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Unit 28: Positive Forms and More on Forms




          is obtained by adding to the rth column of                                            Notes
                                           A 11    A 1k
                                                   
                                           A      A
                                            k 1     kk
          a linear combination of its other columns. Such operations do not change determinants. That
          proves (6), except for the trivial observation that because B is triangular  (B) = B  ... B . Since A
                                                                    k     11   kk
          and P are invertible, B is invertible. Therefore
                                    (B) = B  ... B   0
                                          11  nn
          and so  (A)   0, k = 1, ..., n.
                 k
          Theorem 2: Let f be a form on a finite dimensional vector space V and let A be the matrix of f in
          an ordered basis B. Then f is a positive form if and only if A = A* and the principal minors of A
          are all positive.
          Proof: Suppose that A = A* and  (A)   k   n. By the lemma, there exists an (unique) upper-
                                    k
          triangular matrix P with P  = 1 such that B = AP is lower triangular. The matrix  P* is lower-
                                kk
          triangular, so  that P*B = P*AP is also lower triangular.  Since  A  is self-adjoint,  the matrix
          D = P*AP is self-adjoint. A self-adjoint triangular matrix is necessarily a diagonal matrix. By the
          same reasoning which led to (6),
                                    (D) =  (P*B)
                                   k     k
                                       =  (B)
                                         k
                                       =  (A).
                                          k
          Since D is diagonal, its principal minors are
                                    (D) = D  ... D .
                                   k      11   kk
          From  (D) > 0, 1   k   n, we obtain D  > 0 for each k.
                k                       kk
          If A is the matrix of the form f in the ordered basis B = { , ...,  }, then D = P*AP is the matrix of
                                                       1    n
          f in the basis {  , ...,   } defined by
                       1    n
                                          n
                                      j =   P ij i
                                         i  1
          Since D is diagonal with positive entries on its diagonal, it is obvious that
                                         X*DX > 0.  X   0
          from which it follows that f is a positive form.

          Now, suppose we start with a positive form f. We know that A = A*. How do we show that
            (A) > 0, 1   k   n? Let V  be the subspace spanned by  , ...,   and let f  be the restriction of f to
           k                  k                        1    k      k
          V    V . Evidently f  is a positive form on V  and, in the basis { , ...,  } it is represented by the
           k   k          k                  k               1    k
          matrix.
                                           A 11    A 1k
                                                   
                                           A k 1    A kk
          As a consequence of Theorem 1, we noted that the positivity of a form implies that the determinant
          of any representing matrix is positive.
          There are some comments we should make, in order to complete our discussion of the relation
          between positive forms and matrices. What is it that characterizes the matrices which represent



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