# Min and max values for trace of product of two matrices

Suppose we have freely chosen two matrices A and B, which are 3$\times$3, real and symmetric i.e.:

$A= \left( \begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{array} \right),$

$B= \left( \begin{array}{ccc} \alpha_1 & 0 & 0 \\ 0 & \alpha_2 & 0 \\ 0 & 0 & \alpha_3 \end{array} \right).$

How we can estimate what is the range of values (min, max) for $\text{Tr}(AB)$? I start from Cauchy-Schwarz inequality, where I land on $-\sqrt{\text{Tr}(AA)\text{Tr}(BB)}\leq\text{Tr}(AB)\leq\sqrt{\text{Tr}(AA)\text{Tr}(BB)}$ and how to move next?

• Isn't it $α_1λ_1+α_2λ_2+α_3λ_3$? – Jimmy R. Dec 9 '16 at 10:21
• "freely chosen" $3 \times 3$, real and symmetric matrices need not be diagonal. – Shraddheya Shendre Dec 9 '16 at 10:21
• I'm not sure where you want to "move next". Both inequalities can be attained for certain concrete $A,B$. – Martin Argerami Dec 11 '16 at 18:24

We have $AB= \left( \begin{array}{ccc} \lambda_1 \alpha_1 & 0 & 0 \\ 0 & \lambda_2 \alpha_2& 0 \\ 0 & 0 & \lambda_3\alpha_3 \end{array} \right),$
Hence $Tr(AB)=\lambda_1 \alpha_1+\lambda_2 \alpha_2+\lambda_3 \alpha_3$