Differential of norm with symmetric matrix Let $z: I \rightarrow \mathbb{R}^n$ be a differentiable function. We know that
$$\frac{d}{dt} |z(t)|^2= 2\langle z'(t),z(t) \rangle.$$
Now, let $A$ be any $n\times n$ matrix (maybe symmetric, definite positive...etc). Do we have something similar such that the differential is $2\langle z'(t),Az(t) \rangle$?
Any remark would be helpful (equality, inequality...etc).
 A: Let $I \subset \Bbb{R}$ be an open set, and let $A \in M_{n \times n}(\Bbb{R})$ be a symmetric matrix. If you assume $z:I \to \Bbb{R}^n$ is differentiable, then you can show that the function $f: I \to \Bbb{R}$ defined by
\begin{align}
f(t) = \langle A z(t) , z(t)\rangle
\end{align}
is differentiable and by the chain rule, we have that for every $t \in I$,
\begin{align}
f'(t) &= \langle A z'(t), z(t) \rangle + \langle A z(t) , z'(t)\rangle \\
&= 2 \langle A z(t) , z'(t)\rangle,
\end{align}
the last equality is because $A$ is symmetric.
A: Suppose
$y(t) = \langle z(t), Az(t) \rangle; \tag 1$
then
$\dot y(t) = \langle \dot z(t), Az(t) \rangle + \langle z(t), A\dot z(t) \rangle$
$ = \langle \dot z(t), Az(t) \rangle + \langle A^Tz(t), \dot z(t) \rangle = \langle \dot z(t), Az(t) \rangle + \dot z(t), A^T z(t) \rangle$
$= \langle \dot z(t), (A + A^T)z(t) \rangle; \tag 2$
when $A$ is symmetric, 
$A = A^T, \tag 3$
this becomes
$\dot y(t) = 2\langle \dot z(t), Az(t) \rangle. \tag 4$
Nota Bene: It is also worth observing that the operator $A$ occurring in (2), (4) may be transferred to $\dot z(t)$ via the usual properties of the matrix transpose, viz.
$\dot y(t) = \langle (A + A^T)\dot z(t), z(t) \rangle = \langle z(t), (A + A^T)\dot z(t) \rangle, \tag 5$
and 
$\dot y(t) = 2\langle A\dot z(t), z(t) \rangle = 2 \langle z(t), A\dot z(t) \rangle \tag 4$
in the event that $A = A^T$ is symmetric.  End of Note.
