Math 110, Spring 2015: Homework 10 Solutions

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1 Math 0, Spring 205: Homework 0 Solutions Section 7 Exercise 73: Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let λ, λ 2,, λ k be the distinct eigenvalues of T For each i, let J i be the Jordan canonical form of the restriction of T to K λi Prove that is the Jordan canonical form of T J = J J 2 J k Remark: The statement of the exercise in the book directs you to prove that J is the Jordan canonical form of J Although that statement actually makes sense (it amounts to saying, Prove that the matrix J is in Jordan canonical form ), I m guessing the book intended to state the Jordan canonical form of T The question is also somewhat confusing because it refers to the Jordan canonical form, while the uniqueness of the Jordan canonical form up to ordering of Jordan blocks is not established until the next section of the text Proof We first observe that the matrix J is in Jordan canonical form (cf the remark above) Indeed, by the definition of the direct sum of square matrices on page 320, J is block diagonal with diagonal blocks J i By definition, each J i is in Jordan canonical form; thus, each J i is itself block diagonal with Jordan diagonal blocks So by refining the block-diagonal decomposition of J we see that J is a block-diagonal matrix with Jordan diagonal blocks (Alternatively (and equivalently), note that each J i, being in Jordan canonical form, is a direct sum of Jordan blocks in the sense defined on page 320 Thus J is a direct sum of Jordan blocks and is thus in Jordan canonical form Strictly speaking, one ought to observe that the direct sum of square matrices is an associative operation, so that (A B) C = A B C) It remains to show that J is a matrix representation of T, ie, J = [T] β for some ordered basis β of V For each i k, let β k be a Jordan canonical basis of K λi with respect to which J i is the matrix representation of the restriction of T to K λi We claim β = β β k is the desired basis of V : We first observe that β is in fact a basis of V By Theorem 78 (which was proven in lecture), we have V = k K λi i= Theorem 50(d) then shows that β is a basis of V That J = [T] β then follows from Theorem 525 (Writing out a proof of Theorem 525 in terms of the entry-by-entry definition of page 320 is somewhat instructive but quite tedious At the least, you should be able to explain to someone in words why that theorem is true)

2 Section 72 Exercise 723: Let T be a linear operator on a finite-dimensional vector space V with Jordan canonical form (a) Find the characteristic polynomial of T Solution: Let J denote the Jordan canonical form above; noting that J ti is upper triangular, so that its determinant is simply the product of its diagonal entries, we have p T (t) = det(j ti) = (2 t) 5 (3 t) 2 (b) Find the dot diagram corresponding to each eigenvalue of T Solution: The eigenvalues of T are λ = 2 and λ 2 = 3 The dot diagram for each eigenvalue λ has a column corresponding to each λ-jordan block (in decreasing order of block length from left to right), with each block having a number of dots equal to the length of the corresponding Jordan block In this case, the dot diagrams are λ = 2 λ 2 = 3 (c) For which eigenvalues λ i, if any, does E λi = K λi? Solution: One can read this off from the matrix by looking for eigenvalues whose Jordan blocks are all of length ; that condition is equivalent to the dot diagram s having only one row This holds (only) for the eigenvalue λ = 3 (d) For each eigenvalue λ i, find the smallest positive integer p i for which K λi = N ( (T λ i I) p i) Solution: The integer p i is the maximum length of a λ i -Jordan block in the Jordan canonical form of T; equivalently, it s the number of rows in the dot diagram for λ i (ie the maximum length of a column in the dot diagram) So in this case p = 3 and p 2 = 2

