FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebra. 18th August Anant R. Shastri I. I. T. Bombay

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1 FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebra Anant R. Shastri I. I. T. Bombay 18th August 2012

3 Teachers Enrichment Programme in Linear Algebra at IITB, Aug. 2012

4 Teachers Enrichment Programme in Linear Algebra at IITB, Aug We present a proof of Fundamental Theorem of Algebra through a sequence of easily do-able exercises in Linear Algebra except one single result in elementary real anaylsis, viz., the intermediate value theorem.

5 Based on an article in

6 Based on an article in Amer. Math. Monthly- 110,(2003), pp by Derksen, a German Mathematician.

7 In what follows K will denote any field. However, we need to take K to be either R or C.

8 Ex. 1 Show that every odd degree polynomial P(t) R[t] has a real root.

9 Ex. 1 Show that every odd degree polynomial P(t) R[t] has a real root. This is where Intermediate Value Theorem is used.

10 Ex. 1 Show that every odd degree polynomial P(t) R[t] has a real root. This is where Intermediate Value Theorem is used. We may assume P(t) = t n + a 1 t n 1 + and see that as t +, p(t) +, and

11 Ex. 1 Show that every odd degree polynomial P(t) R[t] has a real root. This is where Intermediate Value Theorem is used. We may assume P(t) = t n + a 1 t n 1 + and see that as t +, p(t) +, and As t, P(t). It follows that there exist t 0 R such that P(t 0 ) = 0.

12 Ex. 1 Show that every odd degree polynomial P(t) R[t] has a real root. This is where Intermediate Value Theorem is used. We may assume P(t) = t n + a 1 t n 1 + and see that as t +, p(t) +, and As t, P(t). It follows that there exist t 0 R such that P(t 0 ) = 0. From now onwards we only use linear algebra.

13 Companion Matrix Let P(t) = t n + a 1 t n a n be a monic polynominal of degree n. Its companian matrix C P is defined to be the n n matrix C P = a n a n 1 a 1.

14 Companion Matrix Let P(t) = t n + a 1 t n a n be a monic polynominal of degree n. Its companian matrix C P is defined to be the n n matrix C P = a n a n 1 a 1 Ex. 2 Show that det(ti n C P ) = P(t)..

15 Ex. 3 Show that every non constant polynomial P(t) K[t] of degree n has a root in K iff every linear map K n K n has an eigen value in K.

16 Ex. 4 Show that every R-linear map f : R 2n+1 R 2n+1 has real eigen value.

17 Ex. 5 Show that the space HERM n (C) of all complex Hermitian n n matrices (i.e., all A such that A = A) is a R-vector space of dimension n 2.

18 Ex. 5 Show that the space HERM n (C) of all complex Hermitian n n matrices (i.e., all A such that A = A) is a R-vector space of dimension n 2. Ex. 6 Given A M n (C), the mappings α A (B) = 1 2 (AB +BA ); β A (B) = 1 2ı (AB BA ) define R-linear maps HERM n (C) HERM n (C). Show that α A, β A commute with each other.

19 Answer: α A β A (B) = 1 4ı α A(AB BA ) = 1 4ı [A(AB BA ) + (AB BA )A ] = 1 4ı [A(AB + BA ) (AB + BA )A ] = 1 4ı β A α A (B).

20 Ex. 7 If α A and β A have a common eigen vector then A has an eigen value in C.

21 Ex. 7 If α A and β A have a common eigen vector then A has an eigen value in C. Answer: If α A (B) = λb and β A (B) = µb then consider AB = (α A + ıβ A )(B) = (λ + ıµ)b.

22 Ex. 7 If α A and β A have a common eigen vector then A has an eigen value in C. Answer: If α A (B) = λb and β A (B) = µb then consider AB = (α A + ıβ A )(B) = (λ + ıµ)b. Since B is an eigen vector, at least one of the column vectors, say u 0.

