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Unit 14: Invariant Subspaces




          In other words, A  = 0 if j  r and i  r.                                            Notes
                        ij
          Schematically, A has the block form

                                        B C 
                                   A =                                           ...(3)
                                        0 D 
          where B is an r × r matrix, C is an r × (n – r) matrix, and D is an (n – r) × (n – r) matrix. The reader
          should note that according to (2) the matrix B is precisely the matrix of the induced operator Tw
          in the ordered basis ’.
          Most often, we shall carry out arguments about T and Tw without making use of the block form
          of the matrix A in (3). But we should note how certain relations between Tw and T are apparent
          from that block form.
          Lemma: Let W be an invariant subspace for T, the characteristic polynomial for the restriction
          operator Tw divides the characteristic polynomial for T. The minimal polynomial for Tw divides
          the minimal polynomial for  T.
          Proof: We have

                                                B C 
                                            A     
                                                0 D 
          where A = [T]  and B = [T ] . Because of the block form of the matrix
                              w ’
                                  det (xI – A) = det (xI – B) det (xI – D)
          That proves the statement about characteristic polynomials. Notice that we used I to represent
          identity matrices of three different sizes.

              th
          The k  power of the matrix A has the block form
                                                B k  C 
                                            k       k
                                           A      k 
                                                 0  D  
          where C  is some r × (n – r) matrix. Therefore, any polynomial which annihilates A also annihilates
                 k
          B (and D too). So, the minimal polynomial for B divides the minimal polynomial for A.


                 Example 5: Let T be any linear operator on a finite-dimensional space  V. Let W be the
          subspace spanned by all of the characteristic vectors of T. Let c ,...,c  be the distinct characteristic
                                                            1  k
          values of T. For each i, let W be the space of characteristic vectors associated with the characteristic
                                i
          value c , and let   be an ordered basis for W . The lemma before Theorem 2 of unit 12 tells us that
                i       i                    i
          ’ = ( ,..., ) is an ordered basis for W. In particular,
                1   k
                                    dim W = dim W  + ... + dim W .
                                                 1          k
          Let ’ = { ,..., } so that the first few ’s form the basis  , the next few  , and so on. Then
                   1   r                                1            2
                                       T  = t ,         i = 1,...,r
                                          i  i  i
          where (t ,..., t ) = (c , c ,..., c ,..., c , c ,..., c ) with c  repeated dim W  times.
                 1   r   1  1  1   k  k   k     i             i
          Now W is invariant under T, since for each  in W we have
                                          = x   + ... + x 
                                             1  1     r  r
                                        T  = t x   + ... + t x 
                                           1 1  1    r r  r




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