Euclidean norm of a solution of matrix differential equation I try to solve a problem I have found in a book where I'm asked to find the solution to a differential equation of the form
$$
x'=Ax
$$
where $A$ is a matrix. The answer is the exponential of $e^{At}$. Then I'm stuck with the second request that is to prove that $\|x(t)\|_2$ is increasing (as a function of $t$) by considering $A+A^T$. What information can give $A+A^T$? I'm not posting the expression for $A$ because I'm looking for a general answer.
Thanks in advance for any suggestion.
Thanks to the suggestion given, one can observe that the square of the norm is increasing if and only if the norm is increasing but the transpose of $x(t)$ is $e^{A^Tt}$. So
$$
\langle x(t),x^T(t)\rangle=\langle e^{At},e^{A^Tt}\rangle=e^{(A^T+A)t}
$$
and if I prove that $A^T+A$ is positive definite I have concluded. It's all correct?
 A: You consider the function $\|x(t)\|_2$. If you want to show that it is increasing, then it's enough to show that $\|x(t)\|_2^2$ is increasing, since $\|x(t)\|_2$ is non-negative and the square-root with this co-domain is increasing.
However, $\|x(t)\|_2^2 = x(t)^T x(t)$. This is a product of differentiable functions of $t$, hence differentiable.
We must use the (dot)product rule to differentiate. Recall that transposes commute with the derivative (they come out in differential quotients, and are continuous so can be exchanged with limits). When we do that , we get
\begin{align}
\frac{d}{dt} x(t)^T x(t) &= \left[\frac{d}{dt} x(t)^T\right]x(t) + x(t)^T \left[\frac{d}{dt} x(t)\right] \\ &=\left[\frac{d}{dt} x(t)\right]^Tx(t) + x(t)^T \left[\frac{d}{dt} x(t)\right] = x(t)^TA^T x(t) + x(t)^TAx(t) \\ &= x(t)^T[A^t+A]x(t)
\end{align}
Now, $A^T+A$ is a symmetric matrix. As long as $A^T+A$ is also positive semi-definite,  it is true that $y^T[A^T+A]y \geq 0$ for any $y \in \mathbb R^n$ (and in particular, for each $x(t)$) , and therefore the function $\|x(t)\|_2$ is increasing (it is strictly increasing if $A^T+A$ is positive definite, since $x(t) \neq 0$ for all $t$ so the derivative is in fact positive).
