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1 Homework ssignment 11 Section 7. pp Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If N 1 nd N 2 re similr, they hve the sme miniml polynomil (cf. pg. 192). Conversely, suppose tht N 1 nd N 2 hve the sme miniml polynomil. The miniml polynomil must be x k for some 1 k. If k = 1 we get tht the mtrix is the zero mtrix, so both N 1 nd N 2 re the zero mtrix. If k = 2 we get tht the Jordn form for N 1 nd N 2 is 1 nd thus, they re similr. If k = then the Jordn form of both mtrices is 1 1 nd thus, they re similr. Exercise. If A is complex 5 5 mtrix with chrcteristic polynomil f = (x 2) (x + 7) 2 nd miniml polynomil p = (x 2) 2 (x + 7), wht is the Jordn form for A? Solution: The block mtrix ssocited to the chrcteristic vlue 2 is mtrix with 2 s long the digonl with n elementry Jordn mtrix of size 2 2 (the multiplicity of 2 in the miniml polynomil) s the first block. i.e.: Anlogously, the block mtrix ssocited to the chrcteristic vlue 7 is 2 2 mtrix with 7 s long the digonl with n elementry Jordn mtrix of size 1 1 s the first block, i.e.: ( 7 ) 7 Hence the Jordn form for A is:

2 Exercise 4. How mny possible Jordn forms re there for 6 6 complex mtrix with chrcteristic polynomil (x + 2) 4 (x 1) 2? Solution: ATTENTION!!! There re 8 possibilities for the miniml polynomil, this implies tht there re t lest 8 different Jordn forms. But the miniml polynomil (x + 2) 2 (x 1) my correspond to TWO different mtrices, nmely nd this corresponds to the fct tht 4 = but lso 4 = Anlogously the miniml polynomil (x + 2) 2 (x 1) 2 corresponds to TWO mtrices. Therefore we hve 1 different Jordn forms. Think of it in this wy: How mny blocks corresponding to the eigenvlue 2 cn we form? This is equivlent to In how mny wys cn we write 4 s sum k with i > nd k? The nswer is 4 = 4 4 = = = = i.e. in 5 different wys. Anlogously, we cn write 2 in only two different wys, nmely 2 = 2 nd 2 = Therefore, multiplying we get 1 different Jordn forms. Exercise 5. The differentition opertor on the spce of polynomils of degree less thn or equl to is represented in the nturl ordered bsis by the mtrix 1 2 Wht is the Jordn form of this mtrix? (F subfield of the complex numbers.)? Solution The chrcteristic polynomil for this mtrix is x 4 nd the miniml polynomil is x 4, therefore the Jordn Form consists of only one block ssocited to the chrcteristic vlue with its first elementry Jordn block of length 4. i.e. the Jordn form of the mtrix is

3 Section 8.1 pp Exercise 2. Let V be vector spce over F. Show tht the sum of two inner products on V is n inner product on V. Is the difference of two inner products n inner product? Show tht positive multiple of n inner product is n inner product. Solution: Let f 1 nd f 2 be two inner products, we will show tht f = f 1 + f 2 stisfies ll the xioms of n inner product. () f( + b, c) = f 1 ( + b, c) + f 2 ( + b, c) = f 1 (, c) + f 1 (b, c) + f 2 (, c) + f 2 (b, c) the first equlity is by definition, the second is becuse f 1 nd f 2 re inner products. But reordering, the lst term of the equlity is equl to f 1 (, c) + f 2 (, c) + f 1 (b, c) + f 2 (b, c) = f(, c) + f(b, c), the lst equlity is gin by definition. (b) f(k, b) = f 1 (k, b) + f 2 (k, b) = kf 1 (, b) + kf 2 (, b) = k(f 1 (, b) + f 2 (, b) = kf(, b) (c) f(b, ) = f 1 (b, ) + f 2 (b, ) = f 1 (, b) + f 2 (, b) = f 1 (, b) + f 2 (, b) = f(, b) (d) If, f 1 (, ) > nd f 2 (, ) > therefore f(, ) = f 1 (, ) + f 2 (, ) > The difference of inner products is NOT n inner product in generl: Let f 1 = f 2 nd f = f 1 f 2, nd let, then f (, ) = f 1 (, ) f 2 (, ) = (which contrdicts xiom d). The proof for the positive multiple of sclr product is nlogous to the prove for the sum. Exercise. Describe ll inner products on R 1 nd on C 1 Solution: Let f be n inner product on R 1, since f is liner on ech vrible we get: f(r, s) = f(r 1, s 1) = rf(1, s 1) = rsf(1, 1) therefore the inner product of the vectors r nd s is just the product of the rel numbers r nd s times f(1, 1). But we know tht f(1, 1) >. So we hve s mny inner products on R 1 s positive rel numbers. Let f be n inner product on C 1, since f is liner on ech vrible we get: f(r, s) = f(r 1, s 1) = rf(1, s 1) = rsf(1, 1) therefore the inner product of the vectors r nd s is just the product of the rel numbers r nd s times f(1, 1). But we know tht f(1, 1) >. So we hve s mny inner products on C 1 s positive rel numbers. Exercise 5. Let ( ) be the stndrd inner product on R 2. () Let α = (1, 2), β = ( 1, 1). If γ is vector such tht (α γ) = 1 nd (β γ) = find γ. (b) Show tht for ny α in R 2 we hve α = (α e 1 )e 1 + (α e 2 )e 2 Solution:

