Limit involving the norm of a Cauchy problem solution Let 
$$A = \begin{bmatrix} 1 & -2009 & 0 \\
 2009 & 1 & 1492 \\
0 & -1492 & 1
 \end{bmatrix}$$
and $X_0 \in \mathbb{R}^3$, $X_0 \ne 0$. Let $X(t)$ be the solution of the Cauchy problem
$$ \begin{cases}X'(t)=AX(t), \quad \forall t \in \mathbb{R} \\
     X(0)=X_0 \, .
   \end{cases}$$
Evaluate, when $k$ varies in $\mathbb{R}$, the limit
$$ \lim_{t \to -\infty} e^{-kt} \lVert X(t) \rVert ^2 \, .$$
I don't really know where to begin. I don't think I'm supposed to solve the cauchy problem explicitly.
 A: Sort of hint
It is not the only possible way to obtain the answer for this question, but it's pretty tempting for me to follow it. Since we are dealing with linear system of equations $\dot{x} = Ax$ we know that the Cauchy problem $x(0) = x_0$ can be solved using formula $x(t) = e^{At} x_0$. The matrix $A$ in question has one remarkable property: off-diagonal elements satisfy $a_{ij} = - a_{ji}$. If it were true for diagonal elements too, we would have a skew-symmetric matrix. A skew-symmetric matrix multiplied by real scalar $t$ is still a skew-symmetric matrix and its exponent is an orthogonal matrix (useful property which is very easy to prove if we calculate $\frac{d}{dt}(x^T x)$, where $x$ satisfies $x' = Ax$, and show that in this case $e^{At} x_0$ preserves the norm of any vector $x_0$). But matrix $A$ is a sum of identity matrix $I$ and skew-symmetric matrix $B$; moreover, $I$ and $B$ commute (this is obvious), so $It$ and $Bt$ commute. In this case $$e^{At} = e^{(I+B)t} = e^{It} \cdot e^{Bt} = e^t \cdot e^{Bt},$$ 
where $e^t$ is an ordinary scalar function. So, it's pretty obvious that $e^{Bt}$ doesn't change the norm of vector (it's an orthogonal transformation) and only $e^t$ multiplier changes the norm. 
