I have an algorithm for calibrating a vector magnetometer. The input is N readings of the x, y, z axes: ( x 1 , x 2 , … , x N ), ( y 1 , y 2 , … , y n ), and ( z 1 , z 2 , … , z N ). The algorithm fits an ellipsoid to the data by estimating a symmetric 3 × 3 matrix A. In order to calibrate the system, it needs to calculate A . I am adapting the algorithm for a microcontroller with very little memory, so cannot load standard matrix manipulation libraries.Is there an explicit formula for calculating the square root of 3 × 3 positive definite matrix?

Question

I have an algorithm for calibrating a vector magnetometer. The input is N readings of the x, y, z axes:   (      x    1    ,      x    2    ,  …  ,      x    N    ),   (      y    1    ,      y    2    ,  …  ,      y    n    ), and   (      z    1    ,      z    2    ,  …  ,      z    N    ). The algorithm fits an ellipsoid to the data by estimating a symmetric   3  ×  3 matrix A. In order to calibrate the system, it needs to calculate       A  . I am adapting the algorithm for a microcontroller with very little memory, so cannot load standard matrix manipulation libraries.Is there an explicit formula for calculating the square root of   3  ×  3 positive definite matrix?

Savanah Hernandez · Accepted Answer

I would rather use an iterative method (e.g. QR iteration or Jacobi&#039;s algorithm) to diagonalise A and find its square root, but there is a semi-closed-form formula for   B  =      A  . By Cayley-Hamilton theorem,                    (1)                              B          3                −        tr        ⁡        (        B        )                  B          2                +        tr        ⁡        (                  B                      −            1                          )        det        (        B        )        B        −        det        (        B        )        I        =        0.            Multiply both sides of (1) by B, we obtain                    (2)                              B          4                −        tr        ⁡        (        B        )                  B          3                +        tr        ⁡        (                  B                      −            1                          )        det        (        B        )                  B          2                −        det        (        B        )        B        =        0.            Substitute (1) into (2) and using the fact that       B    2    =  A, we obtain      A    2    +  tr  ⁡  (  B  )      [    −    tr    ⁡    (    B    )    A    +    tr    ⁡    (          B              −        1              )    det    (    B    )    B    −    det    (    B    )    I    ]    +  tr  ⁡  (      B          −      1        )  det  (  B  )  A  −  det  (  B  )  B  =  0.Therefore                    B                            =                              −                          A              2                        +                          (              tr              ⁡              (              B                              )                2                            −              tr              ⁡              (                              B                                  −                  1                                            )              det              (              B              )              )                        A            +            tr            ⁡            (            B            )            det            (            B            )            I                                              (              tr              ⁡              (              B              )              tr              ⁡              (                              B                                  −                  1                                            )              −              1              )                        det            (            B            )                                              (3)                                  =                              −                          A              2                        +            (            α            +            β            )            A            +            γ            δ            I                                α            γ            −            δ                              where                    α                            =                              λ            1                                λ            2                          +                              λ            2                                λ            3                          +                              λ            3                                λ            1                          ,                            β                            =        tr        ⁡        (        A        )        ,                            γ                            =                              λ            1                          +                              λ            2                          +                              λ            3                          ,                            δ                            =                  det          (          A          )                    and       λ    1    ,      λ    2    ,      λ    3   are the eigenvalues of A.So, to find B using (3), we need the eigenvalues A. Let   s  =  tr  ⁡  (      A    2    ). The characteristic equation of A is then                    (4)                              x          3                −        β                  x          2                +                  1          2                (                  β          2                −        s        )        x        −                  δ          2                =        0            and       λ    1    ,      λ    2    ,      λ    3   are its roots. Now (4) can be solved by radicals.

I have an algorithm for calibrating a vector magnetometer. The input is N readings of the x, y, z ax

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