Show that the transformations $T,R,S \in \mathscr{L}(\mathbb{R}^2)$ given by: $T(x,y)=(x,2y), \ R(x,y)=(x,x+y), \ S(x,y)=(0,x)$ are linearly independent.
By definition $T,R,S$ are l.i. if $\forall (x,y)$, $(\alpha T + \beta R + \gamma S)(x,y)=0 \iff \alpha = \beta = \gamma = 0$. But taking the vector (1,0), for example, this condition is not satisfied. What is wrong here?
The condition for linear independence (correct as you described) leads to the condition: $$ \alpha(x,2y) + \beta(x,x+y) + \gamma(0,x) = ((\alpha + \beta) x,(2 \alpha + \beta) y + (\beta + \gamma) x) = (0,0) $$ To be true for all $x,y$ we require $\alpha + \beta = 0$, $2\alpha + \beta = 0$ which implies that $\alpha = \beta = 0$. This shows that $\gamma = 0$ as well. Hence they are linearly independent.