1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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
    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
  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
    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
    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
    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
    Seems ok!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook