Page 362 - 35Linear Algebra
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362                                                                          Sample Final Exam


                                                               2
                                   (h) When λ = µ and g/λ = R , we get the equation for a circle radius R in
                                       the (z, w)-plane. When λ, µ and g are postive, we have the equation for
                                       an ellipse. Vanishing g along with λ and µ of opposite signs gives a pair
                                       of straight lines. When g is non-vanishing, but λ and µ have opposite
                                       signs, the result is a pair of hyperbolæ. These shapes all come from
                                       cutting a cone with a plane, and are therefore called conic sections.

                              13. We show that L is bijective if and only if M is invertible.
                                   (a) We suppose that L is bijective.

                                         i. Since L is injective, its kernel consists of the zero vector alone.
                                           Hence
                                                                 L = dim ker L = 0.
                                           So by the Dimension Formula,


                                                            dim V = L + rank L = rank L.
                                           Since L is surjective, L(V ) = W. Thus

                                                             rank L = dim L(V ) = dim W.

                                           Thereby
                                                              dim V = rank L = dim W.
                                        ii. Since dim V = dim W, the matrix M is square so we can talk
                                           about its eigenvalues. Since L is injective, its kernel is the zero
                                           vector alone. That is, the only solution to LX = 0 is X = 0 V .
                                           But LX is the same as MX, so the only solution to MX = 0 is
                                           X = 0 V . So M does not have zero as an eigenvalue.
                                        iii. Since MX = 0 has no non-zero solutions, the matrix M is invert-
                                           ible.
                                   (b) Now we suppose that M is an invertible matrix.

                                         i. Since M is invertible, the system MX = 0 has no non-zero solu-
                                           tions. But LX is the same as MX, so the only solution to LX = 0
                                           is X = 0 V . So L does not have zero as an eigenvalue.
                                        ii. Since LX = 0 has no non-zero solutions, the kernel of L is the
                                           zero vector alone. So L is injective.
                                        iii. Since M is invertible, we must have that dim V = dim W. By the
                                           Dimension Formula, we have

                                                                 dim V = L + rank L


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