DE solution's uniqueness and convexity I am lost and don't know how to prove the following:

If $M$ is a positive definite symmetric square matrix and if $\overrightarrow {v}(t)$ is a solution of:
$$\overrightarrow {v'}(t) = M\overrightarrow {v}(t),\qquad t\in[0,T]$$
Then,
1) $\phi(t) = \ln(\|\overrightarrow {v(t)}\|^2) $ is a convex funciton,
2) Solution of the differential equation is unique.

 A: First of all, for any real square matrix of size $n$, symmetric or not, positive definite or not, the equation
$\vec v'(t) = M \vec v(t), \; t \in [0, T], \tag{1}$
with the initial condition $\vec v(0)$ at $t = 0$, always admits the unique solution 
$\vec v(t) = e^{Mt}\vec v(0); \tag{2}$
that (2) solves (1) is easily seen by direct differentiation; (2) yields:
$\vec v'(t) = (e^{Mt})' \vec v(0) = Me^{Mt} \vec v(0) = Mv(t). \tag{3}$
Existence and uniqueness also follow from the standard theory of ordinary differential equations, since the linear vector field $M \vec v$ satisfies a Lipschitz condtion on all of $\Bbb R^n$:
$\Vert M \vec v_1 - M \vec v_2 \Vert = \Vert M (\vec v_1 - \vec v_2) \Vert \le \Vert M \Vert \Vert v_1 - v_2 \Vert. \tag{4}$
Further information on the relationship of Lipschitz continuity to existence and uniqueness may be found in this wikipedia entry, or in any number of standard texts on ordinary differential equations.  Thus, based upon general theoretical considerations, we may consider item (2) in this question as being resolved; we turn to item (1).
To address the convexity of
$\phi(t) = \ln(\Vert \vec v(t) \Vert^2) \tag{5}$
we re-introduce the hypothesized symmetry and positive definiteness of the matrix $M$.  Since $M$ is symmetric, it is possessed of $n$ real eigenvalues $\lambda_i$  and an  orthonormal eigenbasis $\vec e_i$ of $\Bbb R^n$, $1 \le i \le n$, such that
$M \vec e_i = \lambda_i \vec e_i \tag{6}$
for each $i$; since $M$ is positive definite, each $\lambda_i = \langle \vec e_i, M \vec e_i \rangle > 0$; we may expand $\vec v(0)$ in terms of the $\vec e_i$, obtaining
$\vec v(0) = \sum_1^n c_i \vec e_i, \tag{7}$
where as usual $c_i = \langle \vec v(0), \vec e_i \rangle$, $\langle \cdot, \cdot \rangle$ being the ordinary euclidean inner product on $\Bbb R^n$.  We may then express the solution (2) to (1) in the form
$\vec v(t) = e^{Mt} \vec v(0) = e^{Mt}(\sum_1^n c_i \vec e_i) = \sum_1^n c_i e^{Mt} \vec e_i; \tag{8}$
we recall that (6) implies
$e^{Mt} \vec e_i = e^{\lambda _i t} e_i;  \tag{9}$
and thus (8) yields
$\vec v(t) = \sum_1^n c_i e^{\lambda_i t} \vec e_i; \tag{10}$
we compute
$\Vert v(t) \Vert^2 = \langle v(t), v(t) \rangle = \langle \sum_1^n c_i e^{\lambda_i t} \vec e_i, \sum_1^n c_j ^{\lambda_j t} \vec e_j \rangle = \sum_{i,j = 1}^n c_i c_j e^{\lambda_i t} e^{\lambda_j t} \langle \vec e_i, \vec e_j \rangle$
$= \sum_{i,j = 1}^n c_i c_j e^{\lambda_i t} e^{\lambda_j t} \delta_{ij} = \sum_1^n c_i^2 e^{2\lambda_i t}.  \tag{12}$
We thus have
$\phi(t) = \ln (\Vert v(t) \Vert^2) = \ln(\sum_1^n c_i^2 e^{2\lambda_i t});  \tag{13}$
scrutinizing (13), we see that we may, by re-indexing the $c_i$, $\lambda_i$ if necessary, assume that $c_i \ne 0$ for $1 \le i \le m \le n$; then this equation reads
$\phi(t) = \ln (\sum_1^m c_i^2 e^{2\lambda_i t}) \tag{14}$
with each $c_i^2 > 0$.  This being the case, we may write
$\phi(t) = \ln (\sum_1^m e^{\ln c_i^2} e^{2\lambda_i t}) = \ln (\sum_1^m e^{2\lambda_i t + \ln c_i^2}); \tag{15}$
$\phi(t)$ is convex on $[0, T]$ provided that
$\phi(st_1 + (1 - s)t_2) \le s\phi(t_1) + (1 - s) \phi(t_2) \tag{16}$
for any $t_1, t_2 \in [0, T]$; here $s \in [0, 1]$. To the end of estsblishing (16), we introduce the fact that the log-sum-exp functions are themselves convex; that is, letting
$\vec x = (x_1, x_2, \ldots, x_k), \tag{17}$
the log-sum-exp function $f(\vec x)$ of $\vec x$ is defined by the formula
$f(\vec x) = f(x_1, x_2, \ldots, x_k) = \ln (\sum_1^k e^{x_i}); \tag{18}$
if we also set
$\vec y = (y_1, y_2, \ldots, y_k), \tag{19}$
we have
$f(s \vec x + (1 - s) \vec y) \le sf(\vec x) + (1 - s) f(\vec y) \tag{20}$
for all $s$ as above.  If we now take $k = m$ and set
$\vec x(t) = (2\lambda_1 t + \ln c_1^2, 2\lambda_2 t + \ln c_2^2, \ldots, 2\lambda_m t + \ln c_m^2), \tag{21}$
then we may write
$\phi(t) = f(\vec x(t)); \tag{22}$
from the convexity of $f(x)$ we have, for any $t_1, t_2 \in [0, T]$, 
$f(s\vec x(t_1) + (1 - s) \vec x(t_2))$
$\le sf(x(t_1)) + (1 - s)f(\vec x(t_2)) = s\phi(t_1) + (1 - s)\phi(t_2).  \tag{23}$
We conclude our argument with the observation that
$s\vec x(t_1) + (1 - s)\vec x(t_2) = \vec x(st_1 + (1 - s)t_2), \tag{24}$
which may be seen by examining the components of (24):
$s(2\lambda_i t_1 + \ln c_i^2) + (1 - s)(2\lambda_i t_2 + \ln c_i^2) = 2\lambda_i (st_1 + (1 - s) t_2) + \ln c_i^2; \tag{25}$
(23) thus becomes
$f(\vec x(st_1 + (1 - s)t_2) \le s\phi(t_1) + (1 - s)\phi(t_2); \tag{26}$
since
$\phi(st_1 + (1 - s)t_2) = f(\vec x(st_1 + (1 - s)t_2)), \tag{27}$
we find that
$\phi(st_1 + (1 - s)t_2) \le s\phi(t_1) + (1 - s)\phi(t_2)); \tag{28}$
that is, $\phi(t) = \ln (\Vert \vec v \Vert^2)$ is convex on $[0, T]$.
