Discussion of the eigensystem solver on the TI-85.
Also explains the normalization of eigenvectors.
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       TOPIC:   Normalization of eigenvectors
     PRODUCT:   TI-85

     This note is posted to make some general comments about the
     eigensystem solver on the TI-85.

     An eigenvalue of a square matrix A is a value E such that
     A * X = E * X  for some nonzero vector X.  The vector X is an
     eigenvector associated with E.

     For an nxn matrix, there will be n (not necessarily distinct)
     eigenvalues and associated eigenvectors. (These eigenvectors are
     properly called right or column eigenvectors, in contrast to left
     or row eigenvectors.)

     The eigenvalues of A are unique, but any eigenvector can be scaled
     by a nonzero multiplicative constant and remain a solution, so the
     eigenvectors are not unique.  Thus the values of an eigenvector on
     the TI-85 may not agree with another computer or book solution.
     In particular, hand solutions for eigenvectors based on solving
     a set of simultaneous equations from the identity  (A - E*I)*X = 0
     (I is the identity matrix) will usually not be the same as a
     computer solution.  The TI-85 does not normalize the eigenvectors.
     Some computer software normalize the eigenvectors to a maximum
     value of one, some normalize the eigenvectors to a norm of one.
     Either of these normalizations can be done with a few additional
     calculations on the TI-85.

     An example may illustrate:   A = [[2,2,3][1,2,1][2,-2,1]]
     If we calculate eigVl A  and store in list E, we get E={-1,4,2}.
     If we calculated the eigenvectors by hand we might get for the
     eigenvalue (-1) the solution of:

     [[ 3  2  3 ]      [V1      [0	 As:  V1=1
      [ 1  3  1 ]   *	 V2   =   0         V2=0
      [ 2 -2  2 ]]      V3]      0]        V3=-1

     Note that we cannot use a matrix inverse or simultaneous equation
     solver here, because (A - E*I) is always singular.  Also the trivial
     solution V1=V2=V3=0 is not allowed by the definition.  In general, one
     or more of the elements of the eigenvector can be selected arbitrarily
     (another reason there are an infinite number of eigenvectors).  In
     this case we can see from the reduced row echelon form of (A - E*I)
     that V1+V3 = 0 (or, V1= -V3) and V2 = 0.

     On the TI-85, if we calculate eigVc A  and store in matrix V, we get
     (in Fix 3 mode):

     V=[[.667	      1.073    .745 ]
        [1.790e-14  .671     1.118]
        [-.667      .268     -.745]]

     The eigenvector corresponding to the eigenvalue (-1) is the first
     column of V.  Similarly the eigenvector for the eigenvalue (4) is
     the second column.  If we put the eigenvalues on the diagonal of
     a 3x3 matrix D (all off diagonal elements = 0), we can confirm the
     identity  A*V - V*D = 0 (to about 1e-13).	Hint: Edit the matrix D
     in the matrix editor and on the diagonal enter E(1), E(2), E(3).

     To normalize the first eigenvector to a max value of one, try:

     V(1,1,3,1) * (1/V(1,1))   Note: we are picking the maximum absolute
				     value by inspection here, it is
				     not always V(1,1).

     Or to normalize to a norm of one, try:

     V(1,1,3,1) * (1/norm V(1,1,3,1))  for the first eigenvector,
     V(1,2,3,2) * (1/norm V(1,2,3,2))  for the second eigenvector, etc.

     We hope this discussion helps you.  We welcome your questions and
     comments, through GRAPH-TI or by email to the address below.
---------------------------------------------------------------------------
  TI GRAPHIC PRODUCTS TEAM
  Texas Instruments (Consumer Products)
  P.O. Box 650311  M/S 3908		  Internet: ti-cares@lobby.ti.com
  Dallas, Texas  75265			  Fax: 214-917-7103
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