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.

• You don't have to provide exact solution. But since you know what form the solution has, you can compute something for matrix $A$ and from this you can deduce what this limit will be like. And of course you can compute $\exp A$ if you know some its useful properties -- it's easy to do because this matrix is a sum of two commuting matrices. – Evgeny Sep 2 '16 at 7:34
• I know $X(t)=e^{At}X_0$, but how do I estimate $\lVert e^{At}X_0 \rVert^2$? I suspect it has to do something with the eigenvalues of $A$ but I don't really know. – un umile appassionato Sep 2 '16 at 11:02

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.