3 (e) Compute rank(u i ), rank(u 2 i ), nullity(u i ), and nullity(u 2 i ) for each i, where U i denotes the restriction of T λ i I to K λi Solution: By Theorem 79, the number of dots in the first r rows of the dot diagram for λ i is equal to nullity(u r i ) So we have nullity(u ) = 2, nullity(u 2 ) = 2 nullity(u 2 ) = 4, nullity(u 2 2) = 2 Noting that dim(k λi ) is the total number of dots in the dot diagram for λ i, we can apply the Dimension Theorem to obtain rank(u ) = 3, rank(u 2 ) = 0 rank(u 2 ) =, rank(u 2 2) = 0 (Note, however, that it s a bit silly to use dot-diagram reasoning to get the results for U 2 ; indeed, U 2 is the zero operator on K 3 ) Exercise 724: For each of the matrices A that follow, find a Jordan canonical form J and an invertible matrix Q such that J = Q AQ Remark: The solutions for this exercise demonstrate certain shortcut techniques that can be used in the following two scenarios for an eigenvalue λ of an operator T (or a matrix A): dim(e λ ) = dim(k λ ), so that the dot diagram for λ consists of one row 2 dim(e λ ) =, so that the dot diagram for λ consists of one column (b) A = Solution: The characteristic polynomial of A is p A (t) = ( t)(2 t) 2, so the eigenvalues of A are λ = and λ 2 = 2 By Theorem 74(c), we know that the dimension of each generalized eigenspace is the multiplicity of its corresponding eigenvalue as a root of p A (t); that is, dim(k ) =, dim(k 2 ) = 2 Note that dim(k λi ) gives the number of dots in the dot diagram for λ i From this it s clear that the dot diagram of λ = consists of one dot To find the dot diagram of λ 2 = 2, we begin by computing the dimension of the eigenspace E 2 = N(A 2I), which will give us the number of columns in the dot diagram (or equivalently the number of associated Jordan blocks in the Jordan canonical form of A) It s quite easy in this case to show that dim(e 2 ) = 2 = dim(k 2 ) So that tells us that the dot diagram for λ 2 = 2 consists of one row of two dots All in all, we have λ = λ 2 = 2 3

4 From this dot diagram, we can immediately write down a Jordan canonical form of A: 0 0 J = To find a matrix Q such that J = Q AQ, we simply find a Jordan canonical basis β for A Then we can take Q to be the change-of-coordinates matrix Q = [I] γ β, where I : R 3 R 2 is the identity transformation and γ is the standard basis of R 3 In general, to find a Jordan canonical basis, we just have to find a basis for each generalized eigenspace consisting of a disjoint union of cycles; the dot diagram for each eigenvalue guides us by telling us the length of the cycles that will appear In this case, since K λ = E λ for both eigenvalues λ, we just find bases of the eigenspaces E and E 2 Using techniques from Math 54, we can find bases 0 β = 2, β 2 = 2, 0 for E and E 2, respectively Taking the ordered basis β = β β 2, we can take Q to be the changeof-coordinates matrix 0 Q = (c) A = Solution: Once again, p A (t) = ( t)(2 t) 2, so the eigenvalues are λ = and λ 2 = 2, with dim(k ) = and dim(k 2 ) = 2 Again the dot diagram for λ = consists of one dot, and in particular we have K = E Using Math 54, one can find a basis 0 β = for E = K Again appealing to Math 54, one can see that dim(e 2 ) =, so the dot diagram for λ 2 = 2 must be λ 2 = 2 4

5 The basis for K 2 contributing to a Jordan canonical basis of R 3 for A will thus consist of one cycle; any cycle of length 2 will do To find such a cycle, we just find a nonzero 2-eigenvector x of A by solving the equation (A 2I)x = 0 One such eigenvector is x = which we take to be the initial vector of our cycle We can then find the end vector v by solving (A 2I)v = x = = v = 3 2, 5 for example So our cycle basis for K 2 is β 2 = { (A 2I)v, v }, and our Jordan canonical basis is β β 2 = 0 3,, 3, So taking Q to be the change-of-coordinates matrix 0 Q = 2, 3 5 we have J = Q AQ, where J is the Jordan canonical form 0 0 J =

6 Exercise 725: For each linear operator T, find a Jordan canonical form J of T and a Jordan canonical basis β for T (b) T is the linear operator on P 3 (R) defined by T ( f(x) ) = xf (x) Solution: We begin by computing the matrix representation of T with respect to a convenient basis for V = P 3 (R), say the standard basis β 0 = {, x, x 2, x 3 }: A := [T] β0 = From this we can see that λ = 0 is the only eigenvalue of T, with multiplicity m = 4 With a view toward determining the dot diagram for λ = 0, we also note that A 2 = (A 0I) 2 = , A3 = (A 0I) 3 = 0 With all this information in hand, we see that the dot diagram has dim(k 0 ) = m = 4 dots, the number of columns is dim(e 0 ) = nullity(t 0 I) = dim(v ) rank(a) = 4 2 = 2, and the number of dots in the second row is rank(a) rank(a 2 ) = So the dot diagram must be (T 0 I) 2 v (T 0 I)v v w where we have labeled the dots in each column as vectors in a cycle contributing to the desired Jordan canonical basis β To find the cycle corresponding to the first column, it suffices to find a suitable end vector v; note that v is the end vector of a cycle of length 3 if and only if (T 0 I) 3 v = 0 but (T 0 I) 2 v 0 Since we computed above that A 3 = 0, we know that every vector v V = P 3 (R) satisfies (T 0 I) 3 v = 0 One can either compute T 2( f(x) ) = f (x) + xf (x) or examine the matrix A 2 above to see that (T 0 I) 2 v 0 if and only if v P 3 (R) is a polynomial with nonzero cubic term So if we take v = x 3, we obtain a cycle { (T 0 I) 2 v, (T 0 I)v, v } = { 2x, 6x 2, x 3} Finally, to find an appropriate cycle {w} of length, we just need to find a 0-eigenvector w that s linearly independent of the the cycle we just found Either by directly examining the operator T or by solving the matrix equation Aw = 0, we can see that E 0 = ker(t) = span{, x}, 6

