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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Lee if $RT=I$ then both $R,T$ are invertible. $\endgroup$ – 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 $x \not= 0$, $$\begin{bmatrix}0 & 1/x \\ 1/x &0 \end{bmatrix}\begin{bmatrix}0 & x \\ x & 0\end{bmatrix} = I$$

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.