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Real symmetric matrices                    125
                      Example 10.1. Principal axes of a cube
                      Consider a cube of uniform density, mass m and edge length a situated so that
                      three of its edges which meet at a vertex form the coordinate axes x, y, z. The
                      moments and products of inertia (see, for instance, Synge and Griffith 1959,
                      pp 282-93) for the cube in this coordinate frame are
                                                                2
                                               I  = I  = I  = 2ma /3
                                                xx  yy   zz
                                                                 2
                                               I  = I  = I yz  = -ma /4.
                                                xy
                                                    xz
                      These can be formed into the moment-of-inertia tensor





                                                                           2
                      where all the elements have been measured in units 12/(ma ).
                        Algorithm 13 can be used to find the eigensolutions of this matrix. The
                      eigenvalues then give the principal moments of inertia and the eigenvectors give
                      the principal axes to which these moments apply. Algorithm 13 when used on a
                      Data General NOVA operating in 23-bit binary arithmetic finds

                                    2
                             I =2ma /12        v l  = (0·57735, 0·577351, 0·57735) T
                              1
                                     2                                              T
                             I =11ma /12       v  = (0·707107, –0·707107, –4·33488E–8)
                             2                   2
                                     2                                           T
                             I =11ma /12       v  = (0·408248, 0·408248, –0·816496) .
                              3
                                                3
                      (The maximum absolute residual was 3·8147E–6, the maximum inner product
                      4·4226E–7.) The last two principal moments of inertia are the same or
                      degenerate. Thus any linear combination of v 2  and v 3  will give a new vector



                      which forms a new set of principal axes with v 1  and




                      which is orthogonal to  Indeed algorithm 14 on the same system found

                                    2                                        T
                             I =2ma /12        v  1  = (0·57735, 0·57735, 0·57735)
                              1
                                     2                                              T
                             I =11ma /12       v  = (0·732793, –0·678262, –5·45312E–2)
                              2
                                                2
                                     2                                           T
                             I =11ma /12       v  = (0·360111, 0·454562, –0·814674)
                                                3
                              3
                      with a maximum absolute residual of 7·62939E–6 and a maximum inner product
                      2·38419E–7.
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