7 so we can just take w = Thus the Jordan canonical form of T is J = 0 0 0, and a Jordan canonical basis is β = {2x, 6x 2, x 3, } (e) T is the linear operator on M 2 2 (R) defined by ( ) 3 T(A) = (A A t ) 0 3 Solution: Taking β 0 = {E, E 2, E 2, E 22 } to be the usual standard basis of V = M 2 2 (R), we have 0 0 B = [T] β0 = From this we can compute the characteristic polynomial p T (t) = t 3 (t 6), which yields eigenvalues λ = 0 and λ 2 = 6, with multiplicities m = dim(k 0 ) = 3 and m 2 = dim(k 6 ) =, respectively We first deal with the eigenvalue λ = 0 It s fairly easy to see from the matrix B above that rank(t) =, so dim(e 0 ) = dim ( ker(t) ) = 4 = 3 = dim(k 0 ) Thus, every generalized 0-eigenvector of T is in fact a 0-eigenvector; in other words, every cycle corresponding to λ = 0 has length, and so the Jordan canonical form of T will have three Jordan blocks length corresponding to the eigenvalue λ = 0 To find a basis for K 0 consisting of cycles, we may simply take any basis for E 0 Again examining the matrix B and solving the equation Bx = 0, we can see that a basis for E 0 is β = {( ) 0, 0 0 ( ) 0, 0 ( )} ( ) 3 (Alternatively, note that because is invertible, we have T(A) = 0 if and only if A = A 0 3 t So it s not surprising that β is the usual basis for the symmetric matrices) Next, we consider the eigenvalue λ 2 = 6 Since dim(k 6 ) =, we have K 6 = E 6, and the Jordan canonical form of T will contain one Jordan block of length corresponding to λ 2 = 6 To find a basis for K 6 = E 6 consisting of cycles, it suffices to find any 6-eigenvector of T; working in β 0 -coordinates and solving the matrix equation (B 6I)x = 0, we see that we may take {( )} 3 β 2 = 3 0 7

8 as our basis for K 6 Thus, the Jordan canonical form of T is and a Jordan canonical basis is β = β β 2 = J = {( ) 0, ( ) 0, 0, ( ) ( 0 0 3, (f) V is the vector space of polynomial functions in two real variables x and y of degree at most 2, as defined in Example 4, and T is the linear operator on V defined by T ( f(x, y) ) = x f(x, y) + f(x, y) y )} Solution: Take the basis α = {, x, y, x 2, y 2, xy} for V as in Example 4; then A := [T] α = So the only eigenvalue of T is λ = 0; its multiplicity is m = dim(k 0 ) = 6, so the dot diagram will consist of 6 dots Examining the matrix A can show further that the number of columns in the dot diagram is dim(e 0 ) = dim(ker T) = 6 rank A = 3 We can further refine our understanding of the dot diagram by considering (T 0 I) 2 = 2 x x y + 2 y 2 (where we ve used equality of mixed partials), or equivalently its matrix representation A 2 = We see that rank ( (T 0 I) 2) =, so the second row of the dot diagram contains rank(t 0 I) rank(t 0 I) 2 = 3 = 2 8

