Advanced Linear Algebra Homework for Spring 2006 (revised 19 May 2006) [these problems are being modified, so check frequently; the dates refer to when the topic was covered in class; please submit your work, and particularly problems from Curtis, on the Thursday of the following week] 1/31 Resolve the parallelogram conjecture, and do the other exercises suggested in class, including: cross product on R^3 is a non- associative binary operation. [Read Chapter 1 of Curtis] 2/2 Do the exercises suggested in class about abelian groups, rings and fields (division rings), especially: In any ring, r0 = 0r = 0; in a ring with 1, r(-1) = -r = (-1)r, where in each case, r is any element of the ring. Also try this: for what n is the ring Z/nZ a field? Can you describe any other finite fields? Also, assuming the rational numbers Q and the real numbers R are field, verify that the complex numbers C = R + i R is also a field, and that the quaternions H = R + i R + j R + k R is a (noncommutative) field, where i^2 = j^2 = k^2 = ijk = -1; similarly for Q in place of R. [Read Chapter 2 of Curtis] 2/7 Do the module exercises suggested in class, as well as: v0 = O and -v = v(-1) for any vector v in a module; O = Ok for any scalar k in the ring (note the difference between "O" and "0"); also, show the {moon} is a module over any ring. Check the details that any abelian group A is naturally a (right or left) Z-module (as we began to verify in class). For those who know what ideals are: any (right or left) ideal S in a ring T is an example of a (right of left) T-module; explore the possibilities when T = Z (the integers) or T = Z/nZ. Let Z[i] be the ring of Gaussian integers in C, and let M = Z[i,j,k] be the analogous quaternion integers. Show that M is both a (right) Z[i]-module and ring in its own right. Let R[x^2] be the ring of real-valued polynomial functions in x^2, that is, all those polynomials whose terms are only of even degree. Let M_0 and M_1 be all real even and odd functions, respectively. Verify that each M_i is an R[x^2]-module. Can you generalize this? 2/9 Try the submodule and subspace problems suggested in class. Also try these: Let S be any subset of V. Check that span(S) is a subspace. Show that S={1, x, x^2, ... , x^n, ...} is a linearly independent set in the vector space F[x] of all polynomial functions of x, and thus a basis for F[x], since clearly span(S)=F[x]. Also try the related problem, Curtis 2.5#3. Show that the sets S={e_1, e_2, ... , e_n} and T={e_1, e_1+e_2, ... e_1+...+e_n} are bases for K^n. 2/14 Curtis 2.4#3cefg,4cdeg,7,8,10 2/16 Curtis 2.5#1,5 2/21 [No class because of Washington and Lincoln - read rest of Chapter 2 of Curtis] 2/23 In the proof of the "spanning sets are at least as big as linearly independents sets" for vector spaces (over possibly noncommutative fields) we noticed Curtis' proof needed a fix at the first step - provide the fix! Also, the proof used the existence of inverses for nonzero elements in the field, but the result makes sense in any R-module - is the result *true* in any R-module? 2/28 Curtis 2.7#1,3,4 [Hint: use dimension theorem and previous problem] [Read Chapter 3 of Curtis] 3/2 Let L = [.]_B be the linear map V -> F^n which assigns a vector in V its coordinates with respect to B. Let S be the standard basis of F^n. What is the matrix [L]_SB? Show that Hom(V,W) = {linear maps V -> W} is a vector space of dimension dim(Hom(V,W)) = dim(V) dim(W). Curtis 3.11#3,4,5,7,9,10 Curtis 3.12#4,5,8 Curtis 3.13#1,3,7,12 3/7 Let V -L-> W -M-> X be linear maps. Suppose B, C, and D are bases for V, W and X, respectively. Verify that the matrix for the composition ML (w.r.t. bases B and D) is the matrix product of the matrices for M and L (w.r.t....). (Try the case where V, W and X are each 2-dimensional first!) Try to work through the details of rank+nullity theorem proof of the dimension theorem: use the linear map from the direct sum of subspace V and W onto the sum V+W defined by (v,w) -> v-w, and show that the kernel is isomorphic to the intersection of V and W. 3/9 Let V = F^n where F = Z/2Z is the field of two elements. How many ordered bases are there for V? (This is a tricky counting problem: try n=2 case first! How many unordered bases are there? Try the same problem with any prime p instead of p=2. 3/14 Hamming code problems [I'll distribute these in class]. 