23 Ex. 7 If α A and β A have a common eigen vector then A has an eigen value in C. Answer: If α A (B) = λb and β A (B) = µb then consider AB = (α A + ıβ A )(B) = (λ + ıµ)b. Since B is an eigen vector, at least one of the column vectors, say u 0. It follows that Au = (λ + ıµ)u and hence λ + ıµ is an eigen value for A.

24 Ex. 8 Show that any two commuting linear maps α, β : R 2n+1 R 2n+1 have a common eigen vector.

25 Ex. 8 Show that any two commuting linear maps α, β : R 2n+1 R 2n+1 have a common eigen vector. Answer: Use induction and subspaces kernel and image of α λi n where λ is an eigen value of α.

26 Put Ker α λi n = V, and Range(α λi n ) = W.

27 Put Ker α λi n = V, and Range(α λi n ) = W. Then β(v ) V and β(w ) W. Also α(v ) V ; α(w ) W.

28 Put Ker α λi n = V, and Range(α λi n ) = W. Then β(v ) V and β(w ) W. Also α(v ) V ; α(w ) W. If V is of odd dimension then any eigen vector of β will do.

29 Put Ker α λi n = V, and Range(α λi n ) = W. Then β(v ) V and β(w ) W. Also α(v ) V ; α(w ) W. If V is of odd dimension then any eigen vector of β will do. Otherwise W is odd dimension strictly less than 2n + 1. So induction works for α, β : W W.

30 Ex. 9 Every C-linear map A : C 2n+1 C 2n+1 has an eigen value.

31 Ex. 9 Every C-linear map A : C 2n+1 C 2n+1 has an eigen value. Solution: By Ex. 5 and Ex. 8, α A, β A : Herm 2n+1 Herm 2n+1 have a common eigenvector.

32 Ex. 9 Every C-linear map A : C 2n+1 C 2n+1 has an eigen value. Solution: By Ex. 5 and Ex. 8, α A, β A : Herm 2n+1 Herm 2n+1 have a common eigenvector. By Ex. 7, A has an eigenvalue.

33 Ex. 10 Show that the space Sym n (K) of symmetric n n matrices forms a subspace of dimension n(n + 1)/2 of M n (K).

34 Ex. 10 Show that the space Sym n (K) of symmetric n n matrices forms a subspace of dimension n(n + 1)/2 of M n (K). Ex. 11 Given A M n (K), show that φ A : B 1 2 (AB + BAt ); ψ A : B ABA t define two commuting endomorphisms of Sym n (K). Show that if B is a common eigen vector of φ A, ψ A then (A 2 + aa + bid)b = 0 for some a, b K. Further if K = C, conclude that A has an eigen value.

35 Answer: The first part is easy.

36 Answer: The first part is easy. To see the second part, suppose and φ A (B) = AB + BA t = λb ψ A (B) = ABA t = µb

37 Answer: The first part is easy. To see the second part, suppose and φ A (B) = AB + BA t = λb ψ A (B) = ABA t = µb Multiply first relation by A on the left and use the second to obtain which is the same as A 2 B + µb λab = 0 (A 2 λa + µid)b = 0.

38 For the last part, first observe that since B is an eigen vector there is atleast one column vector v which is non zero. Therefore (A 2 + aa + b)v = 0.

39 For the last part, first observe that since B is an eigen vector there is atleast one column vector v which is non zero. Therefore (A 2 + aa + b)v = 0. Now write A 2 + aa + bid = (A λ 1 Id)(A λ 2 Id). If (A λ 2 Id)v = 0 then λ 2 is an eigen value of A. and we are through.

40 For the last part, first observe that since B is an eigen vector there is atleast one column vector v which is non zero. Therefore (A 2 + aa + b)v = 0. Now write A 2 + aa + bid = (A λ 1 Id)(A λ 2 Id). If (A λ 2 Id)v = 0 then λ 2 is an eigen value of A. and we are through. Otherwise u = (A λ 2 Id) 0 and (A λ 1 Id)u = 0 and hence λ 1 is an eigen value of A.

