Injectivity of a Lie Bracket Isomorphism I am having an issue with a Lie bracket isomorphism.

My issue lies with a). I have to show that $\phi$ is a bijective linear map. As a mapping $\Phi: \mathbb{R}^3 \to \Phi(\mathbb{R}^3)$ is certainly surjective. In order to show injectivity I have to take $(a_1,b_1,c_1),(a_2,b_2,c_2) \in \mathbb{R}^3$ and suppose $\Phi(a_1,b_1,c_1)=\Phi(a_2,b_2,c_2)$. This implies
$$
(a_1-a_2)X+(b_1-b_2)Y+(c_1-c_2)Z=0 
$$
If $X,Y,Z$ were linearly independent injectivity would follow immediately. So we have to show linear independence. Consider
$$
0=\lambda_1X+\lambda_2Y+\lambda_3Z 
=(\lambda_3 y-\lambda_2 z) \frac{\partial}{\partial x} 
+(\lambda_1 z-\lambda_3 x) \frac{\partial}{\partial y} 
+(\lambda_2 x-\lambda_1 y) \frac{\partial}{\partial z}   
$$
which leads to the three equations
$$
\lambda_3 y-\lambda_2 z=0 \\
\lambda_1 z-\lambda_3 x=0 \\
\lambda_2 x-\lambda_1 y=0
$$
and therefore
$$
\lambda_1 (z-y)+\lambda_2 (x-z)+\lambda_3 (y-x)=0
$$
Unfortunately, the equation above also holds if $\lambda_1=\lambda_2=\lambda_3=1$, so my strategy fails. So I am confused about how to show injectivity.
 A: The thing is that there is no equivalence between the three equations you have found out and their sum, just an implication. What you have shown is that $(1,1,1)$ is a potential solution, but you do not have shown that it really is one.
Clearly, $\Phi$ is linear. Suppose that $\Phi(a,b,c) = 0$: in other words, assume that
$$
\forall (x,y,z)\in \Bbb R^3, \quad \begin{cases}
cy-bz &= 0\\
az - cx &= 0\\
bx - ay &= 0
\end{cases}
$$
Alternatively setting $(x,y,z) = (1,0,0)$, and $(x,y,z) = (0,1,0)$, we find that a solution $(a,b,c)$ has to satisfy
$$
\begin{cases}
-c &= 0,\\
b &= 0,
\end{cases} \text{ and } \begin{cases} c &= 0,\\
-a &= 0\end{cases}.
$$
Hence, it is mandatory that $a=b=c=0$. The converse is obviously true since $\Phi$ is linear, and therefore, $\Phi(a,b,c) = 0 \iff (a,b,c) = 0$, and $\Phi$ is injective.
A: $\Phi$ is linear so it is enough to show that $\ker \Phi=0.$ Now,
$aX+bY+cZ=0$ leads to
$\tag 1 cy-bz=az-cx=bx-ay=0.$
If the claim is false, then we should be able to find a non-zero vector $(a,b,c)$ such that $(1)$ is true $\textit{for all}$ values of $(x,y,z).$
Toward a contradiction then, without loss of generality, $a=1.$ Then, $y=bx$ and $z=cx$.
But, choosing $\vec x=(1,1,1)$ implies that $b=c=1$ whereas choosing $\vec x=(1,0,0)$ gives $b=c=0,$ so in fact there is no such vector $(a,b,c).$
