Proof solution of $d_{x}\vec{g}(x)=H\vec{g}(x)$

Question

I have:

$d_{x}\vec{g}(x)=H\vec{g}(x)$

where $\vec{g}(x)\in\mathbb{R}^{2}$ and $H\in\mathbb{R}^{2*2}$

The solution is:

$\vec{g}(x)=c_{1}e^{\Lambda_{1}x}\vec{g_{1}}(x)+c_{2}e^{\Lambda_{2}x}\vec{g_{2}}(x)$

where $\vec{g_{1}}(x)$ and $\vec{g_{2}}(x)$ are the eigenvectors of $H$ and $\Lambda_{1}$, $\Lambda_{1}$ the eigenvalues.

I would like to proof this.

How you would do it, is to know that the solution can be written as:

1. $\vec{g}(x)=e^{Hx}\vec{g}(0)$
2. Diagonalize H, take the exponential of each diagonal component (eigenvalues), and project back to orignal space using the eigenvectors.

...but I somehow can not figure it out correctly.

EDIT:

• $H$ is diagonalizable.

• I would like to deduce the form of $\vec{g}(x)$ from the ODE using
  the strategy that I scetched.

Answer 1

IF you knew that $H$ gave rise to an orthonormal basis $\{{\bf v}_1, {\bf v}_2\}$ of eigenvectors for $\mathbb{R}^2$ (such as if $H$ were symmetric), then define $g_1(x) = ({\bf g}(x) \cdot {\bf v}_1)$ and $g_2(x) = ({\bf g}(x) \cdot {\bf v}_2)$. Now, by orthogonality, ${\bf g}(x) = g_1(x) {\bf v}_1 + g_2(x) {\bf v}_2$ and you will find that $$ e^{Hx} {\bf g}(0) = g_1(0) e^{Hx}{\bf v}_1 + g_2(0) e^{H x} {\bf v}_2 = g_1(0) e^{\lambda_1 x}{\bf v}_1 + g_2(0) e^{\lambda_2 x} {\bf v}_2 $$ which, since ${\bf v}_1$ and ${\bf v}_2$ are eigenvectors of $H$, is in the general form you suggest. However, this argument only applies when the eigenvectors are orthogonal.