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Homework Help: Linear transformation

  1. Dec 1, 2015 #1
    1. The problem statement, all variables and given/known data
    R2 to R1 f(x,y)=xy
    Determine if the transformation is linear or not

    2. Relevant equations
    T(V1+V2) = T(V1) + T(V2)
    T(cV1) = Tc(V1)
    3. The attempt at a solution
    If the function f(x,y) = xy we can define another function f(a,b)=ab
    Therefore, f(x,y) = f(a,b), so xy=ab, so all variables could = each other...

    T(x*a + y*b) = T(xy) + T(ab)

    Txa+Tyb = Txy + Tab

    I'm not really sure if I'm on the right track here. Can anyone help me? I need to know the process for finding whether a transformation is linear or not. Usually, I write another function with a vector similar to the original one, in this case it's ab, then i prove it's one-to-one, then try and prove the other 2 equations. Linear transformations are so difficult and it's so hard to find a good resource online that teaches it simply!
  2. jcsd
  3. Dec 1, 2015 #2


    User Avatar
    Homework Helper

    For your first property, you have vectors (x,y) and (a,b). If the transformation f were linear, then you could say that f ( (x,y)+ (a,b) ) = f(x,y) + f(a,b).
    This is not about whether it could be true, it is about whether or not it is always true.
    Remember that the sum of two vectors (x,y) + (a,b) = (x+a, y+b).

    For the scalar multiple, that is c(x,y) = (cx, cy) . So you would test if f(c(x,y) ) = c f(x,y).

    You should find that neither of these hold true.
  4. Dec 1, 2015 #3


    Staff: Mentor

    Why are you doing this? You have not defined another function. All you have done is use different input arguments.
    ?? Use the function you started with -- f. You are doing nothing but confusing yourself by switching to T.
    There is a further confusion in not being clear what x and y represent. When you write T(x*a + y*b), x and y are apparently vectors in ##\mathbb{R}^2##, but in the function definition -- f(x, y) = xy -- x and y are the real number components of a vector in ##\mathbb{R}^2##.

    You absolutely need to be clear on the difference between a vector and its components.
    Let ##\vec{u} = <u_1, u_2>## and ##\vec{v} = <v_1, v_2>## be vectors in ##\mathbb{R}^2##
    What is ##f(\vec{u})##?
    What is ##f(c\vec{u})##?
    What is ##f(a\vec{u} + b\vec{v})##?
  5. Dec 1, 2015 #4
    I like to make things look familiar. I know f and T are the same thing, but I prefer to use T because I'm doing a linear transformation.
    I need to prove f(x,y)=xy is linear

    I do that by showing f(V1 + V2) = f(V1) + f(V2) but in the problem statement I only have one vector: xy, so I created a second vector, ab, so I can prove the equation.
  6. Dec 1, 2015 #5
    The first additive rule for linear transformations states that T(x+y) = T(x) + T(y)

    In just about every problem I've come across I've only been given one vector. x+y, X+1, and I read the textbook and it says to define another vector so you can prove it's one-to-one, and that the additive and cumulative qualities hold.
  7. Dec 1, 2015 #6


    Staff: Mentor

    You're cluttering up your work by using different names than are in the problem. This clutter has the downside of possibly confusing you.

    xy is NOT a vector. It is a number.
    As I said in a previous post, you need to be able to distinguish between the components of a vector (x and y) and the vector itself: ##\begin{bmatrix} x \\ y\end{bmatrix}##
    Are you talking about multiplication of a vector by a scalar?
    See what I wrote at the end of post #3.
  8. Dec 1, 2015 #7
    Let u⃗ =<u1,u2> and v⃗ =<v1,v2> be vectors in R2
    What is f(u⃗ )?
    What is f(cu⃗ )?
    What is f(au⃗ +bv⃗ )?

    f(u) = u1*u2
    f(cu) = f(c(u1*u2))
    f(a(u1*u2) + b(v1*v2))

    I don't understand what the a and b is in the last bit?

    f(x,y)=xy is a function, and NOT a vector.

    Does this mean, for this transformation to be linear:

    f(V1+V2) = f(x,y) = xy
    f(V1)+f(V2) = f(x,y) = xy
    f(ca) = fc(a) = f(x,y) = xy

  9. Dec 1, 2015 #8
    Do I then have to create another function that is equivalent to f(x,y)=xy to prove if it's linear?
    I've stopped using V because it's making me think of vectors and i've started using f!

    W1 = f(x,y) = xy
    W2 = f(a,b) = ab

    f(W1+W2) = f(W1) + f(W2)

    (x+a) * (y+b) ≠ x*y + a*b

    I checked by using a=1 b=2 x=3 y=4

    (3+1)*(4*2) ≠ 3*4 + 1*2
    21 ≠ 14
    Therefore, transformation is not linear.
    Is this it? I think I'm finally starting to get it, maybe...
  10. Dec 1, 2015 #9


    Staff: Mentor

    Yes, assuming that u = <u1, u2>
    No. u is a vector, as is cu. Calculate cu first (IOW, find the components of cu), then apply your function to it.
    f is a function. xy is a number.
    None of these makes any sense.
  11. Dec 1, 2015 #10
    I think my post above this with the original problem makes more sense
  12. Dec 1, 2015 #11


    Staff: Mentor

    No, you only have the one function, f.
    This doesn't make much sense. In part you're saying that W1 = xy and W2 = ab, which means that W1 and W2 are numbers.
    In the line below you have f(W1 + W2), which implies that W1 and W2 are vectors. Which is it? They can't be both vectors and numbers.
    I don't think so.

