Math 110 Professor K. A. RibetMidterm Exam November 3, 2003Please put away all books, calculators, electronic games, cell phones, pagers,.mp3 players, PDAs, and other electronic devices. You may refer to a single2-sided sheet of notes. Please write your name on each sheet of paper thatyou turn in; don’t trust staples to keep your papers together. Explain youranswers in full English sentences as is customary and appropriate. Yourpaper is your ambassador when it is graded.1. Let V be a vector space over a field F and let v be a vector in V . Let T : V →V be a linear transformation. Suppose that Tm(v) = 0 for some positive integerm but that Tm−1(v) is non-zero. Show that the s pan of { v, T (v), . . . , Tm−1(v) }has dimension m.We have to show that the vectors v, T (v), . . . , Tm−1(v) are linearly independent.Assume the contrary, i.e., that some non-trivial linear combination of these vec-tors vanishes. Take a linear combination w ith the fewest poss ible terms and writeit in the form 0 =rXi=0aiTi(v) with r ≤ m − 1. We must have ar6= 0 becauseotherwise we could re-write the sum with fewer terms. Also, some aiwith i < rmust be non-zero because otherwise the whole sum would consist of one term andwe’d end up with Tr(v) = 0, which is contrary to assumption. We apply Tm−rto the linear combination and obtain 0 =rXi=0aiTi+m−r(v) =r−1Xi=0aiTi+m−r(v),with the latter equality coming from the assumption Tm(v) = 0. We thus havea vanishing linear combination with fewer terms than the “minimal” one thatwe started with; this is a contradiction.2. Let T : V → V be a linear map on a non-zero finite-dimensional vector spaceV over a field F . Suppose that the characteristic polynomial of T splits over Finto a product of linear factors. Show that there is a basis B of V such that [T ]Bis upper-triangular.We prove the statement by induction on the dimension of V ; it is trivial ifdim V = 1. The characteristic polynomial of T has a root, and thus T hasan eigenvector v1. The vectors that are multiples of v1form a 1-dimensionalsubspace W of V that is invariant under T (in the sense that T (W ) ⊆ W ). Aswe saw in the homework that was due on October 17, T induces a linear mapU : V/W → V/W whose characteristic polynomial divides that of T . By theinduction hypothesis, there is a basis v2, . . . , vnof V/W in which the matrixof U is upper-triangular. Here, we understand that we are choosing vectorsv2, . . . , vnof V and that the viare their images vi+ W in V/W . The vectorsv1, v2, . . . , vnform a basis of V in which T is upper-triangular.3. Let V be an n-dimensional real or complex inner product space. Let e1, . . . , enbe an orthonormal basis of V . Suppose that T : V → V is a linear transformationand let T∗: V → V be the adjoint of T . Show thatnXj=1kT∗(ej)k2=nXj=1kT (ej)k2.If kT∗vk ≤ kT vk for all v ∈ V , show that kT∗vk = kT vk for all v ∈ V .For the first part, we just compute. Say T (ej) =Xiaijeifor each i. ThenkT (ej)k2=Xi|aij|2, so thatnXj=1kT (ej)k2=nXi,j=1|aij|2. The analogues of theaijfor T∗are the aji, but interchanging i and j and applying a complex conju-gation does not change the double sum that we have just computed. Hence weindeed havenXj=1kT∗(ej)k2=nXj=1kT (ej)k2. For the second part, we suppose thatkT∗vk < kT vk for some v ∈ V . This vector must be non-zero; we can assumethat it has norm 1 by dividing it by its norm. (This preserves the inequality thatwe started with.) Complete the singleton set {v} to a basis v = v1, v2, . . . , vnof V and apply the Gram-Schmidt process to this basis to obtain an orthogonalbasis of V . Divide the resulting vectors by their norms to obtain an orthonormalbasis e1, . . . , enof V . T he first element of this basis, e1, is v. We now look atnXj=1kT∗(ej)k2=nXj=1kT (ej)k2, comparing terms. For each j, the jth term on theleft-hand side is less than or equal to the jth term on the right-hand side. Onthe other hand, the first term on the left is strictly less than the first term onthe right. This is a contradiction: the sums cannot turn out to be equal. Ourhypothesis kT∗vk < kT vk was therefore incorrect, so we are forced to concludethat kT∗vk = kT vk for all v.H110 last midterm—page
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