Page 275 - A Course in Linear Algebra with Applications
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8.1:  Basic  Theory  of  Eigenvectors       259


       The  general  solution  of this  system  is £1  =  |(—1 +  y/—l)  and
           —  d  , where  d  is an  arbitrary  scalar.  Thus  the  eigenvectors
        x 2
        of A  associated  with the  eigenvalue C\  are the non-zero  vectors
        of the  form




        Notice  that  these,  together  with  the  zero  vector,  form  a  1-
                                    2
        dimensional  subspace  of  C .  In  a  similar  manner  the  eigen-
        vectors  for  the  eigenvalue  3 — \/—T are  found  to  be the  vectors
        of the  form




        where  d ^  0.  Again these form with the zero vector  a subspace
            2
        ofC .
            It  should  be  clear  to  the  reader  that  the  method  used  in
        this  example  is  in  fact  a  general  procedure  for  finding  eigen-
        vectors  and  eigenvalues.  This  will  now  be  described  in  detail.

        The  characteristic   equation    of  a  matrix
            Let  A  be  an  n  x n  matrix  over  a field  of scalars  F,  and  let
        X  be  a non-zero  n-column  vector  over  F. The  condition  for  X
        to  be  an  eigenvector  of  A  is  AX  =  cX,  or

                               {A  -  d n)X  =  0,


        where  c  is the  corresponding  eigenvalue.  Hence  the  eigenvec-
        tors  associated  with  c,  together  with  the  zero  vector,  form
        the  null  space  of  the  matrix  A  —  cl n.  This  subspace  is  often
        referred  to  as the  eigenspace  of the  eigenvalue  c.
             Now  (A — d n)X   =  0 is a linear system  of n  equations  in  n
        unknowns.   By  3.3.2 the  condition  for there to  be  a  non-trivial
        solution  of the  system  is that  the  coefficient  matrix  have  zero
        determinant,
                               det(A  -  cl n)  =  0.
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