    Let u = <u1, u2>, and v = <v1, v2>
    Hopefully it's clear that when you see u, we're talking about a vector, but when you see u2, we're talking about one of the components of u.

    What is u + v?
    What is f(u + v)?
    What is cu? (c is a scalar)
    What is f(cu)?
  13. Dec 1, 2015 #12
    u + v = u1*v1 + u2*v2
    f(u + v) = u1*u2 + v1*v2
    cu = c(u1*u2)
    f(cu) = c*u1*u2
  14. Dec 1, 2015 #13


    Staff: Mentor

    NO! This is just plain old vector addition!
    NO! This is nothing more than multiplication of a vector by a scalar.
    You're not going to get this if you don't understand the very basic concepts of vector addition and scalar multiplication.
    If ##\vec{u} = <u_1, u_2>## and ##\vec{v} = <v_1, v_2>## then ##\vec{u} + \vec{v} = <u_1 + v_1, u_2 + v_2>##
    Also, if c is a scalar, then ##c\vec{u} = <cu_1, cu_2>##
    I hope that this looks familiar...
  15. Dec 1, 2015 #14
    where does f(x,y) = xy come into this problem though? I understand the mistakes I made above, but they were because I'm trying to incorporate the original function into what you wrote...
  16. Dec 1, 2015 #15


    Staff: Mentor

    It doesn't come into play yet, because you're still having problems working with vectors -- i.e., adding two vectors and multiplying a vector by a scalar. Until you get these concepts down, you won't have any success working with functions that take vectors as arguments.
  17. Dec 1, 2015 #16
    u + v = (u1+v1 , u2+v2)
    f(u + v) = (u1+u2 + v1+v2)
    cu = c(u1,u2)= (cu1,cu2)
    f(cu) = (cu1,cu2)
  18. Dec 1, 2015 #17


    Staff: Mentor

    No. Look at how f is defined.
    No. Look at how f is defined.
  19. Dec 1, 2015 #18

    f(u+v) = u1*u2+v1*v2

    f(cu) = f(c(u1*u2)
  20. Dec 1, 2015 #19
    Made an edit to the above, which you already said was incorrect -- I'm now sure how though...
    If f(x,y) = xy

    and i substitute the components of vector u (u1,u2) and vector v (v1,v2) into this function, then I get the answer I got.
  21. Dec 1, 2015 #20


    Staff: Mentor

    Both of the above are wrong. You are skipping steps like crazy, and writing stuff without thinking about it.

    You already told me what u + v was; namely u + v = ##<u_1 + v_1, u_2 + v_2>##, and what cu was; namely, cu = ##<cu_1, cu_2>##
    What is f(##<u_1 + v_1, u_2 + v_2>##)?
    What is f(##<cu_1, cu_2>##)?
  22. Dec 1, 2015 #21
    f(x,y) = xy

    f(u1+v1,u2+v2) = (u1+v1)*(u2+v2)
    f(cu1,cu2) = cu1*cu2
  23. Dec 1, 2015 #22

    Ray Vickson

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    Science Advisor
    Homework Helper

    The first one is wrong; the second one just displays the vector cu explicitly, but without actually saying anything.

    You are told that if you have a vector ##\vec{v} = (\text{thing 1}, \text{thing 2})## then ##f(\vec{v}) = \text{thing 1} \times \text{thing 2}##. OK so far?

    Now, what would you get if ##\text{thing 1} = u_1 + v_2## and ##\text{thing 2} = u_2 + v_2##? What would you get if ##\text{thing 1} = c u_1## and ##\text{thing 2} = c u_2##?
  24. Dec 1, 2015 #23


    Staff: Mentor


    So f(u + v) = f(##<u_1 + v_1, u_2 + v_2>) = (u_1 + v_1) \cdot (u_2 + v_2)##
    Is this equal to f(u) + f(v)?

    And f(cu) = f(##<cu_1, cu_2>) = cu_1 \cdot cu_2##
    Is this equal to cf(u)?
  25. Dec 1, 2015 #24


    Staff: Mentor

    Ray, he corrected these in post #21.
  26. Dec 1, 2015 #25
    f(u + v) = f(<u1+v1,u2+v2>)=(u1+v1)⋅(u2+v2)

    f(u) = u1*u2
    f(v) = v1*v2

    No, this is not equal.

    f(cu) = f(<cu1,cu2>)=cu1⋅cu2


    This is not equal either. Therefore it is not a linear transformation.
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