4 () We hve to solve the system of equtions Solving we get γ = ( 7, 2 ) x 1 + 2x 2 = 1 x 1 + x 2 = (b) Writing α in terms of the stndrd bsis we get α = 1 e e 2. On the right hnd of the eqution we get ( 1 e e 2 e 1 )e 1 + ( 1 e e 2 e 2 )e 2 = ( 1 e 1 e 1 )e 1 + ( 2 e 2 e 1 ) + ( 1 e 1 e 2 )e 2 + ( 2 e 2 e 2 ) = 1 e e 2 Section 8.2 pp Exercise 1. Consider R 4 with the stndrd inner product. Let W be the subspce of R 4 consisting of ll vectors which re orthogonl to both α = (1,, 1, 1) nd β = (2,, 1, 2) Solution: We hve to find the solution spce for the system x 1 x + x 4 = 2x 1 + x 2 x + x 2 = Row reducing we get tht the vectors (1, 1, 1, ) nd ( 1,,, 1) form bsis for the solution spce of the system. Exercise 2. Apply the Grm-Schmidt process to the vectors β 1 = (1,, 1), β 2 = (1,, 1), β = (,, 4), to obtin n orthonorml bsis for R with the stndrd inner product. Solution: Notice tht the first two vectors re lredy orthogonl, therefore we only need to find the third vector. The bsis we get is: (1,, 1), (1,, 1), (,, ) Exercise 9. Let V be the subspce of R[x] of polynomils of degree t most. Equip V with the inner product (f g) = 1 f(t)g(t)dt () Find the orthogonl complement of the subspce of sclr polynomils. (b) Apply the Grm-Schmidt process to the bsis {1, x, x 2, x } Solution: (b) The bsis we get is {1, x 1 2, 1 6 x + x2, x 2 x2 + x } 4

5 () Using (b), we obtin the following bsis for the orthogonl complement of the subspce of sclr polynomils, i.e. the orthogonl complement of the subspce spnned by {1}: B = {x 1 2, 1 6 x + x2, x 2 x2 + x } Remrk: When performing the Grm-Schmidt process it is helpful to hve quick ccess to the inner products of the vectors of the bsis. This cn be chieved by writing the mtrix M = (m ij ) where m ij = (e i e j ). In this wy, if we write ny two vectors α nd β in terms of the bsis {e i } we get the following: (α β) = [α]m[β] T. In the previous exmple we get the mtrix 1 1/2 1/ 1/4 M = 1/2 1/ 1/4 1/5 1/ 1/4 1/5 1/6 1/4 1/5 1/6 1/7 so, for exmple, the inner product of α = 2 + x + x 2 nd β = 8x 2 x is equl to [2,, 1, ]M[,, 8, 1] T. With this nottion, if A = {1 = b 1, x = b 2, x 2 = b, x = b 4 } the Grm-Schmidt process becomes: 1 = b 1 2 = b 2 b 2M T 1 1 M T 1 1 = b b M T 1 1 M T 1 = b 4 b 4M T 1 1 M T 1 1 b M T 2 2 M T b 4M T 2 2 M T 2 2 b 4M T M T 5

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