For symmetric matrices $A,B$ $\in$ $\mathbb{R}_{2\times2}$, their eigenvalues are $\lambda_1$,$\lambda_2$ ($0\leq\lambda_1 \leq \lambda_2$) and $\mu_1$,$\mu_2$ ($0\leq\mu_1 \leq \mu_2$), respectively.
Prove
$tr(AB)$$\leq$$\lambda_1$$\mu_1$+$\lambda_2$$\mu_2$.
When I tried to prove it, it may be solved with complex calculation.
I think there is much more simple or more intuitive way. I think rearrangement inequality would be used.
Best Answer
Here's a way of doing things: we can write $$ A = \lambda_1 x_1x_1^T + \lambda_2 x_2x_2^T\\ B = \mu_1 y_1y_1^T + \mu_2 y_2y_2^T $$ where the $x_i$ and $y_i$ are unit vectors. We then calculate $$ \operatorname{Tr}(AB) = \lambda_1 \mu_1 (x_1^Ty_1)^2 + \lambda_1 \mu_2 (x_1^Ty_2)^2 + \lambda_2 \mu_1 (x_2^Ty_1)^2 + \lambda_2 \mu_2 (x_2^Ty_2)^2 $$ Now, define $t = (x_1^Ty_1)^2$. We can rewrite the above as $$ \operatorname{Tr}(AB) = t \lambda_1 \mu_1 + (1-t)\lambda_1 \mu_2 + (1-t)\lambda_2 \mu_1 + t\lambda_2 \mu_2 =\\ t [\lambda_1 \mu_1 + \lambda_2 \mu_2] + (1-t)[\lambda_1\mu_2 + \lambda_2\mu_1] $$ Where $0 \leq t \leq 1$. It suffices to maximize this quantity.
Also, note that $$ [\lambda_1 \mu_1 + \lambda_2 \mu_2] - [\lambda_1\mu_2 + \lambda_2\mu_1] = (\lambda_2 - \lambda_1)(\mu_2 - \mu_1) \geq 0 $$