9 dots Thus, the dot diagram for λ = 0 must be (T 0 I) 2 v (T 0 I)w u (T 0 I)v v w Proceeding as in part (b), we first find v such that T 2 v 0; we may take v = x 2, which gives the cycle β = { (T 0 I) 2 v, (T 0 I)v, v } = { 2, 2x, x 2} Next, to find a cycle {(T 0 I)w, w} of length two, we find w such that (T 0 I) 2 w = 0 but (T 0 I)w 0, taking care to ensure that the resulting cycle is linearly independent of the previously constructed cycle One might be tempted to guess w = y, which will generate a cycle of length 2; however, the initial vector of the resulting cycle would be, which is not linearly independent of the previous cycle However, taking w = x 2 y 2 yields the cycle β 2 = { (T 0 I)w, w } = { 2x 2y, x 2 y 2}, which does the job Finally, to get our last cycle of length we just need a 0-eigenvector of T that is linearly independent of the previously collected vectors Examining the matrix A (or using Math 54 techniques to find its nullspace), we see that we can take β 3 = {u} = {2xy x 2 y 2 } Putting all of this together, we obtain the Jordan canonical form and a Jordan canonical basis J = β = β β 2 β 3 = {2, 2x, x 2, 2x 2y, x 2 y 2, 2xy x 2 y 2 } 9

10 Exercise 727: Let A be an n n matrix whose characteristic polynomial splits, γ be a cycle of generalized eigenvectors corresponding to an eigenvalue λ, and W be the subspace spanned by γ Define γ to be the ordered set obtained from γ by reversing the order of the vectors in γ (a) Prove that [T W ] γ = ( [T W ] γ ) t Solution: (Presumably here T = L A : F n F n ) Informally, [T W ] γ is obtained by first reversing the order of the columns of [T W ] γ and then reversing the order of the rows of the resulting matrix More formally, write C = [T W ] γ and B = [T W ] γ Write γ = {v,, v k } and γ = {u,, u k }, so that u j = v k+ j for all j k What we want to prove is that B ij = C ji for all i, j k But since γ is a cycle, A is a Jordan block; ie, if j = i + C ji = λ if j = i 0 otherwise To phrase this in terms of T, note that for l k we have while T(v ) = λv But this means v l = (T λ I)v l+ T(v l+ ) = v l + λv l+, T(u j ) = T(v k+ j ) = { uj+ + λu j if j k λu j if j = k By the definition of the matrix representation B = [T W ] γ of T W with respect to the basis γ if i = j + B ij = λ if i = j 0 otherwise = C ji (b) Let J be the Jordan canonical form of A Use (a) to prove that J and J t are similar Solution: Again, let T = L A : F n F n (where F is the base field over which the characteristic polynomial of A splits) It follows from part (a) that if β is a Jordan canonical basis for T and β is the basis of F n obtained by reversing the ordering of β, then [T] β = ( [T] β ) t = J t But then taking Q to be the change-of-coordinates matrix Q = [I] β β, we have so J is similar to J t (c) Use (b) to prove that A and A t are similar Q JQ = Q [T] β Q = [T] β = J t, Solution: Let P be the matrix whose columns are the vectors of a Jordan canonical basis for T = L A Then P JP = A, which also shows J t = (P t ) A t (P t ) So with denoting similarity, we have A J J t A 2, where the middle similarity is part (b) Since similarity is an equivalence relation, A A t 0

11 Exercise 728: Let T be a linear operator on a finite-dimensional vector space, and suppose that the characteristic polynomial of T splits Let β be a Jordan canonical basis for T (a) Prove that for any nonzero scalar c, {cx x β} is a Jordan canonical basis for T Solution: Write β = {cx x β}; it s easy to see that because β is a basis, so is β Indeed, the change-of-coordinates matrix [I] β β is just ci But then if J is the Jordan canonical form of T, so that [T] β = J, we have [T] β = [I] β β [T] β[i] β β = (ci) J(cI) = (c c)j = J So the matrix representation of T with respect to β is in Jordan canonical form; ie, β is a Jordan canonical basis for T (b) Suppose that γ is one of the cycles of generalized eigenvectors that forms β, and suppose that γ corresponds to the eigenvalue λ and has length greater than Let x be the end vector of γ, and let y be a nonzero vector in E λ Let γ be the ordered set obtained from γ by replacing x by x + y Prove that γ is a cycle of generalized eigenvectors corresponding to λ, and that if γ replaces γ in the union that defines β, then the new union is also a Jordan canonical basis for T Solution: We first prove that γ is a cycle of generalized λ-eigenvectors Suppose γ has length k, so that γ = {(T λ I) k x,, (T λ I)x, x} Since y is an eigenvector of T with eigenvalue λ, we have (T λ I) q y = 0 for all q N Thus (T λ I) q (x + y) = (T λ I) q x for all q N, in particular, k is the minimal natural number such that (T λ I) k (x + y) = 0 Thus, γ = {(T λ I) k x,, (T λ I)x, x + y} = {(T λ I) k (x + y),, (T λ I)(x + y), x + y} is a cycle Note that this also shows γ is disjoint from all of the other cycles appearing in β: If x + y appeared in some other cycle δ, then we would also have (T λ I)(x + y) = (T λ I)x δ γ, which would mean γ and δ were not disjoint to begin with Now let β denote the new union Thus far we have shown that β is a disjoint union of cycles; therefore, if β is a basis, then it must be a Jordan canonical basis for T By construction, β contains dim(v ) vectors, so it suffices to show that β spans V Because β is a basis of V, it further suffices to show that β span(β ), and because x is the only vector in β not contained in β, we may reduce further still to showing that x span(β ) But since the initial vectors of the cycles corresponding to λ in β (and hence in β ) form a basis of E λ, we must have y span(β ), and so (x + y) y = x span(β )