3/16 Think of the following problems as take-home midterm: As a warm-up, try the problem I adapted from the "car guys": Consider the Roman alphabet a basis for a vector space (or a free module over the integers Z), and any word is regarded as a vector sum of its letters. If we define a linear map so that ONE -> 2, TWO -> 3 and ELEVEN -> 5, then where does TWELVE -> ?? Is this enough information to determine where TEN -> ?? What about TWENTY -> ?? (Give details!) Let L: V -> W be a linear map between finite dimensional vector spaces. Let k = dim(im(L)) = rank(L), m = dim(W) and n = dim(V). Then there are bases B for V and C for W such that the m x n matrix for L w.r.t these bases is | I_k 0 | [L]_CB = | | | 0 0 | where I_k is the k x k identity matrix and where 0 denotes a matrix of all 0's of the appropriate size. (Prove this!) In other words, any m x n matrix A can be put in the above form by multiplying it on the left and right by invertible m x m and n x n matrices X and Y: | I_k 0 | XAY = | | . | 0 0 | This means that the only invariant of a linear map when we are free to choose bases for range and domain independently is rank(L), which is simply an integer between 0 and m. (Also prove this!) Try carrying this out explicitly for the matrix | 1 2 3 | A = | 4 5 6 | . | 7 8 9 | (We sketched this last part in class - write up the details!) [You should also read Chapters 5, 6 and 7 of Curtis over the break (we'll return to ideas from 4 later).] 3/28 Curtis 7.22#6,7,8,9,10 3/30 Curtis 7.22#2(cf. hw from 3/2 above),4,5,11,12 4/4 Show that the space of k-forms on F^n is a vector space over F. Try to compute the dimension of this vector space. This is useful for the following: Verify that my definition of det (an n-form \omega with \omega(e_1,...,e_n) = 1) agrees with Curtis's. Also verify this gives an algorithm for computing det (Theorem 16.9 in Curtis 5.16). Curtis 5.16#4 and 5.18#4 4/6 Curtis 5.19#7,10 4/11 Curtis 7.23#3,4,6 4/13 Curtis 7.24#4,10 4/18 Curtis 7.24#1,2[skip part g in each],3 Also show that if N:W -> W is nilpotent of order e (N^e = 0, but N^{e-1} \= 0) and w is a nonzero vector in W which is not in ker(N^{e-1}), then {w, N(w), N^2(w),...,N^{e-1}(w)} is linearly independent in W. 4/20 Curtis 7.25#2,3,4,8[read/think these through, at least] In class we defined the exponential of an n x n matrix (over a subfield of the complex numbers for the series to converge) exp(A) = I + A + ... + A^k/k! + .... Use the triangular form theorem A = D + N, where D is diagonalizable and N is nilpotent of some order e at most n (and, as is clear once in triangular form, DN = ND) to get a simpler formula for exp(A). [Hint: in general, exp(A+B) = exp(A)exp(B) *only* when A and B commute.] Carry this out to compute exp(A) for the 2 x 2 matrix |-14 -25| A = | | , | 9 16| which is similar (conjugate) to the matrix |1 1| B = | | . |0 1| ===================================================================== Let's summarize our "streamlined" alternative approach to the canonical form theorems, avoiding some of the ring theory in Curtis, and using instead "Fitting's Lemma": First we discussed finding a basis in which a nilpotent linear map N:V -> V has d x d matrix (d = dim V) in the canonical form 0 0 0 ... 0 0| 1 0 0 ... 0 0| 0 1 0 ... 0 0| 0 0 0 1 ... 0 0| : : : : : 0 0 0 ... 1 0| ------------------- | 0 | 0 | where the e x e block corresponds to order of nilpotency e of N. This is what I call Fitting's Lemma. We proved that if v is a nonzero vector in V which is not in ker(N^{e-1}), then {v, N(v), N^2(v),..., N^{p-1}(v)} is linearly independent in V, and that it can be extended to a basis with the properties desired above. (This is essentially the problem that I suggested for 4/18.) Consequently, e is at most d. (Note: the argument in class actually only dealt with the case rank(N) = e-1; the general case is a little more complicated, since the rank of N can be as large as d-1, and the canonical form of N can in fact have several blocks of the above form; the idea is that there could be a larger linearly independent set {v1,...,vs} not in ker(N^{e-1} which would give rise to s blocks of size e; and it's even more complicated, because there could be additional vectors, independent of N(v1),...,N(vs), and these will give rise to more blocks of size e-1; so the ultimate canonical form of a nilpotent N may have several blocks of size e, several more of size e-1, and so forth....) Note that this works over any field F, not necessarily algebraically closed. I emphasized this result because it is also at the heart of rational canonical form, also valid over any F. Next step: