Page 358 - 35Linear Algebra
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358                                                                          Sample Final Exam


                                       Hence, to compute L(v) we use linearity of L

                                                             2
                                                      1
                                                                                              3
                                                                           1
                                                                   3
                                                                                     2
                                            L(v) = L(b v 1 + b v 2 + b v 3 ) = b L(v 1 ) + b L(v 2 ) + b L(v 3 )
                                                                                    1
                                                                                  
                                                                                   b

                                                        = L(v 1 ) L(v 2 ) L(v 3 )   2   .
                                                                                   b
                                                                                   b 3
                                   (f) From the notation of the previous part, we see that we can list linear
                                                            3
                                       transformations L : B → B by writing out all possible bit-valued row
                                       vectors

                                                                    0 0 0 ,

                                                                    1 0 0 ,

                                                                    0 1 0 ,

                                                                    0 0 1 ,

                                                                    1 1 0 ,

                                                                    1 0 1 ,

                                                                    0 1 1 ,

                                                                    1 1 1 .
                                                  3
                                                                                           3
                                       There are 2 = 8 different linear transformations L : B → B, exactly
                                                                              3
                                       the same as the number of elements in B .
                                   (g) Yes, essentially just because L 1 and L 2 are linear transformations. In
                                                                                   3
                                       detail for any bits (a, b) and vectors (u, v) in B it is easy to check the
                                       linearity property for (αL 1 + βL 2 )
                                              (αL 1 + βL 2 )(au + bv) = αL 1 (au + bv) + βL 2 (au + bv)

                                                   = αaL 1 (u) + αbL 1 (v) + βaL 1 (u) + βbL 1 (v)

                                                   = a(αL 1 (u) + βL 2 (v)) + b(αL 1 (u) + βL 2 (v))
                                                     = a(αL 1 + βL 2 )(u) + b(αL 1 + βL 2 )(v) .
                                       Here the first line used the definition of (αL 1 + βL 2 ), the second line
                                       depended on the linearity of L 1 and L 2 , the third line was just algebra
                                       and the fourth used the definition of (αL 1 + βL 2 ) again.
                                   (h) Yes. The easiest way to see this is the identification above of these
                                       maps with bit-valued column vectors. In that notation, a basis is
                                                       n                               o

                                                         1 0 0 , 0 1 0 , 0 0 1           .

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