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How can we find the maximum value of $\left( \vec{k}^TT_1\vec{m} \right) \left( \vec{n}^TT_2\vec{l} \right) +\left( \vec{k}_{\bot}^{T}T_1\vec{m}_{\bot} \right) \left( \vec{n}_{\bot}^{T}T_2\vec{l}_{\bot} \right) $ with $k$, $l$, $m$ and $n$ changable unit vectors, notation $\bot$ stands for perpendicular, and $T_1$ and $T_2$ are two 3 by 3 matrices? This can be formally stated in formula as $\underset{k,l,m,n}{\max}\left( \vec{k}^TT_1\vec{m} \right) \left( \vec{n}^TT_2\vec{l} \right) +\left( \vec{k}_{\bot}^{T}T_1\vec{m}_{\bot} \right) \left( \vec{n}_{\bot}^{T}T_2\vec{l}_{\bot} \right) $.

Here are some of my thinkings: I think that we can use singular value decomposition of $T_1$ and $T_2$, then by changeable of unit vectors, we can take $T_1$ and $T_2$ as diagonal vectors. Then I guess the maximum value is $\xi _1\eta _1+\xi _2\eta _2$ where $\xi_1\ge\xi_2\ge\xi_3$ are singular values of matrix $T_1$, and the same meaning for $\eta_i$, with $i=1,2,3$. I did an intense numerical search by generating random vectors to verify this. But I don't know how analytically prove it. I find that when $\vec m$ stands for unit vector set, i.e. unit sphere, then $T_1 \vec m$ is the set of an ellipsoid, but the hard thing for me to prove it geometrically is that I found $T_1 \vec m$ no longer perpendicular to $T_1 \vec m_\bot$ all the time, or else we can take the maximum value of term $\vec{k}^TT_1\vec{m}$ simply as the length of the vector after the action of $T_1$.

Any ideas? Thanks in advance!

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