I'm currently in need of some help. I have the following problem:
I've found my eigenvalues, which are:
$$\lambda = \frac{-ahu\pm\sqrt{a^2h^2u^2+4ahu}}{2u},~~~~~~a,h,u>0$$
For clarity, let $\lambda_1$ be the positive, and $\lambda_2$ be the negative. Eigenvalues are:
$$\begin{bmatrix} \dfrac{ah}{ah+\lambda_i}\\1 \end{bmatrix}, ~~~ i\in\{1,2\}$$
My problem is now, that if we change $h$ to $h+\Delta h$ we get:
$$\lambda = \frac{-a(h+\Delta h)u\pm\sqrt{a^2(h+\Delta h)^2u^2+4a(h+\Delta h)u}}{2u},~~~~~~a,h+\Delta h,u>0$$
I would like to show how this change affects the eigenvectors direcion. We can form a line from the eigenvector, and I would like to show the slope of this line is steeper (given that $\Delta h>0$). I've shown in numerically with an example but would like to show that this holds true generally.
I don't know the formal name for this kind of evaluation (English is not my native language), so I apologize if there's a previous post containing the answer. Thank you in advance! :-)
Using that $a,h,u>0$ we can tidy up the expression for the eigenvalues \begin{equation} \lambda_i=\frac{ah}{2}\left(-1\pm\sqrt{1+4/ahu}\right) \end{equation}
and the eigenvectors \begin{equation} \vec v_i=\begin{pmatrix}\frac{ah}{ah+\lambda_i}\\1\end{pmatrix}=\begin{pmatrix}\frac{2}{1\pm\sqrt{1+4/ahu}}\\1\end{pmatrix} \end{equation}
Let us take a look at the first component of this vector. For $h\to0$ we have $v_{1,x}=v_{2,x}=0$, and the derivative can be found through the chain rule to be \begin{equation} \frac{\partial v_{i,x}}{\partial h}=\pm\frac{4}{ah^2u\sqrt{1+4/ahu}\left(1\pm\sqrt{1+4/ahu}\right)^2} \end{equation}
Again using $a,h,u>0$ we see that $\frac{\partial}{\partial h} v_{1,x}>0$ and $\frac{\partial}{\partial h}v_{2,x}<0$. So $v_{1,x}$ starts out zero and strictly increases, whereas $v_{2,x}$ starts out zero and strictly decreases.
Since $v_{i,y}=1$ is constant, this means both vectors start out pointing in the $(0,1)$ direction and will increasingly tilt one of two ways, asymptotically approximating the following limits
\begin{equation} \lim_{h\to\infty}\frac{\vec v_1}{||\vec v_1||}=\frac{1}{2}\begin{pmatrix}\sqrt{3}\\1\end{pmatrix}\qquad \lim_{h\to\infty}\frac{\vec v_2}{||\vec v_2||}=\begin{pmatrix}-1\\0\end{pmatrix}\quad \end{equation}