Show that T preserves scalar multiplication - Linear Transformations

[NewtonianAlch]

Homework Statement

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

Show that T preserves scalar multiplication, i.e T(λx) = λT(x) for all λ \in ℝ and all x \in ℝ^{2}

The Attempt at a Solution

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

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.

 

[HallsofIvy]

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 "x_2= 0" and "x_2\ne 0" separately. You have "x_2= 0" in two different places which makes it hard to read.

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

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