41 Let S 1 (K, r) denote the following statement: Any endomorphism A : K n K n has an eigen value for all n not divisible by 2 r. Let S 2 (K, r) denote the statement: Any two commuting endomorphisms A 1, A 2 : K n K n have a common eigen vector for all n not divisible by 2 r.

42 Let S 1 (K, r) denote the following statement: Any endomorphism A : K n K n has an eigen value for all n not divisible by 2 r. Let S 2 (K, r) denote the statement: Any two commuting endomorphisms A 1, A 2 : K n K n have a common eigen vector for all n not divisible by 2 r. Ex. 12 Prove that S 1 (K, r) = S 2 (K, r).

43 Let S 1 (K, r) denote the following statement: Any endomorphism A : K n K n has an eigen value for all n not divisible by 2 r. Let S 2 (K, r) denote the statement: Any two commuting endomorphisms A 1, A 2 : K n K n have a common eigen vector for all n not divisible by 2 r. Ex. 12 Prove that S 1 (K, r) = S 2 (K, r). Answer: Exactly as in Ex. 8.

44 Ex. 13 Prove that S 1 (C, r) = S 1 (C, r + 1).

45 Ex. 13 Prove that S 1 (C, r) = S 1 (C, r + 1). Answer: Let n = 2 k m where m is odd and k r. Let A M n (C). Then φ A, ψ A are two mutually commuting operators on Sym n (C) which is of dimension n(n + 1)/2 = 2 k 1 m(2 k m + 1) which is not divisible by 2 r.

46 Ex. 13 Prove that S 1 (C, r) = S 1 (C, r + 1). Answer: Let n = 2 k m where m is odd and k r. Let A M n (C). Then φ A, ψ A are two mutually commuting operators on Sym n (C) which is of dimension n(n + 1)/2 = 2 k 1 m(2 k m + 1) which is not divisible by 2 r. Therefore S 1 (C, r) together with Ex. 12 implies that φ A, ψ A have a common eigen vector.

47 Ex. 13 Prove that S 1 (C, r) = S 1 (C, r + 1). Answer: Let n = 2 k m where m is odd and k r. Let A M n (C). Then φ A, ψ A are two mutually commuting operators on Sym n (C) which is of dimension n(n + 1)/2 = 2 k 1 m(2 k m + 1) which is not divisible by 2 r. Therefore S 1 (C, r) together with Ex. 12 implies that φ A, ψ A have a common eigen vector. By Ex. 11, this implies A has an eigen value in C.

48 Ex. 14 Conclude that every non constant polynomial over complex numbers has a root.

49 Ex. 14 Conclude that every non constant polynomial over complex numbers has a root. Answer:

50 Ex. 14 Conclude that every non constant polynomial over complex numbers has a root. Answer: By Ex.3, it is enough to show that every n n complex matrix has an eigen value.

51 Ex. 14 Conclude that every non constant polynomial over complex numbers has a root. Answer: By Ex.3, it is enough to show that every n n complex matrix has an eigen value. By Ex.9 this holds for odd n. This implies S 1 (C, 1). By Ex. 13 applied repeatedly, we get S 1 (C, r) for all r. Write n = 2 r m where m is odd. Then S 1 (C, r + 1) implies every endomorphism A : C n C n has an eigenvalue.

52 THANK YOU FOR YOUR ATTENTION

A Proof of Fundamental Theorem of Algebra Through Linear Algebra due to Derksen. Anant R. Shastri. February 13, 2011

A Proof of Fundamental Theorem of Algebra Through Linear Algebra due to Derksen February 13, 2011 MA106-Linear Algebra 2011 MA106-Linear Algebra 2011 We present a proof of the Fundamental Theorem of Algebra