Math 110 Professor K. A. RibetFinal Exam December 16, 2003Please put away all books, calculators, electronic games, cell phones, pagers,.mp3 players, PDAs, and other electronic devices. You may refer t o 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 A be an n × n matrix. Suppose that there is a non-zero row vector y suchthat yA = y. Prove that there is a non-zero column vector x such that Ax = x.(Here, A, x and y have entries in a field F .)This is a restatement of problem 6 on HW #14. What is given is that 1 is aneigenvalue of the transpose of A. It follows that 1 is an e igenvalue of A; thisgives the conclusion.2. Let A and B be n × n matrices over a field F . Suppose that A2= Aand B2= B. Prove that A and B are similar if and only if they have the sam erank.This is problem 10 in HW #14. If A and B are similar, then they certainly havethe same rank. Indeed, we saw early on in the course that the rank of a matrixdoes not change if you multiply it on either side by an invertible matrix. Theharder direction is the converse. If T2= T , where T is a linear operator on avector space V , then we know well that V is the direct sum of the null space of Tand the space of vectors that are fixed by T . (See, e.g., problem 17 on page 98of the textbook.) The dimension of this latter space is clearly the rank of T .Choose a basis v1, . . . , vrfor the range of T and a basis vr+1, . . . , vnfor the nullspace of T . The matrix of T with respect to the basis v1, . . . , vnis the directsum of the r × r identity matrix and the (n − r) × (n − r) zero-matrix. Takingnow T = LA, we see that A is similar to a matrix that depends only on its rank.If A and B have the same rank, they are each similar to a common matrix, sothey’re similar to each other.3. Suppose that T : V → V is a linear transformation on a finite-dimensionalreal inner product space. Let T∗be the adjoint of T . Show that V is the directsum of the null space of T and the range of T∗.The rank of T∗coincides with the rank of T for various reasons. For example,in matrix terms, this equality is the statement that a square matrix and itstranspose have the same rank. Hence the dimensions of the null space of T andthe range of T∗add up to the dimension of V . This necessary condition forV to be the indicated direct sum is a good sign! Also, it means that V is thedirect sum of the two spaces if and only if V is the sum of the two spaces andthat V is the direct sum of the two spaces if and only if the spaces have zerointersection in V . Let us prove the latter statement. Suppose that T (v) = 0 andthat v = T∗(w) for some w. We need to prove that v = 0. It is enough to showthat hv, vi = 0. But hv, vi = hv, T∗(w)i = hT (v), wi = h0, wi = 0.4. Let A be a symmetric real matrix whose square has trace 0. Show that A = 0.Use the fact that A is similar to a diagonal matrix. If B is similar to A, then Bhas the same trace as A; also, B = 0 if and only if A = 0. Hence we can, and do,assume that A is a diagonal matrix. Say that the diagonal entries are a1, . . . , an.The hypothesis is thatXa2i= 0. Since the aiare real numbers, they all mustbe 0. Hence A = 0.5. Let T : V → W be a linear transformation between finite-dimensional vectorspaces. Let X be a subspace of W . Let T−1(X) be the set of vectors in Vthat map to X. Show that T−1(X) is a subspace of V and that dim T−1(X) ≥dim V − dim W + dim X.This seems to be problem 2 of the “further review problems.” As I write thisanswer, I have the impression that the problem is harder than I thought, butperhaps there’s an easier way to say what I’m about to explain. Let Y be therange of T , so that X and Y are both subspaces of W . A third subspace isX ∩ Y . Consider the quotient space W/X and the natural map ι: Y → W/Xthat sends y ∈ Y to y + X. The null space of this map is Y ∩ X. Hence dim Y =dim(Y ∩X)+rank(ι) ≤ dim(Y ∩X)+dim(W/X) = dim(Y ∩X)+dim W −dim X.Now let U be the restriction of T to T−1(X). Since T−1(X) contains the nullspace of T , the nullity of U is the same thing as the nullity of T . The rangeof U is Y ∩ X. We have dim T−1(X) = nullity(T ) + dim(Y ∩ X) ≥ nullity(T ) +dim X + dim Y − dim W = dim V + dim X − dim W , where we have used theequality dim V = nullity(T ) + rank(T ) = nullity(T ) + dim Y .6. Suppose that V is a real finite-dimensional inner product space and thatT : V → V is a linear transformation with the property that hT (x), T (y)i = 0H110 final—page 2whenever x and y are elements of V such that hx, yi = 0. Assume that there isa non-zero v ∈ V for which kT (v)k = kvk. Show that T is orthogonal.This is a slightly friendlier version of problem 8 on HW #14. After scaling v, wemay assume that kvk = 1. Complete v to a basis of V and then apply the Gram-Schmidt process. We emerge with an orthonormal basis e1, . . . enof V with v =e1. Let i be greater than 1 and let w = ei. Then hv+w, v+w i = hv, vi+hw, w i = 2because v ⊥ w. Similarly, hT (v + w), T (v + w)i = kT (v)k2+ kT (w)k2becauseT (v) ⊥ T (w) by hypothesis. It follows that kT (w)k2= 1 because we knewalready that kT (v)k2was 1. In other words, we have kT (ei)k = 1 for all i;equivalently hT (ei), T (ej)i = δijfor all i and j. It follows by linearity thathT (x), T (y)i = hx, yi for x, y ∈ V . Thus T is orthogonal.7. Let T be a nilpotent operator on a finite-dimensional complex vector space.Using the tablei 0 1 2 3 4 5 · · ·nullity(Ti) 0 4 7 9 10 10 · · · ,find the Jordan canonical form for T .I …