Eigenvalues Problem with eigenvalue variable substitution I have the following eigenvalues problem ($a,\ b,\ c,\ d,\ x,\ y$ are known values, with $x \neq y$):
$$\left| \begin{array}{cc}
a-\alpha x& b \\
c & d-\alpha y
\end{array} \right| = 0$$
so I thought it was possible to solve it like this:
$$\left| \begin{array}{cc}
a-\lambda& b \\
c & d-\lambda
\end{array} \right| = 0$$
And at the end I could do this:
$$ \lambda_1 = \alpha_1 \ x $$
$$ \lambda_2 = \alpha_2 \ y $$
Is this correct?
PS: I tried it with some values and it worked. But at first it seems that it doesn't make much sense. And as I was trying to find some kind of answer to this I arrived at this expression:
$$\lambda^2-\lambda(a+d) = \alpha^2 x y - \alpha(ay+dx)$$
$$\Leftrightarrow \ \  \frac{\lambda}{\alpha}=\frac{\alpha xy-(ay+dx)}{\lambda-(a+d)}$$
If I could get rid of the $a$ and $d$ it would be great.
 A: The approach cannot be correct.

*

*The original determinant equation leads to a quadratic in $\alpha$, so there are two solutions. The eigenvalue equation leads to a quadratic in $\lambda$, so there are two solutions. If $x \neq y$ and $\lambda_1 \neq \lambda_2$ and $\alpha_1 \neq \alpha_2$: there is no good reason why $\alpha_1$ should be paired with $x$ and not with $y$. (In other words, to get from $\lambda_1$ and $\lambda_2$ to $\alpha_1, \alpha_2$ as you suggested by dividing by $x$ and $y$, the solutions you get depend on a choice on how to pair the values $x,y$ with the values $\lambda_1,\lambda_2$, so the scheme cannot possibly be correct.


*In the case $x = y$, you can just divide the entire matrix by $x$, and you will have that indeed $\lambda = x \alpha$ gives you a scaling of the solutions. But this is not true in general.
Remember that one interpretation of eigenvalues and eigenvectors, in the case $b = c$, is that the eigenvectors allow you to simultaneous diagonalize two symmetric matrices, one being $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ and the other being the identity. Turns out your first equation is the generalization where you try to simultaneously diagonalize the matrices $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ and $\mathrm{diag}(x,y)$. The relation between the two problems are complicated geometrically, and you can't really expect a good way to map solutions of one to the other.


*Explicit counterexample: let $a = b = c = 1$, $d = 2$, $x = \frac13$, and $y = 1$. Then you find
$$ \lambda = \frac{3 \pm \sqrt{5}}{2} $$
and
$$ \alpha = \frac{5 \pm \sqrt{13}}{2} $$
For your approach to hold $\lambda$ and $\alpha$ must be rationally related, which they are not.
