,0) = l(f1+f2) = t1+t2 = (2,1) = 2e1+e2 ---> t1+t2 = 2e1+e2
    l(0,1,2) = l(f2+2f3) = t2+2t3 = (1,1) = e1 + e2 --> t2+2t3 = e1 + e2
    l(0,0,1) = l(f3) = t3 = (-1,3) = -1e1 +3e2 --> t3 = -1e1+3e2
     
  7. Jan 5, 2016 #6

    HallsofIvy

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    Any linear transformation, L, from R3 to R2 can be written as L(x, y, z)= (ax+ by+ cz, dx+ ey+ fz). Because
    L(1,1,0) = (2,1) we must have a+ b= 2, d+ e= 1
    Because L(0,1,2) = (1,1) we must have b+ 2c= 1, e+ 2f= 1
    Because L(2,0,0) = (-1,-3) we must have 2a= -1, 2d= -3.
    Solve those 6 equations for a, b, c, d, e, f.

    This is exactly the same as writing the linear transformation as the matrix [tex]\begin{bmatrix}a & b & c \\ d & e & d\end{bmatrix}[/tex].
     
  8. Jan 5, 2016 #7

    Ray Vickson

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    l(1,1,0) = (2,1) ==> t1+t2 = 2e1+e2 --- (1)
    l(0,1,2) = (1,1) ==> t2+2t3 = e1+e2 ---- (2)
    l(2,0,0) = (-1,-3) ==> 2 t1 = -e1-3 e2 --- (3)
    Thus, t1 = -(1/2)e1 - (3/2)e2 from (3). Putting this into (1) gives t2 = 2e1 + e2 -[-(1/2)e1 - (3/2) e2] = (5/2)e1 + (5/2) e2 = (5/2)(e1+e2)
    Now (2) gives 2t3 = e1+e2 - (5/2)(e1+e2) = -(3/2)(e1+e2) ==> t3 = -(3/4)(e1+e2)

    Aside from getting one of the equations wrong, what you are doing is simplicity itself: you are just using the basic properties (actually, definitions) of linear transformations: l(a*v) = a*l(v) for a number 'a' and a vector 'v'; and l(v1+v2) = l(v1) + l(v2) for two vectors v1 and v2. That's all there is to it!
     
  9. Jan 5, 2016 #8
    I guess the problem I'm having is visualizing it. I know we can think of linear transformations as say stretching or shrinking something, but it's hard as well to think of something in two dimensions being changed to something in three dimensions, or vise versa. Don't get me started on four dimensions...
     
  10. Jan 5, 2016 #9

    Mark44

    Staff: Mentor

    Well, you need five dimensions to fully visualize the transformation of this problem: three dimensions for the domain, and two more dimensions for the codomain. The transformation maps a vector in space (##\mathbb{R}^3##) to one in the plane (##\mathbb{R}^2##). The only way I can think of to visualize this is with a small three-D region for the domain, and a separate two-D region for the codomain, with a curved arrow going from the first to the second.
     
  11. Jan 6, 2016 #10

    vela

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    A linear transformation maps vectors in the domain to vectors in the codomain. You could express this as ##\vec{y} = L(\vec{x})##, where ##\vec{x}## is an element of the domain and ##\vec{y}## is an element in the codomain. Writing "L=AX" doesn't make sense.

    It turns out that you can associate the linear transformation L with a matrix A where ##\vec{y} = A\vec{x}##.

    Here you have the domain and codomain backwards. You were given, for example, that L(1,1,0) = (2,1). In terms of the matrix A, that means
    $$\begin{pmatrix} 2 \\ 1 \end{pmatrix} = A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}.$$ To allow the matrix multiplication to work out and get a 2x1 result, A has to be 2x3.

    If you express A the way HallsofIvy did, the one matrix equation above is equivalent to two regular equations, namely
    \begin{align*}
    2 &= a\cdot 1 + b \cdot 1 + c \cdot 0 \\
    1 &= d \cdot 1 + e \cdot 1 + f \cdot 0
    \end{align*} Do the same thing with the other pairs of vectors, and you'll end up with six equations in total. What you want to show is that these six equations have only one solution.
     
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