Givens rotations versus Euler angles

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SUMMARY

This discussion focuses on the implementation of QR decomposition using Givens rotations and the comparison of the resulting rotation matrix Q with Euler angles. The user successfully calculated G1, G2, and G3 to achieve an upper triangular residual matrix R, confirming the correctness of their code. However, they encountered discrepancies when testing the relationship between Givens rotations and Euler angles, particularly noting that R was non-diagonal for certain rotations. The user seeks to identify flaws in their approach or implementation.

PREREQUISITES
  • Understanding of QR decomposition
  • Familiarity with Givens rotations
  • Knowledge of rotation matrices
  • Basic concepts of Euler angles
NEXT STEPS
  • Research the mathematical properties of Givens rotations
  • Explore the relationship between Givens rotations and Euler angles
  • Learn about the implications of non-diagonal residual matrices in QR decomposition
  • Investigate the conditions under which Givens rotations approximate Euler angles
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Mathematicians, computer scientists, and engineers involved in numerical linear algebra, particularly those working with QR decomposition and rotation matrices.

makc
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Trying to implement QR decomposition using Givens rotations, I calculate G1 to zero n32 of original matrix A, then G2 to zero n31 of G1 * A, then G2 to zero n21 of G2 * G1 * A. Residual matrix, R = G3 * G2 * G1 * A comes upper triangular as expected, so I believe my code is correct.

Looking at rotation matrix Q = G1^T * G2^T * G3^T against Euler rotation matrix X * Y * Z, I thought that I could use angles of Givens rotations as Euler angles up to sign correction at max.

To test this idea, I ran QR decomposition on pure rotation matrix A, expecting R to be identity matrix. This really worked out for certain rotataions, but for other cases, R was non-diagonal and, as you guessed, Givens rotations were nowhere near Euler angles.

I wonder where is a flaw could be in my idea or implementation?
 
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##Q^{-1}=(G_1\cdot G_2 \cdot G_3)^{-1} = G_3^{-1}\cdot G_2^{-1} \cdot G_1^{-1} \neq G_1^\tau\cdot G_2^\tau \cdot G_3^\tau##
 

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