Missing "trivial step" in the proof of the fundamental theorem of the local theory of curves. 
Given $k_o,\tau_o: I \to \mathbb R, \mathcal C^\infty, k_o>0,$ there exists a curve $\alpha:I\to \mathbb R^3$ parameterized by arc length, such that $k(s)=k_o(s)$ and $\tau(s)=\tau_o(s)$ that is unique up to direct isometries of $\mathbb R^3.$

Consider $$x'(s) = A_o(s) x(s)\tag{*}$$ with
$$A_o=\begin{bmatrix}
\mathcal O_3 & k_o (s)\bf I_3 & \mathcal O_3\\
-k_o(s)\bf I_3 & \mathcal O_3 & -\tau_o(s)\bf I_3\\
\mathcal O_3 & \tau_o (s)\bf I_3 & \mathcal O_3
\end{bmatrix}$$
corresponding to a $9\times 9$ block matrix. We choose $\bf a\in \mathbb R^9,$ such that
$$\bf a=\begin{bmatrix}\bf t_o,\bf m_o,\bf b_o\end{bmatrix}^\top=\begin{bmatrix}(a_1,\dots,a_3),(a_4,\dots,a_6),(a_7,\dots,a_9)\end{bmatrix}^\top$$ form an oriented orthonormal basis of $\mathbb R^3.$ We define $f:I\to \mathbb R^9$ to be a solution of $(*)$ as ${\bf f}=[f_1,f_2,f_3]^\top,$ ${\bf m}=[f_4,f_5,f_6]^\top$ and ${\bf b}=[f_7,f_8,f_9]^\top,$ functions of $s,$ with initial values $\bf a.$
The proof proceeds by proving that $\bf f, m$ and $\bf b$ forms a positive orthonomal basis of $\mathbb R^3$ for any $s$ by looking at the matrix
$$M(s)=
\begin{bmatrix}
\langle \bf t, t \rangle & \langle \bf t, n \rangle & \langle \bf t, b \rangle \\
\langle \bf t, n \rangle & \langle \bf n, n \rangle & \langle \bf n, b \rangle \\
\langle \bf t, b \rangle & \langle \bf n, b \rangle & \langle \bf b, b \rangle
\end{bmatrix}= \bf I[3\times 3]$$
And after all this "it is a simple computation"...
$$2M'(s) = A(s) M(s) - M(s)A(s),$$ where 
$$A=\begin{bmatrix}
0 & k_o (s) &  O\\
-k_o(s) & O & -\tau_o(s)\\
 O & \tau_o (s) &  O
\end{bmatrix}$$
Can I get a hint as how this simple computation would take shape?
 A: First note that
$$
M'(s) = \begin{bmatrix}
2\langle \mathbb{t}',\mathbb{t}\rangle & \langle \mathbb{t}',\mathbb{n}\rangle + \langle \mathbb{t},\mathbb{n}'\rangle & \langle \mathbb{t}',\mathbb{b}\rangle+\langle \mathbb{t},\mathbb{b}'\rangle\\
\langle \mathbb{t}',\mathbb{n}\rangle + \langle \mathbb{t},\mathbb{n}'\rangle & 2\langle \mathbb{n}',\mathbb{n}\rangle & \langle \mathbb{n}',\mathbb{b}\rangle + \langle \mathbb{n},\mathbb{b}'\rangle\\
\langle \mathbb{t}',\mathbb{b}\rangle + \langle \mathbb{t},\mathbb{b}'\rangle & \langle \mathbb{n}',\mathbb{b}\rangle + \langle \mathbb{n},\mathbb{b}'\rangle & 2\langle \mathbb{b}',\mathbb{b}\rangle
\end{bmatrix},
$$
then use the Frenet formulas
\begin{align*}
\mathbb{t}' &= k_0(s)\mathbb{n},\\
\mathbb{n}' &= -k_0(s)\mathbb{t} + \tau_0(s)\mathbb{b},\\
\mathbb{b}' &= -\tau_0(s)\mathbb{n}
\end{align*}
to write $M'(s)$ in terms of $\mathbb{t}$, $\mathbb{n}$ and $\mathbb{b}$ (and not of their derivatives). By the other hand, compute explicitely $A(s)M(s)-M(s)A(s)$ and compare. It is a little bit tedious, but it is a simple calculation.
