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Showing a transformation is not linear

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img526.imageshack.us/img526/4926/screenshot20120128at941.png [Broken]

    3. The attempt at a solution
    A linear transformation must satisfy the property: [itex]T(a \vec{X} ) = a T(\vec{X} )[/itex] [itex] \forall X \in V, a \in R[/itex]. However, it is not in general true that [itex](a^2 x^2,a^2 y^2 = a(x^2,y^2)[/itex]. In particular, we can see this fails for [itex]a = 3, x= 4,y=5[/itex]. Indeed,

    [itex](3^2 4^2,3^2 5^2) ≠ 3(4^2,5^2)[/itex]
    [itex](9 \ast 16,9 \ast 25) ≠ 3(16,25)[/itex]
    [itex](144,225) ≠ (48,75)[/itex]

    Thus, this is not a linear transformation.

    Am I doing this right?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 28, 2012 #2

    micromass

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    This is perfect!!
     
  4. Jan 28, 2012 #3
    Good to hear I did the first one right. Thanks! There are four of these problems in total and I want to make sure I know how to do them correctly.

    http://img267.imageshack.us/img267/844/screenshot20120128at956.png [Broken]

    A linear transformation must satisfy the property that [itex]T(\vec{A} + \vec{B}) = T(\vec{A}) + T(\vec{B})[/itex]. Let two vectors in [itex]R^3[/itex] be denoted by [itex](x_1,y_1,z_1),(x_2,y_2,z_2)[/itex]. Thus,

    [itex]T((x_1,y_1,z_1) + (x_2,y_2,z_2)) = T(x_1 +x_2,y_1 + y_2, z_1 + z_2)[/itex]
    [itex]T(x_1 +x_2,y_1 + y_2, z_1 + z_2) = (x_1 +x_2+y_1 + y_2+z_1 + z_2,1)[/itex]

    But, [itex]T(x_1,y_1,z_1) + T(x_2,y_2,z_2) = (x_1+y_1+z_1,1) + (x_2+y_2+z_2,1)[/itex]
    [itex](x_1+y_1+z_1,1) + (x_2+y_2+z_2,1) = (x_1 +x_2+y_1 + y_2+z_1 + z_2,2)[/itex]

    However, [itex](x_1 +x_2+y_1 + y_2+z_1 + z_2,1) ≠ (x_1 +x_2+y_1 + y_2+z_1 + z_2,2)[/itex]. Thus, this is not a linear transformation.
     
    Last edited by a moderator: May 5, 2017
  5. Jan 28, 2012 #4

    micromass

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    Also good!!
     
  6. Jan 28, 2012 #5
    http://img192.imageshack.us/img192/7066/screenshot20120128at100.png [Broken]

    We said earlier that a linear transformation has the property: [itex]T(a \vec{X} ) = a T(\vec{X})[/itex]. In this problem,

    [itex]T(a x) = (1,-1)[/itex]. However, [itex]a T(x) = a (1,-1) = (a,-a)[/itex].

    The equation [itex](1,-1) = (a,-a)[/itex] holds only in the case [itex]a =1[/itex]. Because the equation does not hold for any other number, this is not a linear transformation.
     
    Last edited by a moderator: May 5, 2017
  7. Jan 28, 2012 #6

    micromass

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    Fine!!
     
  8. Jan 28, 2012 #7
    http://img85.imageshack.us/img85/6540/screenshot20120128at101.png [Broken]
    We'll use the same property as in the first and third problems to show this is not a linear transformation.

    [itex]T[a(x,y)] = T(ax,ay) = (a^2 x y,ay,ax)[/itex]. However, [itex]aT(x,y) = a(xy,y,x) = (axy,ay,ax)[/itex]. Again, we observe that [itex](a^2 x y,ay,ax) = (axy,ay,ax)[/itex] will be true only in two particular cases. These are [itex]a = 1,a= 0[/itex]. Because the equation is not true in general, we conclude this is not a linear transformation.
     
