FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebraars/TEP-fta.pdf · Linear Algebra at IITB,...
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FUNDAMENTAL THEOREM of ALGEBRAthrough Linear Algebra
Anant R. ShastriI. I. T. Bombay
18th August 2012
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Teachers’ Enrichment Programme inLinear Algebra at IITB, Aug. 2012
I We present a proof ofFundamental Theorem of Algebrathrough a sequence of easily do-able exercises inLinear Algebra except one single result inelementary real anaylsis, viz.,the intermediate value theorem.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Teachers’ Enrichment Programme inLinear Algebra at IITB, Aug. 2012
I We present a proof ofFundamental Theorem of Algebrathrough a sequence of easily do-able exercises inLinear Algebra except one single result inelementary real anaylsis, viz.,the intermediate value theorem.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Based on an article in
I Amer. Math. Monthly- 110,(2003), pp.620-623. by Derksen, a GermanMathematician.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Based on an article in
I Amer. Math. Monthly- 110,(2003), pp.620-623. by Derksen, a GermanMathematician.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I In what follows K will denote any field.However, we need to take K to be either R or C.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 1 Show that every odd degree polynomialP(t) ∈ R[t] has a real root.
I This is where Intermediate Value Theorem isused.
I We may assume P(t) = tn + a1tn−1 + · · · and
see that as t → +∞, p(t)→ +∞, and
I As t → −∞,P(t)→ −∞. It follows that thereexist t0 ∈ R such that P(t0) = 0.
I From now onwards we only use linear algebra.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 1 Show that every odd degree polynomialP(t) ∈ R[t] has a real root.
I This is where Intermediate Value Theorem isused.
I We may assume P(t) = tn + a1tn−1 + · · · and
see that as t → +∞, p(t)→ +∞, and
I As t → −∞,P(t)→ −∞. It follows that thereexist t0 ∈ R such that P(t0) = 0.
I From now onwards we only use linear algebra.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 1 Show that every odd degree polynomialP(t) ∈ R[t] has a real root.
I This is where Intermediate Value Theorem isused.
I We may assume P(t) = tn + a1tn−1 + · · · and
see that as t → +∞, p(t)→ +∞, and
I As t → −∞,P(t)→ −∞. It follows that thereexist t0 ∈ R such that P(t0) = 0.
I From now onwards we only use linear algebra.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 1 Show that every odd degree polynomialP(t) ∈ R[t] has a real root.
I This is where Intermediate Value Theorem isused.
I We may assume P(t) = tn + a1tn−1 + · · · and
see that as t → +∞, p(t)→ +∞, and
I As t → −∞,P(t)→ −∞. It follows that thereexist t0 ∈ R such that P(t0) = 0.
I From now onwards we only use linear algebra.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 1 Show that every odd degree polynomialP(t) ∈ R[t] has a real root.
I This is where Intermediate Value Theorem isused.
I We may assume P(t) = tn + a1tn−1 + · · · and
see that as t → +∞, p(t)→ +∞, and
I As t → −∞,P(t)→ −∞. It follows that thereexist t0 ∈ R such that P(t0) = 0.
I From now onwards we only use linear algebra.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Companion Matrix LetP(t) = tn + a1t
n−1 + · · ·+ an be a monicpolynominal of degree n. Its companian matrixCP is defined to be the n × n matrix
CP =
0 1 0 · · · 0 00 0 1 · · · 0 0...
......
...0 · · · · · · 0 1−an −an−1 · · · · · · −a1
.
I Ex. 2 Show that det(tIn − CP) = P(t).
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Companion Matrix LetP(t) = tn + a1t
n−1 + · · ·+ an be a monicpolynominal of degree n. Its companian matrixCP is defined to be the n × n matrix
CP =
0 1 0 · · · 0 00 0 1 · · · 0 0...
......
...0 · · · · · · 0 1−an −an−1 · · · · · · −a1
.I Ex. 2 Show that det(tIn − CP) = P(t).
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
Ex. 3 Show that every non constant polynomialP(t) ∈ K[t] of degree n has a root in K iff everylinear map Kn → Kn has an eigen value in K.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 4 Show that every R-linear mapf : R2n+1 → R2n+1 has real eigen value.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 5 Show that the space HERMn(C) of allcomplex Hermitian n × n matrices (i.e., all Asuch that A∗ = A) is a R-vector space ofdimension n2.
