Homework 6 Solutions. Solution. Note {e t, te t, t 2 e t, e 2t } is linearly independent. If β = {e t, te t, t 2 e t, e 2t }, then

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1 Homework 6 Solutions 1 Let V be the real vector space spanned by the functions e t, te t, t 2 e t, e 2t Find a Jordan canonical basis and a Jordan canonical form of T on V dened by T (f) = f Solution Note {e t, te t, t 2 e t, e 2t } is linearly independent If β = {e t, te t, t 2 e t, e 2t }, then 1 1 [T ] β = A = Thus, p T (x) = (x 1) 3 (x 2) We look at N((A I) 3 ) We have 0 1 A I = 2 0 which has rank 3, so V 1 = N((A I)) has dimension d 1 = 1 From this we must have d 2 = 2 and d 3 = 3 (as there is only one Jordan block) We see 2 0 (A I) 2 = and (A I) 3 = Hence, N((A I) 2 ) = {(a, b, c, d) c = d = 0} and N((A I) 3 ) = {(a, b, c, d) d = 0}, so {(0, 0, 1, 0), (A I)(0, 0, 1, 0), (A I) 2 (0, 0, 1, 0)} works as a basis for N((A I)) This is {(0, 0, 1, 0), (0, 2, 0, 0), (2, 0, 0, 0)}, or {t 2 e t, 2te t, 2e t } For N((A 2I)), we see e 2t works, so our Jordan canonical basis is {t 2 e t, 2te t, 2e t, e 2t } with corresponding matrix Find a Jordan canonical basis and a Jordan canonical form of the operator T on P 3 (R) given by T (f(x)) = xf (x) Solution In the basis β = {1, x, x 2, x 3 }, we have A = [T ] β = Thus, p T (x) = x 4 We see A 2 = 2 1

2 and A 3 = 0, so we have d 1 = dim N(A) = 2, d 2 = 3, and d 4 = 4 We see N(A) = {(a, b, c, d) c = d = 0}, N(A 2 ) = {(a, b, c, d) d = 0} Hence, we may choose (0, 0, 0, 1) to start our rst chain We get A(0, 0, 0, 1) = (0, 0, 6, 0) and A 2 (0, 0, 0, 1) = (0, 12, 0, 0) Now, in N(A) we already have (0, 12, 0, 0) Hence, to complete a basis of N(A) we take (1, 0, 0, 0) Thus, our Jordan canonical basis is {(0, 0, 0, 1), (0, 0, 6, 0), (0, 12, 0, 0), (1, 0, 0, 0)} which is {x 3, 6x 2, 12x, 1} with corresponding matrix Find a Jordan canonical basis and a Jordan canonical form of the operator T on P 3 (R) given by T (f) = f + 2f Solution In the basis β = {1, x, x 2, x 3 }, we have A = [T ] β = so p T (x) = (x 2) 4 We have 2 0 (A 2I) = 0 6 and (A 2I) 2 = 0 We see N((A 2I)) = {(a, b, c, d) c = d = 0} and so a basis of N((A 2I) 2 ) relative to N((A 2I)) is given by {(0, 0, 1, 0), (0, 0, 0, 1)} Applying A 2, we get the chains {(0, 0, 1, 0), (2, 0, 0, 0)} and {(0, 0, 0, 1), (0, 6, 0, 0)} Hence, our Jordan basis is {(0, 0, 1, 0), (2, 0, 0, 0), (0, 0, 0, 1), (0, 6, 0, 0)} which is 2 0 {x 2, 2, x 3, 6x} and which has corresponding matrix Find [ a ] Jordan canonical basis and a Jordan canonical form of the operator T on M 2 2 (R) given by 3 1 T (A) = A A 0 3 t Solution We'll use B as the matrix of T in the basis β = {e 1, e 2, e 3, e 4 } where e 1 = ] [ ] [ ], e 3 =, and e = We see 0 1 [ B = [T ] β = [ ] 1 0, e 2 = so p T (x) = (x 2)((x 3)(x 3)(x 2) 1(x 2)) = (x 2) 2 ((x 3) 2 1) This is (x 2) 2 (x 2 6x + 8) = (x 2) 3 (x 4) For λ = 2, we have 1 0 (B 2I) =

