Killing Fields on Euclidean Spaces Let $K$ be a set of all Killing vector fields on $\mathbf R^n$ (with the Euclidean metric $\bar g$) which vanish at the origin.
(A vector field $V$ on a Riemannian manifold $(M, g)$ is said to be a Killing vector field if the flow of $V$ acts by isometries of $M$. This is equivalent to saying that $\mathcal L_Vg=0$).
If $V\in K$, then by using $\mathcal L_V\bar g=0$, we get that the matrix $[\partial V^i/\partial x^j]$ is anti-symmetric, where $V^i$ are the components of $V$ in the standard coordinates.
Define a map $T:K\to \mathfrak o(n)$ as
$$T(V)= \left[\frac{\partial V^i}{\partial x^j}(0)\right]$$
where $\mathfrak o(n)$ is the Lie algebra of $O(n)$, which is "same" as the space of $n\times n$ real anti-symmetric matrices.

Problem. To show that $T$ is injective.

I am quite lost here.
 A: We need the following lemma, whose proof will be at the bottom of this answer. 
Lemma: If $\phi, \psi$ are Riemannian isometries on open subsets $U \to V$ of $\mathbb R^n$, with $\phi(p) = \psi(p)$ and $d\phi_p = d\psi_p$ for some $p \in U$, then $\phi = \psi$. 
Now, suppose $V$ is a Killing vector field vanishing at $0$ in the kernel of the map $T : K \to \mathfrak o(n)$. Since $V$ vanishes at $0$, if $\theta$ is the flow of $V$, $\theta_t(0) = 0$ for every $t$ in the flow domain. We claim $d(\theta_t)_0$ is independent of $t.$ For then, by the lemma, $\theta_t = \mathrm{Id}$ on all of $\mathbb R^n$, and so $V = 0$. 
Let $\theta_t^i$ be the $i^\textrm{th}$ component of the flow $\theta_t$. Then $\dfrac{d}{dt}\bigg|_{t=0} \theta^i_t(x) = V^i(x)$ for $x \in \mathbb R^n$, where $V^i$ are the component functions of $V$. To prove the claim, it suffies to show that $\dfrac{\partial \theta_t^i}{\partial x^j}(0)$ is independent of $t$ for every $i,j$. This follows from Clairaut's theorem: for every $t_0$ in the flow domain, we have
$$
\frac{d}{dt}\bigg|_{t=t_0} \frac{\partial}{\partial x^j} \theta_t^i(0) = \frac{\partial}{\partial x^j} \frac{d}{dt}\bigg|_{t=t_0} \theta_t^i(0) = \frac{\partial}{\partial x^j} \frac{d}{dt}\bigg|_{t=0} \theta^i_t(\theta_{t_0}(0)) = \frac{\partial}{\partial x^j} V^i(\theta_{t_0}(0)) = \frac{\partial V^i}{\partial x^j}(0) = 0
$$
since $V \in \ker(T)$. Therefore $\dfrac{\partial \theta_t^i}{\partial x^j}(0)$ is independent of $t$ for every $i,j$, hence so is $d(\theta_t)_0$, so $V=0$ by the lemma, so $T$ is injective. QED. 
Proof of lemma: Assume $U$ is convex. Isometries take line segments to line segments, so $V$ is also convex. For $q \in U$, let $\gamma(t) = p + t(q-p)$. Then $\phi \circ \gamma$ and $\psi \circ \gamma$ are both line segments from $r := \phi(p) = \psi(p)$ to $\phi(q)$ and to $\psi(q)$ respectively; that is, $\phi \circ \gamma(t) = r + t(\phi(q)-r)$ and $\psi \circ \gamma(t) = r + t(\psi(q) - r)$. Since $d\phi_p = d\psi_p$, we get $$\phi(q) - r = \frac d{dt}\bigg|_{t=0} (\phi \circ \gamma)(t) = d\phi_p(\dot\gamma(0)) = d\psi_p(\dot\gamma(0)) = \frac{d}{dt}\bigg|_{t=0} (\psi \circ \gamma)(t) = \psi(q)-r
$$
so $\phi(q) = \psi(q)$. If $U$ is not convex, $U$ is open, hence a union of convex open sets, on each of which $\phi = \psi$. QED. 
A: Here is a slightly simpler, non-geometric, proof:
Consider the coordinate derivatives
$$ a_{ijk} = \frac{\partial}{\partial x^i} \frac{\partial}{\partial x^j} V^k $$
from calculus we know that this expression is symmetric in $i$ and $j$. Killing's equation implies that this expression is antisymmetric in $j$ and $k$. 
So
$$ -a_{ikj} = a_{ijk} = a_{jik} = -a_{jki} = -a_{kji} = a_{kij} = a_{ikj} $$
where each of the equals signs comes from swapping the first two indices (which is symmetric) or the last two indices (which is antisymmetric). Note that this says $-a_{ikj} = a_{ikj} \implies a_{ikj} = 0$. 
This shows that $V$ is necessarily linear in $\vec{x}$. Since $V(0) = 0$ it is uniquely determined by its derivative $\partial_j V^k$ at one point.  
