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Show that T preserves scalar multiplication - Linear Transformations

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Let T:ℝ[itex]^{2}[/itex]→ℝ be defined by
    T[tex]\left(\begin{array}{c} x_{1} \\x_{2}\end{array}\right)[/tex] = (0 if x[itex]_{2}[/itex] = 0. [itex]\frac{x^{3}_{1}}{x^{2}_{2}}[/itex] otherwise.)

    Show that T preserves scalar multiplication, i.e T(λx) = λT(x) for all λ [itex]\in[/itex] ℝ and all x [itex]\in[/itex] ℝ[itex]^{2}[/itex]
    3. The attempt at a solution

    T(λx) = T[tex]\left(\begin{array}{c} (λx_{1}) \\(λx_{2})\end{array}\right)[/tex] = (λ0 = 0 if x[itex]_{2}[/itex] = 0, or [itex]\frac{(λx_{1})^{3}}{(λx_{2})^{2}}[/itex])
    = λT[tex]\left(\begin{array}{c} x_{1} \\x_{2}\end{array}\right)[/tex] = λ0 = 0 if x[itex]_{2}[/itex] = 0, or
    λ*[tex]\left(\begin{array}{c} (x_{1})^{3} \\(x_{2})^{2}\end{array}\right)[/tex]

    Is that a correct proof?

    It's a bit hard to read because whenever I try to put a vector, it puts it into a new line.
     
  2. jcsd
  3. Oct 5, 2011 #2

    HallsofIvy

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    It's putting new lines wherever you try to put a vector because you are putting "bits and pieces" inside [ tex ] tags. Don't do that. Put entire equations inside [ tex ] tags.

    Also do "[itex]x_2= 0[/itex]" and "[itex]x_2\ne 0[/itex]" separately. You have "[itex]x_2= 0[/itex]" in two different places which makes it hard to read.

    If [itex]x_2= 0[/itex], then [itex]\lambda x_2= 0[/itex] for any [itex]\lambda[/itex] so [itex]T(\lambda x)= 0= \lambda T(x)[/itex].

    If [itex]x_2\ne 0[/itex], then [itex]\lambda x_2\ne 0[/itex] for any [itex]\lambda[/itex] except [itex]\lambda= 0[/itex] so you should do two separate cases, [itex]\lambda= 0[/itex] and [itex]\lambda\ne 0[/itex], here.
     
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