# Homework Help: Find out if it's linear transformation

1. Oct 16, 2016

### Kernul

1. The problem statement, all variables and given/known data
Does a linear transformation $g : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ so that $g((2, -3)) = (5, -4)$ and $g((-\frac{1}{2}, \frac{3}{4})) = (0, 2)$ exist?

2. Relevant equations

3. The attempt at a solution
For a linear transformation to exist we need to know if those two satisfy the following conditions:
$g(\vec v_1 + \vec v_2) = g(\vec v_1) + g(\vec v_2)$ for any $\vec v_1, \vec v_2$ in $\mathbb{R}^2$
$g(\alpha \vec v) = \alpha g(\vec v)$ for any scalar $\alpha$
So I thought I should do like this:
I sum the two vectors $(2, -3)$ and $(-\frac{1}{2}, \frac{3}{4})$ like this $g((2, -3) + (-\frac{1}{2}, \frac{3}{4}))$ and put it equal to the sum of $(5, -4)$ and $(0, 2)$. I will then have something like this:
$g((2, -3) + (-\frac{1}{2}, \frac{3}{4})) = (5, -4) + (0, 2)$
$g((\frac{3}{2}, -\frac{9}{4})) = (5, -2)$
But at this point I actually don't know if it really exist or not. Am I missing something?

2. Oct 16, 2016

### PeroK

Do you notice anything about the two vectors being acted on by $g$?

3. Oct 16, 2016

### Kernul

Oh! If I multiply by $-4$ the second vector acted on by $g$, I'll get the first vector! But I have to multiply the vector that is is equal to by $-4$ too. And since $(0, -8)$ is not equal to $(5, -4)$ means that the linear transformation doesn't exist, right?

4. Oct 16, 2016

Exactly.