Do Carmo: Linear Killing field anti-symmetric? In Exercise 3.5a of Riemannian Geometry, do Carmo defines a vector field $v$ on $\mathbb{R}^n$ to be linear if it's linear as a map $v\colon \mathbb{R}^n \to\mathbb{R}^n$. He then asks the reader to prove that such a $v$, defined by a matrix $A$, is a Killing field iff $A$ is anti-symmetric. However, I've run into a problem.
Suppose $v$ is a linear Killing field with matrix $A$. The flow of $v$ is given by $\phi_t(x)=tAx + x$ and is defined for all $t,x$. Its derivative is $d\phi_t=tA + I$. Then since $\phi_t$ is (by do Carmo's definition) an isometry, for all $x\in \mathbb{R}^n$, we have
\begin{align}
\langle x, x \rangle
&= \langle d\phi_tx, d\phi_tx \rangle\\
&= \langle tAx + x, tAx + x\rangle\\
&= t^2\langle Ax,Ax\rangle + 2t\langle Ax,x\rangle + \langle x,x\rangle\\
0 &= t^2\langle x,A^\top Ax\rangle + 2t\langle Ax,x\rangle.
\end{align}
But this is true for all $t$, so the coefficients of $t$ and $t^2$ on the rhs must be identically zero. Hence, $\langle Ax,x\rangle=0$ for all $x$, which means $A$ is anti-symmetric, and $\langle x,A^\top Ax\rangle =0$ for all $x$, which means $A^\top A$ is anti-symmetric. But $A^\top A$ is also symmetric, so $A^\top A=0$. Hence, $-A^2 = A^\top A = 0$. Since a general anti-symmetric matrix is not nilpotent (are there any nilpotent nonzero anti-symmetric matrices?), do Carmo's statement is false.
What have I missed?
 A: Your equation for the flow is wrong.  What's going wrong is that if $\gamma_x(t) = \phi_t(x)$ you need $\frac{d}{dt}\phi_t(\gamma(t)) = \gamma'(t)$ for all $t$, while yours just works for $t=0$.
For example, on $\mathbb{R}^2$, consider the function $$\phi_t(\vec{x}) = \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t\end{bmatrix} \vec{x}.$$
One can easily check that this really is an isometry, hence $X=\frac{d}{dt}\phi_t |_{t=0}$ is a Killing vector field.  Computing, we, in fact, get $$X(\vec{x}) = \begin{bmatrix} 0 & 1\\ -1&0 \end{bmatrix}\vec{x}.$$
Notice that in my flow, for fixed $\vec{x}$, $\phi_t(\vec{x})$ has a bounded image whereas for your formula, so long as $A\neq 0$, the image of $tA\vec{x} + \vec{x}$ has unbounded image, so your formula must have a mistake.
What is true is the following:  If $A$ is linear, then the flow is $$\phi_t(\vec{x}) = \exp(tA)\vec{x} = (I + tA + \frac{t^2}{2!}A^2 + ...)\vec{x}.$$  This follows because \begin{align*} \phi_0(\vec{x}) &= \exp(0)\vec{x} \\ &= I\vec{x}\\ &= \vec{x}\end{align*} and, if $\gamma=\gamma_\vec{x}(t) = \phi_t(x)$, then \begin{align*} A(\gamma(t)) &= A\phi_t(\vec{x})\\  &= A\exp(tA) \vec{x} \\ &= (\exp(tA)\vec{x})' \\ &= \gamma'(t).\end{align*}
In terms of your inner product calculation, we have $d\phi_{t_0} v = \exp(t_0 A)v$ for $v$ a tangent vector in $\mathbb{R}^n$.  Then, they key fact for $\exp$ that you'll need is that $\exp(A)\exp(-A) = I$, or said another way, $\exp(A)^{-1} = \exp(-A)$.  (Note that, in general, we don't have $\exp(A)\exp(B) = \exp(AB)$ unless $A$ and $B$ commute.)  So, if $A$ is antisymmetric, we have \begin{align*} \exp(A)^\top &= \exp(A^\top) \\ &= \exp(-A) \\ &= \exp(A)^{-1}\end{align*} and this should make your calculation work out.
