Marka ciyaar kubbadda cagta bilaabato, kubaddu waxay dhex jiifataa bartamaha garoonka ka dibna waxaa lagu wareegaa garoonka muddo 45 daqiiqo ah iyadoo la beddelayo oo la beddelayo. Bilowga qeybta labaad kubaddu mar kale ayey ku taal bartamaha garoonka. Waxaan ku tusineynaa qaab fudud aljebrada toosan oo midkoodna dhibco aan la koobi karayn oogada dusheeda ay had iyo jeer isku mid yihiin isla meeshii hore ama sida saxda 2
Marka hore, dhaqdhaqaaqyada kubadda ee la sameeyay qeybtii hore si fudud ayay isugu daraan vector eber ah. Sidaa darteed waa la dayaci karaa. Tani waxay ka tagaysaa tiro xaddidan oo wareegyo \(A_1, ..., A_n \in \mathbb{R}^{3 \times 3}\) oo leh \(A_k\) orthogonal iyo \(\det(A_k) = 1 \,\, \forall \,\, k \in \{1,...,n\}\) . Laba wareeg oo kasta \(A_i, A_j\) kuwa soo socda ayaa qabanaya::
$$ (A_i A_j)^T (A_i A_j) = A_j^T A_i^T A_i A_j = A_j^T (A_i^T A_i) A_j = A_j^T E_3 A_j = A_j^T A_j = E_3 $$
sida
$$ \det(A_i A_j) = \det(A_i) \cdot \det(A_j)=1 \cdot 1 = 1. $$
Tan macnaheedu waxa weeye \( A_i A_j \) markale waa meerto, waana sababta \( A_1 ... A_n \) waliba u tahay hal wareeg.
Hadday hadda tahay \( A_1 ... A_n = E_n \) , markaa markaa si cad dhammaan qodobbada dusha sare ee kubbadda waxay ku yaalliin barta bilowga - tan kale (suuragalnimada badan) kiiska eigenvector ee \( A_1 ... A_n \) mid yahay \( A_1 ... A_n \) wareegga eigenvalue \(1\) . Tan macnaheedu waxa weeye in si hufan labadan qodob, oo ku jiifa salka wareegga, laftooda lagu sawirayo.