I Ex. 6 Given A ∈ Mn(C), the mappings
αA(B) =1
2(AB+BA∗); βA(B) =
1
2ı(AB−BA∗)
define R-linear maps HERMn(C)→ HERMn(C).Show that αA, βA commute with each other.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 5 Show that the space HERMn(C) of allcomplex Hermitian n × n matrices (i.e., all Asuch that A∗ = A) is a R-vector space ofdimension n2.
I Ex. 6 Given A ∈ Mn(C), the mappings
αA(B) =1
2(AB+BA∗); βA(B) =
1
2ı(AB−BA∗)
define R-linear maps HERMn(C)→ HERMn(C).Show that αA, βA commute with each other.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Answer:
αA ◦ βA(B) = 14ıαA(AB − BA∗)
= 14ı[A(AB − BA∗) + (AB − BA∗)A∗]
= 14ı[A(AB + BA∗)− (AB + BA∗)A∗]
= 14ıβA ◦ αA(B).
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 7 If αA and βA have a common eigen vectorthen A has an eigen value in C.
I Answer: If αA(B) = λB and βA(B) = µB thenconsider
AB = (αA + ıβA)(B) = (λ + ıµ)B .
I Since B is an eigen vector, at least one of thecolumn vectors , say u 6= 0.
I It follows that Au = (λ+ ıµ)u and hence λ+ ıµis an eigen value for A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 7 If αA and βA have a common eigen vectorthen A has an eigen value in C.
I Answer: If αA(B) = λB and βA(B) = µB thenconsider
AB = (αA + ıβA)(B) = (λ + ıµ)B .
I Since B is an eigen vector, at least one of thecolumn vectors , say u 6= 0.
I It follows that Au = (λ+ ıµ)u and hence λ+ ıµis an eigen value for A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 7 If αA and βA have a common eigen vectorthen A has an eigen value in C.
I Answer: If αA(B) = λB and βA(B) = µB thenconsider
AB = (αA + ıβA)(B) = (λ + ıµ)B .
I Since B is an eigen vector, at least one of thecolumn vectors , say u 6= 0.
I It follows that Au = (λ+ ıµ)u and hence λ+ ıµis an eigen value for A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 7 If αA and βA have a common eigen vectorthen A has an eigen value in C.
I Answer: If αA(B) = λB and βA(B) = µB thenconsider
AB = (αA + ıβA)(B) = (λ + ıµ)B .
I Since B is an eigen vector, at least one of thecolumn vectors , say u 6= 0.
I It follows that Au = (λ+ ıµ)u and hence λ+ ıµis an eigen value for A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 8 Show that any two commuting linearmaps α, β : R2n+1 → R2n+1 have a commoneigen vector.
I Answer: Use induction and subspaces kernel andimage of α− λIn where λ is an eigen value of α.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 8 Show that any two commuting linearmaps α, β : R2n+1 → R2n+1 have a commoneigen vector.
I Answer: Use induction and subspaces kernel andimage of α− λIn where λ is an eigen value of α.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Put Ker α− λIn = V , andRange(α− λIn) = W .
I Then β(V ) ⊂ V and β(W ) ⊂ W . Alsoα(V ) ⊂ V ;α(W ) ⊂ W .
I If V is of odd dimension then any eigen vectorof β will do.
I Otherwise W is odd dimension strictly less than2n + 1. So induction works for α, β : W → W .
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Put Ker α− λIn = V , andRange(α− λIn) = W .
I Then β(V ) ⊂ V and β(W ) ⊂ W . Alsoα(V ) ⊂ V ;α(W ) ⊂ W .
I If V is of odd dimension then any eigen vectorof β will do.
I Otherwise W is odd dimension strictly less than2n + 1. So induction works for α, β : W → W .
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Put Ker α− λIn = V , andRange(α− λIn) = W .
I Then β(V ) ⊂ V and β(W ) ⊂ W . Alsoα(V ) ⊂ V ;α(W ) ⊂ W .
I If V is of odd dimension then any eigen vectorof β will do.
I Otherwise W is odd dimension strictly less than2n + 1. So induction works for α, β : W → W .
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Put Ker α− λIn = V , andRange(α− λIn) = W .
I Then β(V ) ⊂ V and β(W ) ⊂ W . Alsoα(V ) ⊂ V ;α(W ) ⊂ W .
I If V is of odd dimension then any eigen vectorof β will do.
I Otherwise W is odd dimension strictly less than2n + 1. So induction works for α, β : W → W .