we applied this to an arbitrary linear map L: V -> V under the assumption that its minimal polynomial m(x) factors into powers of linear factors m(x) = (x - a_1)^{e_1} ... (x - a_k)^{e_k} (which will be true automatically when F is closed field). A key point is that the map N_i:V_i -> V_i, defined as (L - a_i) restricted to V_i = ker((L - a_i)^{p_i}), is nilpotent of order e_i (by minimality of m). This L restricted to V_i has the block form indicated above, except with a_i along the main diagonal. A lemma which I referred to in class verifies that V_i is in fact an L-invariant subspace (try as an exercise): Suppose L, M are commuting linear maps V -> V, that is LM = ML. Then ker(M) and im(M) are L-invariant subspace (similarly for the roles of M and L reversed). We applied this with M=(L-a_i)^{e_i}, which clearly commutes with L. Finally, we deduce the Hamilton-Cayley theorem from this: 1) m(x) divides the characteristic polynomial p(x) = det(x-L) 2) L satisfies p(x), i.e. p(L)=0 at least when F is closed (since then clearly the factors of m divide those of p); but a little thought shows that this must really be true for arbitrary F as well (exercise, using the fact we can extend to algebraic closure K of F to get all the roots, but coefficients of polynomials still remain in F)! One final note: problem 23#6 is not correct if the characteristic of F is 2, since then the matrix |1 0| | | |1 1| satisfies the hypotheses, but not the conclusion - you might think about this in terms of canonical form theorem above (the square of this matrix is I, so it has minimal polynomial x^2 - 1 = (x-1)^2 in characteristic 2! ===================================================================== 4/25 Curtis 4.15#2,9[this relies on the exercise suggested in class, i.e. V = W \directsum W^\perp for any subspace W of V],12,13 4/27 Carry out the KAN factorization of your favorite 2 x 2 matrices, including my favorite: |1 2| C = | | . |3 4| Determine the symmetries of the capital letters of the Roman alphabet, assuming these are written in the maximally symmetric form (in particular, L and Q have nontrivial symmetries, and O has an infinite family), and express as matrices with respect to the standard basis of R^2. Find the QR factorization of the (singular) 3 x 3 matrix |1 2 3| | | A = |4 5 6| . | | |7 8 9| 5/2 Back to KAN factorization. For small matrices, this isn't very practical, but the factorization C = KAN gives a nice way to compute the inverse of C: in fact, the inverses of K and A are very easy, being K* and 1/A (meaning the diagonal matrix with reciprocal entries); the inverse of N is also easy: since N = I - P where P is nilpotent, its inverse is I + P + P^2 + ... + P^k where k+1 is the order of nilpotency (verify this). Carry this out for the 2 x 2 example from 4/27. [Here and forever more, A* denotes the (conjugate, if complex entries) transpose of a matrix A.] 5/4 Let W be the subpace of R^4 perpendicular to [1 1 1 1]*. Compute matrices for orthogonal projections onto W^\perp and onto W. Find the closest points on W and W^\perp to the vector [1 2 3 4]*. Curtis 9.30#1,2,3 5/9 Curtis 9.31#1,3,5-8[really one extended problem] THE FOLLOWING MATERIAL WILL BE ADAPTED INTO YOUR TAKE HOME FINAL EXAM: Suppose we have an endomorphism L: V -> V of a real vector space V, and we hope to find eigenspace and eigenvalues, or at least invariant subspaces. Today we discussed how to do this using complex eigenspaces and eigenvalues for L, by regarding V as a real subspace of a complex vector space V_C. To make sense of this in general requires the notion of tensor product, but we can be rather concrete: if V has basis B (over R) then V_C also has basis B (over C); in other words, we simply extend the coefficient field from R to C. (Note that the same trick works for any field extension F to K: a vector space V_F over F can be viewed as an F-subspace of V_K). Equivalently (and even more concretely) we know that we may simply regard V as R^n and V_C as C^n, by choosing some basis B as above. Note that complex conjugation defines a real (but not complex) linear mapping #:C^n -> C^n which fixes R^n, that is, R^n is the 1-eigenspace of # (in class I used a "bar" instead of #, but can't here without TeX). In fact, if U is any subspace of C^n which is fixed by # then the vectors in U have all entries real. Consider any complex subspace W of C^n and its image #W under complex conjugation. Inside W+#W is the (real) subspace U of vectors of the form u=w+#w for