I am having trouble wrapping my mind against a simple problem: Suppose we have the following eigenvector equation for $A\in\mathbb{R}^{n\times n}$ and $\alpha \in \mathbb{R}$.
$$ \left(\alpha A\right)u = \lambda u $$
Where $u$ is an eigenvector for $A$ and $\lambda$ is its eigenvalue.
I am trying to compute $$ \frac{d\lambda}{d\alpha } $$
Since, if $\lambda$ is an eigenvalue for $A$, then $\alpha \lambda$ is an eigenvalue for $\alpha A$, for $\alpha \in \mathbb{R}$. It follows that
$$ \lambda = \alpha c_0; \ c_0\in \mathbb{C} $$
Rearranging terms and letting $c_1 = \log c_0$, we have that
$$ \log\lambda = \log \alpha + c_1 $$
Which can be written as
$$ \int \frac{d\lambda}{\lambda} = \int \frac{d\alpha}{\alpha} + c_1 $$
From this last equation, is it logical to conclude, therefore, that
$$ \frac{d\lambda}{\lambda} = \frac{d\alpha}{\alpha} $$
and hence,
$$ \frac{d\lambda}{d\alpha} = \frac{\lambda}{\alpha} \ ? $$
I have never solved an equation of this form. What I did realize is that is that, solving this equation was the inverse of solving an ODE.
$$\lambda(\alpha)=\alpha\lambda(1)\implies\frac{d\lambda(\alpha)}{d\alpha}=\lambda(1).$$