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$$\begin{array}{ll} \text{minimize} & R_1^2+R_2^2\\ \text{subject to} & RT=I\end{array}$$ where $R\in\mathbb{R^{2\times2}}$, $T\in\mathbb{R^{2\times2}}$, $R=\begin{bmatrix}R_1 & R_2\\R_3 &R_4\end{bmatrix}$, $T=\begin{bmatrix}T_1 & T_2\\T_3 &T_4\end{bmatrix}$.

My attempt: I know that if $RT=I$ then $\min R^{}R^*=(T^*T^{})^{-1}$. Then $R_1^2+R_2^2 \geq (T^*T^{})^{-1}_{1\times 1}$, where $(T^*T^{})^{-1}_{1\times1}$ is the first diagonal element of $(T^*T^{})^{-1}$. Can we get tighter bound?

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  • $\begingroup$ Do you know if $T$ is invertible? $\endgroup$ – Michael McGovern Jan 3 at 4:19
  • $\begingroup$ @MichaelMcGovern actually no, but I saw that I have divided by $\det(T)$, so my attempt works only for invertible $T$ $\endgroup$ – Lee Jan 3 at 4:21
  • $\begingroup$ What is $T^*$ in this problem? $\endgroup$ – Michael McGovern Jan 3 at 4:34
  • $\begingroup$ it is transpose $\endgroup$ – Lee Jan 3 at 4:37
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    $\begingroup$ @Lee if $RT=I$ then both $R,T$ are invertible. $\endgroup$ – user376343 Jan 3 at 6:03
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We can make $R_1^2 + R_2^2$ arbitrarily close to $0$. To show this, consider the identity, for any $x \not= 0$, $$\begin{bmatrix}0 & 1/x \\ 1/x &0 \end{bmatrix}\begin{bmatrix}0 & x \\ x & 0\end{bmatrix} = I$$

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  • $\begingroup$ this is same as my attempt: $R_1^2+R_2^2 \geq (T^*T^{})^{-1}_{1\times 1}$, where you have reached equality. But I believe there must be even tighter bound $\endgroup$ – Lee Jan 3 at 5:11
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    $\begingroup$ Maybe I misunderstand. Is $T$ fixed and we are minimizing $R_1^2 + R_2^2$ in terms of fixed $T_i$? Or do we have control over $T$? I interpreted the second way, in which case I'm saying $0 < R_1^2 + R_2^2 < \epsilon$ is achievable for every $\epsilon > 0$, and it doesn't get any tighter than that $\endgroup$ – Badam Baplan Jan 3 at 5:18
  • $\begingroup$ the first way is correct, sorry for not mentioning it in the question. We need to minimize $R_1^2+R_2^2$ in terms of fixed $T$. $\endgroup$ – Lee Jan 3 at 5:28
  • $\begingroup$ So I think I will close this question and mark your answer as accepted. Thanks for @Lee for his hint as well. The lower bound is $(T^*T^{})^{-1}_{1\times 1}=(T_2^2+T_4^2)/\det(T)$ $\endgroup$ – Lee Jan 3 at 8:44

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