some w in W; again, these are just the vectors fixed by #. Return now to our R-linear map L: R^n -> R^n, which we may extend to a C-linear map C^n -> C^n. Suppose w is an eigenvector of L with eigenvalue m (both may be complex), and let W be the 1-dimensional complex subspace spanned by w. Then W+#W is an L-invariant subpace, and because L is real, its real subspace U is also L-invariant. There are two possibilities: 1) If the eigenvalue m is real, then L(w+#w)=mw+#(mw)=m(w+#w) so u=w+#w is a real eigenvector, and W=#W. Conversely, if W=#W, then U is 1-dimensional over R, spanned by a real eigenvector u, and in this case the eigenvalue m must be real. 2) When m is not real, W and #W are distinct, U is 2-dimensional over R, and L restricted to U can be put in the form |(m+#m)/2 -(m-#m)/2i| | | |(m-#m)/2i (m+#m)/2 | in a suitable basis for U (and conversely). Note that in the special case m is a unit complex number (as in class), 2) is just the corresponding rotation matrix with cosines and sines of the argument (angle) of m; and 1) occurs for a reflection with m=-1 or 1 (cf. Curtis Theorems 30.2 and 30.3). 5/11 The following is part of your final exam: Let V be a finite dimensional vector space over C, with hermitian inner product (think of C^n with (x,x)=x*x if you wish). The key lemma for a linear map A: V -> V which lets us show that it can be diagonalized with respect to an orthonormal (unitary) basis when self-adjoint (symmetric or hermitian) is this: if a subspace W is A-invariant, then its orthogonal complement W^\perp is A*-invariant. Prove this lemma, and apply it to show we can diagonalize skew-adjoint, unitary, or, more generally, normal (A*A=AA*) maps in the same way. To do this you should prove and use the lemma (cf. Curtis 32.13): Suppose L, M are commuting linear maps V -> V, that is LM = ML. Then L and M have a common 1-dimensional invariant subspace W. [Compare to the lemma from 4/25, and contrast with the following: commuting linear maps need not have the same invariant subspaces, since obviously a scalar multiple of the identity (for which any subspace of V is invariant) commutes with any L (which may have only particular invariant subspaces).] To show that a normal linear map can be diagonalized, let L=A and M=A*, and observe that A**=A. Thus the n-1 dimensional subspace W^\perp is also invariant under under both A and A*, so we can use the dimension reduction argument as before (induction on n if you like). In applications, one of the most useful ways to compute eigenvectors and eigenvalues of a symmetric matrix A is to find the maximum of the associated quadratic form Q(x) = x*Ax among unit vectors x. Show that the maximum point v is an eigenvector of A belonging to the largest eigenvalue, the maximum value Q(v). More generally, the critical points and values of Q correspond to the eigenvectors and eigenvalues of A. To find such maximum value, please justify the following algorithm: pick a random unit vector (say, v_0 = e_1) to begin, then apply A to get v_1 = Av_0, then rescale to get unit vector u_1; we can iterate this to get a sequence of unit vectors u_n (in practice, simply use v_n = Av_{n-1} and rescale only at the end to get u_n) which converge to the eigenvector v. (In general, the eigenvectors of A are fixed points [up to "+/-" sign] of this rescaled action of A on the set of unit vectors.) Carry out this algorithm for the 2 x 2 symmetric matrix | 1 2 | A = | | . | 2 3 | Fibonacci numbers will appear, and the u_n should converge to a multiple of [1 T]* where T is the "golden mean" (1 + \sqrt5)/2. By this method, you should find the largest eigenvalue of A is T^3. Check algebraicly that this is a root of the characteristic polynomial. What happens when you begin the iteration at v_0 = e_2? Actually, a better algorithm for the largest eigenvalue/vector of A is to iterate v -> (1-h)v + hAv , rescaling to the sphere of unit vectors as appropriate. For h small this is a discrete approximation to the gradient flow of Q, which will generically converge to the largest eigenvalue/vector. (When h=1, this coincides with the above algorithm.) PLEASE NOTE: I EXPECT TO BE AROUND GANG (LGRT 1535) IN THE AFTERNOONS NEXT WEEK (MTW, 5/22-24) - IF YOU COMPLETE YOUR EXAM AND PASS IT TO ME BEFORE NOON ON THURSDAY 5/25, I WILL BE GRATEFUL - I'LL THEN BE OUT OF TOWN TILL TUESDAY 5/30 AND WILL ONLY ACCEPT EXAMS TILL NOON ON THAT DAY WITH ADVANCE (BEFORE 5/24) EMAIL NOTICE FROM YOU.... HAVE A MATHEMATICALLY WONDERFUL