12 Exercise 720: Let T be a linear operator whose characteristic polynomial splits, and let λ be an eigenvalue of T (a) Prove that dim(k λ ) is the sum of the lengths of all the blocks corresponding to λ in the Jordan canonical form of T Solution: It follows from Theorem 79 that the number of dots in the dot diagram of T Kλ is dim(k λ ) But the total number of dots in the diagram is the sum of the numbers of dots in each column, which in turn is the sum of the lengths of cycles corresponding to λ in a Jordan canonical basis for T, which in turn is the sum of the lengths of the Jordan blocks corresponding to λ in the Jordan canonical form of T (b) Deduce that E λ = K λ if and only if all the Jordan blocks corresponding to λ are matrices Solution: By the Corollary to Theorem 79 on page 499, dim(e λ ) is equal to the number of Jordan blocks corresponding to λ So E λ = K λ if and only if dim(e λ ) = dim(k λ ), which in turn occurs if and only if the number of Jordan blocks corresponding to λ is equal to the sum of the lengths of all the Jordan blocks corresponding to λ, which in turn occurs if and only if the length of each Jordan block corresponding to λ is, which finally in turn occurs if and only if each such Jordan block is a matrix Exercise 729: Let F = C and J = λ λ λ λ be the m m Jordan block corresponding to λ, and let N = J λi m Prove the following results: (a) N m = 0, and for r < m, N r ij = { if j = i + r 0 otherwise Solution: In my opinion, the cleanest way to phrase this and to think about it is by considering the action of N by left multiplication on the standard basis vectors in C m Note that the j th column of N is just Ne j ; thus, we see that { 0 if j = Ne j = e j if 2 j m Repeating this r times, we see N r e j = { 0 if j r e j r if r + j m Since N r ij is the ith entry in the column vector N r e j, the desired formulas follow immediately 2

13 (b) For any integer r m, J = r(r )(r m+2) λ r m+ r(r ) (r m+3) λ r m+2 λ r rλ r r(r ) 2! λ r 2 (m )! 0 λ r rλ r (m 2)! λ r Solution: Recall the binomial formula for powers of sums of two numbers (real or complex): here C(n, k) = (a + b) n = n C(n, k)a k b n k ; k=0 n! n(n ) (n k + ) = k!(n k)! k! is ( the binomial coefficient, read as n choose k (You may well be more familiar with the notation n ) k ; the only reason I ve avoided that notation here is to prevent confusion with matrices) If you think about the proof of this formula (you can prove it by induction), all it relies upon are purely combinatorial counting arguments and the associative, distributive, and commutative properties of multiplication and addition of numbers The point is that although matrix multiplication is not commutative in general, if A and B are two commuting m m matrices, ie AB = BA, then we can apply the binomial formula to take powers of their sum: (A + B) r = r C(n, k)a k B n k, k=0 with the convention that A 0 = I, the m m identity matrix Now note that we can write J = N + λi Both N and λi behave nicely with respect to taking powers, and they commute (everything commutes with the identity); the prospect of using the binomial formula motivates making this decomposition So, noting that N k = 0 for k m, we have J r = (N + λi) r = r C(r, k)n k (λi) r k = k=0 m k=0 C(r, k)λ r k N k Noting that, from part (a), the matrix N k has all entries zero except for s on the k th superdiagonal, we see that J r has the desired form 3