    Last edited by a moderator: May 5, 2017
  9. Jan 28, 2012 #8

    micromass

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    Good!! (maybe it's also good to find explicit values of a,x and y such that the equation doesn't hold)
     
  10. Jan 28, 2012 #9
    Okay. I'll add a specific case to show that the equation does not hold. Thanks! My other homework problems are a little bit different, but I'd like to talk about this one in particular:

    http://img220.imageshack.us/img220/7180/screenshot20120128at104.png [Broken]

    First, I considered the idea of multiplication as a binary operator. That is, for well-chosen [itex]a_1,a_2,a_3 \in R[/itex] it is conceivable that [itex]a_1 * a_2 = a_3[/itex]. That is, I am able to input two numbers and get one number as a result. With regard to this question, I said let [itex]T(x) = y[/itex] where [itex]y \in R[/itex]. Thus, using the idea of multiplication as a binary operator, I would have [itex]x * t = y[/itex]. In the cases that [itex]x ≠ 0[/itex] I can solve to find [itex]t = \frac{y}{x} \in R[/itex]. In the case [itex] x=0[/itex], we know that any linear transformation of the zero vector of the domain must map to the zero vector of the codomain. In this case, [itex]t[/itex] can be any such number because [itex]t * \vec{0} = \vec {0}[/itex].

    First of all, is this a proof? It seems conceivable to me that this could work but I'm not entirely sure.
     
    Last edited by a moderator: May 5, 2017
  11. Jan 28, 2012 #10

    micromass

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    Hmmm, I don't quite understand the proof :( So I guess it's not correct.

    Here's a way to prove it: let T(1)=t. Can you now use linearity to show that T(x)=tx??
     
  12. Jan 28, 2012 #11
    :,(
    Alright, letting [itex]T(1) = t[/itex]. We know that [itex]x \in R[/itex]. Thus, we may treat [itex]x[/itex] as a kind of scalar. We can write,

    [itex]T(x) = T(1*x) = xT(1) = xt[/itex]. Thus, the existence of such a [itex]t[/itex] is shown.
     
  13. Jan 28, 2012 #12

    micromass

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    Tht's good!!
     
  14. Jan 29, 2012 #13
    Okay. That's much shorter than what I had previously, so that's an improvement (and it's right, too!). Below is another problem that I'd like my work checked on.

    http://img337.imageshack.us/img337/1043/screenshot20120129at101.png [Broken]

    Let [itex]\left\{ A_1,...,A_n \right\}[/itex] be a basis for V, and [itex]\left\{ B_1,...,B_n \right\}[/itex] be a basis for W. If [itex]T[/itex] is injective, then [itex]T(A_i) ≠ T(A_k)[/itex] where [itex]1≤i,k≤n[/itex]. If [itex]T[/itex] is injective, then [itex]ker(T) = \left\{ 0 \right\} [/itex]. We know, [itex]dim(V) = dim(ker(T)) + dim(Im(T))[/itex]. We write [itex]n = 0 + dim(Im(T))[/itex]. This implies [itex]dim(Im(T)) = n [/itex], and thus, [itex]Im(T) = V[/itex], which means [itex]T[/itex] is surjective.

    Suppose [itex]T[/itex] is surjective. Then [itex]Im(T) = V[/itex]. Thus, [itex]dim(Im(T)) = dim(V)[/itex]. We write, [itex]dim(Im(T)) = n[/itex]. We know, [itex]dim(V) = dim(ker(T)) + dim(Im(T))[/itex]. We write, [itex]n = dim(ker(T)) + n[/itex]. This implies [itex]dim(ker(T)) = 0[/itex], and thus [itex]ker(T) = \left\{ 0 \right\} [/itex]. This means that [itex]T[/itex] is injective.
     
    Last edited by a moderator: May 5, 2017
  15. Jan 29, 2012 #14

    micromass

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    Seems ok!!
     
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