SUMMER!!!!!!!!! ====================================================================== ==========================TAKE-HOME FINAL============================= ====================================================================== PART A. Suppose we have an endomorphism L: V -> V of a real vector space V, and we hope to find eigenspaces and eigenvalues, or at least invariant subspaces. Here's a variation of what we did in class, using complex eigenspaces and eigenvalues for L acting on a complex vector space V_C, regarding V as a real subspace of V_C. To make sense of this in general requires the notion of tensor product, but we can be rather concrete: if V has basis B (over R) then V_C also has basis B (over C); in other words, we simply extend the coefficient field from R to C. (Note that the same trick works for any field extension F to K: a vector space V_F over F can be viewed as an F-subspace of V_K). PROBLEM A.1. In the situation above, show that V_K = span_K(B) is indeed a vector space over F, that V=V_F is a subspace of V_K, and that dim_F(V_K) = dim_F(K) dim_F(V), where we regard K itself as a vector space over F. Check the case F=R and K=C=R+iR: if B is a basis for V over R, then the union of B and iB is a basis for V_C over R, so dim_R(V_C) = 2 dim_R(V). Equivalently (and even more concretely) we know that we may simply regard V as R^n and V_C as C^n, by choosing some basis B as above. We will assume this perspective for the rest of the problem Note that complex conjugation defines a real (but not complex) linear mapping #:C^n -> C^n which fixes R^n, that is, R^n is the 1-eigenspace of #. In fact, if U is any subspace of C^n which is fixed by # then the vectors in U have all entries real. Consider any complex subspace W of C^n and its image #W under complex conjugation. Inside W+#W is the (real) subspace U of vectors of the form u=w+#w for some w in W; again, these are just the vectors fixed by #. PROBLEM A.2. What is the matrix for # in the standard R basis {e1, ie1, e2, ie2, ... , en, ien} for C^n? Return now to our R-linear map L: R^n -> R^n, which we may extend to a C-linear map C^n -> C^n. Suppose w is an eigenvector of L with eigenvalue m (both may be complex), and let W be the 1-dimensional complex subspace spanned by w. PROBLEM A.3. Show that W+#W is an L-invariant subpace, and because L is real, its real subspace U is also L-invariant. There are two possibilities for U: 1) If the eigenvalue m is real, then L(w+#w)=mw+#(mw)=m(w+#w) so u=w+#w is a real eigenvector, and W=#W. Conversely, if W=#W, then U is 1-dimensional over R, spanned by a real eigenvector u, and in this case the eigenvalue m must be real. 2) When m is not real, W and #W are distinct, U is 2-dimensional over R, and L restricted to U can be put in the form |(m+#m)/2 -(m-#m)/2i| | | |(m-#m)/2i (m+#m)/2 | in a suitable basis for U (and conversely). PROBLEM A.4. Verify statements 1) and 2) above, and check that in the special case m is a unit complex number (as mentioned in class), 2) is just the corresponding rotation matrix with cosines and sines of the argument (angle) of m; and 1) occurs for a reflection with m=-1 or 1 (cf. Curtis Theorems 30.2 and 30.3). PART B. Let V be a finite dimensional vector space over C, with hermitian inner product (,) - think of C^n with (x,y)=x*y if you wish. The key lemma for any linear map A: V -> V which lets us show that it can be diagonalized with respect to an orthonormal (or unitary) basis when A is symmetric (or hermitian) is this: If a subspace W of V is A-invariant, then its orthogonal complement W^\perp is A*-invariant, where A*:V->V is defined by the property (A*x,y) = (x,Ay). PROBLEM B.2. Prove this lemma. In the following, we will apply this lemma to show we can diagonalize not only symmetric (or hermitian) maps, but also skew-symmetric (or skew-hermitian), unitary, or, more generally, normal (A*A=AA*) maps in the same way. PROBLEM B.3. Check that hermitian (A*=A), skew-hermitian (A*=-A), and unitary (A*A=I=AA*) maps are all normal. To apply this do this you will need another lemma (cf. the first part of Curtis 32.13, but this is more general): Suppose L, M are commuting linear maps V -> V, that is LM = ML. Then L and M have a common 1-dimensional invariant subspace W, i.e. a common eigenvector w with span({w}) = W. PROBLEM B.4. Prove this lemma. [Compare to the lemma from 4/20, and contrast with the following: commuting linear maps need not have the same invariant subspaces, since obviously a scalar multiple