Solution of the differential equation? First of all, I am sorry that I could not pick up a better title for the question :)
Today, I am working with DE questions involving symmetric positive definite matrices, and this is the second question I came up today, and could not find a way to prove it:
If $A$ is an symmetric positive definite square matrix, and $m$ is a positive number, then for each nonzero solution of the equation:
$$ \overrightarrow {v''}(t) +m\overrightarrow {v'}(t)+ A\overrightarrow {v}(t)=0,$$
$\phi(t),$  which is defined as: $$\phi(t)=||v'(t)||^2 + ||v(t)||^2$$
tends to $0$, as $t \rightarrow+\infty$ with an exponential rate, and also when $t \rightarrow-\infty $,   $\phi(t)$ grows exponentially.
 A: Hint. As $A$ is positive definite, then it can written as $A=U^*DU$, where $U$ orthogonal and $D$ diagonal with positive elements. Let $D=\mathrm{diag}(d_1,\ldots,d_n)$, and $w=Uv=(w_1,\ldots,w_n)$. Clearly, the $w_i$'s satisfy the equations
$$
w''_i+mw_i'+d_iw_i=0.
$$
Remember that 
if $\lambda_1,\lambda_2$ are the zeroes of $\lambda^2+m\lambda+d_i=0$, then
$w_i$ has to be of the form
$$
w_i(t)=c_1\mathrm{e}^{\lambda_1 t}+c_2\mathrm{e}^{\lambda_1 t},\quad\text{if}
\quad\lambda_1\ne\lambda_2,
$$
or
$$
w_i(t)=(c_1t+c_2)\mathrm{e}^{\lambda_1 t},,\quad\text{if}
\quad\lambda_1=\lambda_2.
$$
Clearly, either both $\lambda_1,\lambda_2$ are negative or they are complex conjugate with negative real parts.
Prove that $w_i(t)\to 0$, exponentially, and so does $w'(t)$, and next observe that
$\varphi(t)=\|w(t)\|^2+\|w'(t)\|^2$.
A: Define $x = \begin{bmatrix}v\\ v^\prime\end{bmatrix}$ and note that
$$x^\prime = \begin{bmatrix}0 & I\\-A & -mI\end{bmatrix}x$$
Using the fact that $A$ is positive definite and $m$ is positive, you can show that the matrix above has eigenvalues only in the left half part of the complex plane. This gives you all you need I think. 
