Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear transformation being onto or not (Three problems I need help understanding)

  1. Nov 14, 2011 #1
    Which of the transformations are onto?

    1) T:R2 -> R2, where T(x,y) = (5x-y, 0)

    I don't know if I'm understanding this correctly but this transformation is NOT onto because if I let
    5x-y = a
    0 = b

    this means that b doesn't cover all the range of T? Could someone explain it better if I'm wrong.

    2) T:R3 -> R2 where T(x,y,z) = (x+y, x-z)
    so I equate this to
    x+y = a
    x-z = b
    which makes the matrix
    row1 = (1 1 0 a)
    row 2 = (1 0 -1 b)
    then once I reduce it to row echelon form, I ultimately get
    row 1 = (1 0 -1 b)
    row 2 =( 0 1 1 a-b)

    then I get stuck because I don't understand what that means..I'm going to guess it's NOT onto because for any value a and b, I can't get and x,y, or z?

    Please correct me if I'm wrong.

    3) T:R2 -> R3, where T(x,y) = (y, x, x-y)
    I equate this again to
    y = a
    x= b
    x-y = c
    which forms the matrix
    row 1 = (0 1 a)
    row 2 =(1 0 b)
    row 3 = (1 -1 c)
    then perform row echelon to ultimately get
    row 1 = (1 0 b)
    row 2 = (0 1 a)
    row 3 = (0 -1 -b+c)
    I think this is NOT onto again because of the last row but I can't be 100% sure. Any guidance would help a lot.

    I know this is much to ask but if you can even help me with just one, it would mean a lot. Thank you for any help.
  2. jcsd
  3. Nov 15, 2011 #2


    User Avatar
    Science Advisor

    Re: Linear transformation being onto or not (Three problems I need help understanding

    there's different ways to approach these problems.

    1) your reasoning here is correct, because the second coordinate of T(x,y) is always 0, so if we have T(x,y) = (a,b), and b ≠ 0, then T(x,y) = (a,b) has no solution. for example, there is no (x,y) such that T(x,y) = (0,1).

    2) we can use the same approach as (1):

    x+y = a
    x-z = b

    from the first equation, y = a-x.
    from the second, z = x-b, so (x,a-x,x-b) = x(1,-1,1) + (0,a,-b) should be a solution, for any x. we only need one, so let's choose x = 0.

    T(0,a,-b) = (0+a,0-(-b)) = (a,b), no matter what a and b are. we have specifically exhibited a pre-image (in fact, a lot of them, depending on what we choose for x) for (a,b), so T is onto.

    if we want to use matrices, we need to write T in terms of some basis. the basis {(1,0,0),(0,1,0), (0,0,1)} works pretty well. the first column for T is T(1,0,0) = (1,1). the second column is T(0,1,0) = (1,0), and the third column is T(0,0,1) = (0,-1). so the augmented matrix is A' =

    [tex]\begin{bmatrix}1&1&0&a\\1&0&-1&b \end{bmatrix}[/tex]

    as you correctly stated. however, this reduces to:

    [tex]\begin{bmatrix}1&0&-1&b\\0&1&1&a-b \end{bmatrix}[/tex]

    what does this mean? it means that any solution (x,y,z) to T(x,y,z) = (a,b),

    also satisfies:

    [tex]\begin{bmatrix}1&0&-1\\0&1&1\end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}b\\a-b \end{bmatrix}(*)[/tex]

    that is: x-z = b, and y+z = a-b. since we only have 2 rows, we can eliminate z from our equations (say, assigning it the value t), so that we get:

    x = b+z = b+t
    y = a-b-t, that is: (x,y,z) = (b+t,a-b-t,t) = t(1,-1,1) + (b,a-b,0).

    and, indeed T(t(-1,1,1) + (b,a-b,0)) = tT(1,-1,1) + T(b,a-b,0) = t(0,0) + (b+a-b,b-0) = (a,b).

    but (*) tells us more. it tells us that rank(T) = 2, and since dim(R2) = 2 = rank(T) = dim(im(T)), im(T) is a 2-dimensional subspace of R2, that is, im(T) is ALL of R2, so T is onto (we don't even have to actually calculate the solutions to see this).

    we also see that the nullspace of A = ker(T) is dependent only on the single parameter t, and that it has the basis {(1,-1,1)} (this is the solution when a = b = 0), so dim(ker(T)) = nullity(A) = 1. this illustrates a general principal of linear algebra:

    general solution = homogeneous solutions + specific solution.

    note that our two methods give the same basis for ker(T), but two different "specific" solutions:

    (b,a-b,0) and (0,a,-b). but these are both examples of the general solution:

    t(1,-1,1) + (b,a-b,0). for t = 0, we get (b,a-b,0). for t = -b, we get (-b,b,-b) + (b,a-b,0) = (0,a,-b).

    look at problem 3 again, in light of this.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook