Let
- $n\ge2$, be a natural number
- $R \in \mathbb{R}^{d \times d}$
- $R R^{T}=I_{d}$.
Prove or disprove that $R^TR=I_d$.
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Michael
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1Hi @Michael , Do you know about the result that for any $A,B$ square matrices we have, $$(A . B)^{T} = B^{T} . A^{T}$$ – vishalnaakar25 Jun 14 '25 at 09:35
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1What is $n$, exactly? – Rodrigo de Azevedo Jun 14 '25 at 09:41
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1@vishalnaakar25 applying your suggestion here means $(RR^T)^T=({R^T}^TR^T)=RR^T$. So, I'm not sure how this helps. – Michael Jun 14 '25 at 09:45
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@RodrigodeAzevedo n is a natural number. Edited the question. – Michael Jun 14 '25 at 09:46
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Yes that seems useless in this regard. But, it is easy to see that $R^{-1} = R^{T}$ and $R^{-1}.R = R R^{-1}. = I$ and .......... – vishalnaakar25 Jun 14 '25 at 09:52
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We know that $AB=I$ implies $BA=I$, se the duplicates. – Dietrich Burde Jun 14 '25 at 09:52
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@Michael And is $n$ the number of rows / columns of $R$? Where is it used? – Rodrigo de Azevedo Jun 14 '25 at 10:23