of the identity (for which any subspace of V is invariant) commutes with any L (which may have only particular invariant subspaces). You might also think about whether the following is true for a vector space V over an arbitrary field K: Suppose L, M are commuting linear maps V -> V, that is LM = ML. Then L and M have a common irreducible invariant subspace W. In particular, when K is algebraically closed, W is 1-dimensional, and when K is a real field, 1- or 2-dimensional; for example any pair of real 2-by-2 rotation matrices commute, but generally R^2 is the only invariant subspace.] Now, to show that a normal linear map can be diagonalized, let L=A and M=A*, and observe that A**=A. Thus the n-1 dimensional subspace W^\perp is also invariant under under both A and A*, so we can use the dimension reduction argument as before (induction on n if you like). PROBLEM B.5. Carry out the details of the above paragraph. PART C. In applications, one of the most useful ways to compute eigenvectors and eigenvalues of an n-by-n symmetric (or Hermitian) matrix A is to find the maximum of the associated quadratic form Q(x) = x*Ax among unit vectors x, which being a closed, bounded set - indeed, a sphere of real dimension n-1 (or 2n-1) - must have a maximum by a calculus argument. This is known as the Rayleigh-Ritz method. Problem C.1. Show that the maximum point v is an eigenvector of A belonging to the largest eigenvalue, the maximum value Q(v). More generally, the critical points and values of Q correspond to the eigenvectors and eigenvalues of A. (Hint: Use calculus to verify the gradient vector gradQ satisfies gradQ|v = 2Av, so that at any critical point v, the method of Lagrange multipliers forces grad Q to be a multiple of grad(x,x)|v=2v, that is Av=av for some a in R. Try the 2-by-2 case first to see how the partial derivatives work before you do the n-by-n case.) To find such maximum value, pick a random unit vector (say, v_0 = e_1) to begin, then apply A to get v_1 = Av_0, then rescale to get unit vector u_1; we can iterate this to get a sequence of unit vectors u_n (in practice, simply use v_n = Av_{n-1} and rescale only at the end to get u_n) which converge to the eigenvector v. (In general, the eigenvectors of A are fixed points [up to "+/-" sign] of this rescaled action of A on the set of unit vectors, and thus you want to be sure your random vector is not already an eigenvector of A.) PROBLEM C.2. Explain why the above algorithm should work in general, and carry it out explicitly for the 2 x 2 symmetric matrix | 1 2 | A = | | . | 2 3 | What happens when you begin the iteration at v_0 = e_2? In the example, Fibonacci numbers will appear, and the u_n should converge to multiple of the column vector [1 T]* where T is the "golden mean" (1 + \sqrt5)/2. By this method, you should find the largest eigenvalue of A is T^3. PROBLEM C.3. Check algebraicly that this is a root of the characteristic polynomial. What is the other root? [Actually, a better algorithm for the largest eigenvalue/vector of A is to iterate v -> (1-h)v + hAv , rescaling to the sphere of unit vectors as appropriate. For h small this is a discrete approximation to the gradient flow of Q, which will generically converge to the largest eigenvalue/vector. When h=1, this coincides with the above algorithm.] PROBLEM C.4. In the symmetric 2-by-2 case, how might you find the smallest eigenvalue and its eigenvector? (Compare with C.3. Hint: The eigenspaces are perpendicular, so it's easy to find that other eigenvector in this case. In the n-by-n case there are eigenvalues intermediate to the largest and smallest, so these are obtained by a similar method: after finding the largest eigenvalue and its eigenvector w, restrict to the subsphere of unit vectors perpendicular to w and repeat to find the next largest eigenvalue and its eigenvector....) PLEASE NOTE: I EXPECT TO BE AROUND GANG (LGRT 1535) IN THE AFTERNOONS NEXT WEEK (MTW, 5/22-24) - IF YOU COMPLETE YOUR EXAM AND PASS IT TO ME (IN PERSON, OR UNDER MY OFFICE DOOR, IN AN ENVELOPE IF POSSIBLE) BEFORE NOON ON THURSDAY 5/25, I WILL BE GRATEFUL - AND YOU'LL GET A BONUS POINT! I AM OUT OF TOWN THEN TILL TUESDAY 5/30: ONLY WITH ADVANCE (BEFORE 5/24) EMAIL NOTICE FROM YOU, WILL I ACCEPT EXAMS TILL NOON ON 5/30.... ====================================================================== ==========================TAKE-HOME FINAL============================= ======================================================================