# Minimizing $R_1^2+R_2^2$ subject to $RT=I$

$$\begin{array}{ll} \text{minimize} & R_1^2+R_2^2\\ \text{subject to} & RT=I\end{array}$$ 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?

• Do you know if $T$ is invertible? – Michael McGovern Jan 3 at 4:19
• @MichaelMcGovern actually no, but I saw that I have divided by $\det(T)$, so my attempt works only for invertible $T$ – Lee Jan 3 at 4:21
• What is $T^*$ in this problem? – Michael McGovern Jan 3 at 4:34
• it is transpose – Lee Jan 3 at 4:37
• @Lee if $RT=I$ then both $R,T$ are invertible. – user376343 Jan 3 at 6:03

We can make $$R_1^2 + R_2^2$$ arbitrarily close to $$0$$. To show this, consider the identity, for any $$x \not= 0$$, $$\begin{bmatrix}0 & 1/x \\ 1/x &0 \end{bmatrix}\begin{bmatrix}0 & x \\ x & 0\end{bmatrix} = I$$
• 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 – Lee Jan 3 at 5:11
• 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 – Badam Baplan Jan 3 at 5:18
• 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 $T$. – Lee Jan 3 at 5:28
• 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 $(T^*T^{})^{-1}_{1\times 1}=(T_2^2+T_4^2)/\det(T)$ – Lee Jan 3 at 8:44