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4.3 Linear Independence                                                            189
                   Example 4.3.6

                                    Let V be the vector space of real-valued functions of a real variable, and let S =
                                    {f 1 (x),f 2 (x),...,f n (x)} be a set of functions that are n−1 times differentiable.
                                                 28
                                    The Wronski    matrix is defined to be
                                                           f 1 (x)   f 2 (x)  ···   f n (x)
                                                                                          



                                                          f (x)     f (x)    ···   f (x)  
                                                                      2
                                                            1
                                                                                     n
                                                                                          
                                               W(x)=        .         .      .       .      .
                                                             .         .              .
                                                            .         .       . .    .    
                                                                                          
                                                          (n−1)      (n−1)         (n−1)
                                                         f     (x) f     (x) ··· f n    (x)
                                                          1          2
                                    Problem: If there is at least one point x = x 0 such that W(x 0 ) is nonsingular,
                                    prove that S must be a linearly independent set.
                                    Solution: Suppose that
                                                     0= α 1 f 1 (x)+ α 2 f 2 (x)+ ··· + α n f n (x)  (4.3.18)
                                    for all values of x. When x = x 0 , it follows that
                                              0= α 1 f 1 (x 0 )+ α 2 f 2 (x 0 )+ ··· + α n f n (x 0 ),


                                              0= α 1 f (x 0 )+ α 2 f (x 0 )+ ··· + α n f (x 0 ),

                                                               2
                                                                              n
                                                     1
                                                .
                                                .
                                                .
                                                     (n−1)         (n−1)              (n−1)
                                              0= α 1 f    (x 0 )+ α 2 f  (x 0 )+ ··· + α n f  (x 0 ),
                                                     1             2                  n
                                                            
                                                           α 1
                                                           α 2

                                    which means that v =  .  ∈ N W(x 0 ) . But N W(x 0 ) = {0} because
                                                             
                                                         
                                                            .
                                                            .
                                                           α n
                                    W(x 0 ) is nonsingular, and hence v = 0. Therefore, the only solution for the
                                    α ’s in (4.3.18) is the trivial solution α 1 = α 2 = ··· = α n = 0 thereby insuring
                                    that S is linearly independent.
                                 28
                                                                                                 e
                                    This matrix is named in honor of the Polish mathematician Jozef Maria H¨oen´ Wronski
                                    (1778–1853), who studied four special forms of determinants, one of which was the deter-
                                    minant of the matrix that bears his name. Wronski was born to a poor family near Poznan,
                                    Poland, but he studied in Germany and spent most of his life in France. He is reported to have
                                    been an egotistical person who wrote in an exhaustively wearisome style. Consequently, almost
                                    no one read his work. Had it not been for his lone follower, Ferdinand Schweins (1780–1856)
                                    of Heidelberg, Wronski would probably be unknown today. Schweins preserved and extended
                                    Wronski’s results in his own writings, which in turn received attention from others. Wronski
                                    also wrote on philosophy. While trying to reconcile Kant’s metaphysics with Leibniz’s calculus,
                                    Wronski developed a social philosophy called “Messianism” that was based on the belief that
                                    absolute truth could be achieved through mathematics.
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