3 which has rank 3 We see which has rank 2 Lastly (B 2I) 2 = (B 2I) 3 = which has rank 1 Thus we have N((B 2I) 2 ) = {(a, b, c, d) b + c = 0 and 2b 2c + d = 0} and N((B 2I) 3 ) = {(a, b, c, d) 2b + 2c d = 0} Hence, a basis for N((B 2I) 3 ) relative to N((B 2I) 2 ) is {(0, 0, 1, 2)} We have (B 2I)(0, 0, 1, 2) = (1, 1, 1, 0) and (B 2I) 2 (0, 0, 1, 2) = (1, 0, 0, 0) We have (B 4I) = so N((B 4I)) = {(a, b, c, d) 2a + c = b c + d = b c = d = 0} which is {(a, b, c, d) 2a + c = b + c = d = 0} Hence, a nonzero vector in N((B 4I)) is (1, 2, 2, 0) Adding this to our basis for N((B 2I) 3 ), we get {(0, 0, 1, 2), (1, 1, 1, 0), (1, 0, 0, 0), (1, 2, 2, 0)} as a Jordan basis with corresponding matrix Find [ a ] Jordan canonical basis and a Jordan canonical form of the operator T on M 2 2 (R) given by 3 1 T (A) = (A A 0 3 t ) Solution With β as in 4 we have B = [T ] β = so p T (x) = x 2 ((x 3) 2 9) = x 2 (x 2 6x) = x 3 (x 6) We see B has rank 1, so dim N(B) = 3 We see N(B) = {(a, b, c, d) b + c = 0}, so a basis for this space is {(0, 1, 1, 0), (1, 0, 0, 0), (0, 0, 0, 1)} Now, B 6I = so N((B 6I) = {(a, b, c, d) 6a b + c = 3b 3c = 6d = 0}, or {(a, b, c, d) 6a b + c = b + c = d = 0} Hence, a nonzero vector in this space is (1, 3, 3, 0) With the other vectors, our Jordan basis is thus {(0, 1, 1, 0), (1, 0, 0, 0), (0, 0, 0, 1), (1, 3, 3, 0)} with corresponding matrix 0 6 3

4 6 Let V be the real vector space of polynomials in 2 variables of degree at most 2 Find a Jordan basis and a Jordan canonical form for the operator T on V given by T (f) = f x + f y (here subscripts denote dierentiation with respect to those variables) Proof Let β = {1, x, y, x 2, xy, y 2 } be a basis for V Then, A = [T ] β = 1 2 so p T (x) = x 6 We see 2 2 A 2 = and so A 3 = 0 Thus, d 1 = 3, d 2 = 5, and d 3 = 6 For a basis of N(A 3 )/N(A 2 ), we may take any vector (a, b, c, d, e, f) with d + e + f 0, such as (0, 0, 0, 0, 1, 1) Then, A(0, 0, 0, 0, 1, 1) = (0, 1, 3, 0, 0, 0) and A 2 (0, 0, 0, 0, 1, 1) = (4, 0, 0, 0, 0, 0) Now, in N(A 2 )/N(A), we have (0, 1, 3, 0, 0, 0) + N(A) We wish to nd a basis relative to N(A), so rst we nd a basis of N(A) We have N(A) = {(a, b, c, d, e, f) b + c = 2d+e = e+2f = 0}, so a basis for this space is {(1, 0, 0, 0, 0, 0), (0, 1, 1, 0, 0, 0), (0, 0, 0, 1, 2, 1)} We extend this to a basis of N(A 2 ) by including (0, 1, 3, 0, 0, 0) from our chain and some other vector (a, b, c, d, e, f) with d + e + f = 0 We see (0, 0, 0, 0, 1, 1) works, and we use it to get a chain with A(0, 0, 0, 0, 1, 1) = (0, 1, 1, 0, 0, 0) Now, in N(A), we have (4, 0, 0, 0, 0, 0) and (0, 1, 1, 0, 0, 0) To extend this to a basis of N(A), we include (0, 0, 0, 1, 2, 1) Hence, our Jordan basis is {(0, 0, 0, 0, 1, 1), (0, 1, 3, 0, 0, 0), (4, 0, 0, 0, 0, 0), (0, 0, 0, 0, 1, 1), (0, 1, 1, 0, 0, 0), (0, 0, 0, 1, 2, 1)} which is {xy + y 2, x + 3y, 4, xy + y 2, x + y, x 2 2xy + y 2 } with corresponding matrix Let A M n n (F ) be such that p A splits Prove that A and A t are similar Proof 1 Let J A be a Jordan canonical form for A, which exists as p A splits We note A J A We claim J A JA t Indeed, suppose β = m k=1 β k is a Jordan canonical basis for which each block in J A corresponds to some β k For β k = {x 1, x 2,, x l }, let γ k = {x l,, x 1 } Then, for W k = span(β k ), we see that W k is T -invariant with λ λ 0 [T ] βk = 0 0 λ 4

