Relation between eigenvalues of symmetric $2\times2$ non-singular matrices The eigenvalues of a symmetric $2\times2$ non-singular matrix $$A=\left(\begin{matrix} a_{11} & a_{12}\\ a_{12} & a_{22}\end{matrix}\right)$$ are $\alpha_1$ and $\alpha_2$. How are the eigenvalues of
$$B=\left(\begin{matrix} b\cdot a_{11} & c\cdot a_{12}\\ c\cdot a_{12} & b\cdot a_{22}\end{matrix}\right)$$ and $$C=\left(\begin{matrix} c\cdot a_{11} & c\cdot a_{12}\\ c\cdot a_{12} & b\cdot a_{22}\end{matrix}\right)$$
in which $b$ and $c$ are positive constants, related with $\alpha_1$ and $\alpha_2$.

Let $\alpha_{1,2}(b,c)$ denote the eigenvalues of $B$. You have
$$
\alpha_1(b,c) + \alpha_2(b,c) = \operatorname{tr}(B) = b\operatorname{tr}(A) = b(\alpha_1+\alpha_2)
$$
and (which is the more messy part)
\begin{align*}
\alpha_1(b,c)\alpha_2(b,c)
&= \det(B) = b^2a_{11}a_{22} - c^2a_{12}^2 = b^2\left(\det(A)+a_{12}^2\right) - c^2a_{12}^2\\
&= b^2\alpha_1\alpha_2 + (b^2-c^2)a_{12}^2.
\end{align*}
Now, we solve the first equation for $\alpha_2(b,c)$ and insert it into the second one:
$$
\alpha_1(b,c)\left(b(\alpha_1+\alpha_2)-\alpha_1(b,c)\right) = b^2\alpha_1\alpha_2 + (b^2-c^2)a_{12}^2.
$$
By symmetry, the same equation holds with $\alpha_1(b,c)$ replaced by $\alpha_2(b,c)$. Now, we solve for it and obtain
$$
\alpha_{1,2}(b,c) = \frac b 2\left[\alpha_1+\alpha_2\pm\sqrt{(\alpha_1-\alpha_2)^2 + 4(t^2-1)a_{12}^2}\,\right],
$$
where $t = c/b$.
