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.