5 and so λ λ 0 [T ] γk = 1 0 λ Thus, J A J t A We claim J A t J t A Indeed, if J A = C 1 AC, then J t A = Ct A t (C 1 ) t and (C 1 ) t = (C t ) 1, so J t A At Proof 2 We note that for arbitrary λ F and k N 0, dim N((A t λi) k ) = dim N(((A λi) k ) t ) = dim V columnrank(((a λi) k ) t ) which, as column rank = row rank, is dim V rowrank(((a λi) k ) t ) = dim V columnrank((a λi) k ) = dim N((A λi) k ) As J A is determined by the values dim N((A λi) k ), we conclude J A = J A t up to rearrangement Hence, A J A = J A t A t 8 Let T be a linear operator on a nite dimensional vector space such that p T splits and let λ 1,, λ k be the eigenvalues of T Dene S : V V by S(v) = k j=1 λ jv j where v = k j=1 v j is the (unique) expansion of v as a sum of generalized eigenvectors Claim S is diagonalizable and T S is nilpotent Proof Let β be a Jordan basis for T, so [T ] β = J where J is a Jordan block matrix Then, if v β, we have v W λj (the generalized eigenspace corresponding to λ j ) for some 1 j k Hence, S(v) = λ j v, so [S] β is a diagonal matrix Moreover, [S] β [T ] β is a matrix consisting of all zeros except possibly some ones immediately below the diagonal Thus, [S T ] β is strictly lower triangular and hence nilpotent (eg by (S T ) t being strinctly upper triangular and hence nilpotent and ((S T ) t ) m = ((S T ) m ) t for all m) 9 Let P 1, P 2 be two projectors in the vector space V such that P 1 P 2 = P 2 P 1 Show that P 1 P 2 is also a projector and that R(P 1 P 2 ) = R(P 1 ) R(P 2 ) Proof We see (P 1 P 2 ) 2 = P 1 P 2 P 1 P 2 = P 1 P 1 P 2 P 2 = P 1 P 2, so P 1 P 2 is a projector Now, suppose x R(P 1 P 2 ), say x = P 1 P 2 (v) Then, x = P 1 (P 2 (v)), so x R(P 1 ) Moreover, x = P 2 (P 1 (v)) by hypothesis, so x R(P 2 ) Thus, x R(P 1 ) R(P 2 ) Now, let x R(P 1 ) R(P 2 ), so x = P 1 (v) = P 2 (u) for some v, u V Then, P 1 (x) = P 1 (P 1 (v)) = P 1 (v) = x and P 2 (x) = P 2 (P 2 (u)) = P 2 (u) = x Thus, P 1 (P 2 (x)) = P 1 (x) = x, so x R(P 1 P 2 ) 10 Let P [ : V V ] be a projector in a vector space V of dimension n Prove there is a basis β of V such Ik 0 that [P ] β = n k Proof P 2 P = 0 so if f(x) = x(x 1), we see m p f Thus, m P is a product of distinct linear factors, so P is diagonalizable As 0 and 1 are the only zeros of f, they are the only eigenvalues of P, so by choosing a basis β of eigenvectors in which those with eigenvalue 1 come rst, we get the desired result 5