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 $\alpha_2$.

  • $\begingroup$ What an ugly exercise. I don't see any nice factorization. The dependencies look messy. $\endgroup$ – Friedrich Philipp Mar 24 '16 at 15:54
  • $\begingroup$ Are you sure of the coefficients ? Especialy for $C$ where coefficient $c$ is present three times ? $\endgroup$ – Jean Marie Mar 24 '16 at 17:06
  • $\begingroup$ Yes, JeanMarie I`m pretty sure. $\endgroup$ – Asatur Khurshudyan Mar 24 '16 at 20:18

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 $t = c/b$.

  • $\begingroup$ Everything is clear, but in my problem $a_{12}$ shouldn`t appear in the relation. $\endgroup$ – Asatur Khurshudyan Mar 25 '16 at 11:44
  • $\begingroup$ Well, but it does. If there was an expression without $a_{12}$, $a_{12}$ would depend on $a_{11}$ and $a_{22}$. Please check the answer as done. $\endgroup$ – Friedrich Philipp Mar 25 '16 at 14:00

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