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 9 Every C-linear map A : C2n+1 → C2n+1 hasan eigen value.
I Solution: By Ex. 5 and Ex. 8,αA, βA : Herm2n+1 → Herm2n+1 have a commoneigenvector.
I By Ex. 7, A has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 9 Every C-linear map A : C2n+1 → C2n+1 hasan eigen value.
I Solution: By Ex. 5 and Ex. 8,αA, βA : Herm2n+1 → Herm2n+1 have a commoneigenvector.
I By Ex. 7, A has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 9 Every C-linear map A : C2n+1 → C2n+1 hasan eigen value.
I Solution: By Ex. 5 and Ex. 8,αA, βA : Herm2n+1 → Herm2n+1 have a commoneigenvector.
I By Ex. 7, A has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 10 Show that the space Symn(K) ofsymmetric n × n matrices forms a subspace ofdimension n(n + 1)/2 of Mn(K).
I Ex. 11 Given A ∈ Mn(K), show that
φA : B 7→ 1
2(AB + BAt); ψA : B 7→ ABAt
define two commuting endomorphisms ofSymn(K). Show that if B is a common eigenvector of φA, ψA then (A2 + aA + bId)B = 0 forsome a, b ∈ K. Further if K = C, conclude thatA has an eigen value.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 10 Show that the space Symn(K) ofsymmetric n × n matrices forms a subspace ofdimension n(n + 1)/2 of Mn(K).
I Ex. 11 Given A ∈ Mn(K), show that
φA : B 7→ 1
2(AB + BAt); ψA : B 7→ ABAt
define two commuting endomorphisms ofSymn(K). Show that if B is a common eigenvector of φA, ψA then (A2 + aA + bId)B = 0 forsome a, b ∈ K. Further if K = C, conclude thatA has an eigen value.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Answer: The first part is easy.
I To see the second part, suppose
φA(B) = AB + BAt = λB
andψA(B) = ABAt = µB
I Multiply first relation by A on the left and usethe second to obtain
A2B + µB − λAB = 0
which is the same as
(A2 − λA + µId)B = 0.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Answer: The first part is easy.I To see the second part, suppose
φA(B) = AB + BAt = λB
andψA(B) = ABAt = µB
I Multiply first relation by A on the left and usethe second to obtain
A2B + µB − λAB = 0
which is the same as
(A2 − λA + µId)B = 0.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Answer: The first part is easy.I To see the second part, suppose
φA(B) = AB + BAt = λB
andψA(B) = ABAt = µB
I Multiply first relation by A on the left and usethe second to obtain
A2B + µB − λAB = 0
which is the same as
(A2 − λA + µId)B = 0.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I For the last part, first observe that since B is aneigen vector there is atleast one column vector vwhich is non zero. Therefore(A2 + aA + b)v = 0.
I Now writeA2 + aA + bId = (A− λ1Id)(A− λ2Id). If(A− λ2Id)v = 0 then λ2 is an eigen value of A.and we are through.
I Otherwise u = (A− λ2Id) 6= 0 and(A− λ1Id)u = 0 and hence λ1 is an eigen valueof A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I For the last part, first observe that since B is aneigen vector there is atleast one column vector vwhich is non zero. Therefore(A2 + aA + b)v = 0.
I Now writeA2 + aA + bId = (A− λ1Id)(A− λ2Id). If(A− λ2Id)v = 0 then λ2 is an eigen value of A.and we are through.
I Otherwise u = (A− λ2Id) 6= 0 and(A− λ1Id)u = 0 and hence λ1 is an eigen valueof A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I For the last part, first observe that since B is aneigen vector there is atleast one column vector vwhich is non zero. Therefore(A2 + aA + b)v = 0.
I Now writeA2 + aA + bId = (A− λ1Id)(A− λ2Id). If(A− λ2Id)v = 0 then λ2 is an eigen value of A.and we are through.
I Otherwise u = (A− λ2Id) 6= 0 and(A− λ1Id)u = 0 and hence λ1 is an eigen valueof A.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Let S1(K, r) denote the following statement:Any endomorphism A : Kn → Kn has an eigenvalue for all n not divisible by 2r . Let S2(K, r)denote the statement: Any two commutingendomorphisms A1,A2 : Kn → Kn have acommon eigen vector for all n not divisible by 2r .
I Ex. 12 Prove that S1(K, r) =⇒ S2(K, r).
I Answer: Exactly as in Ex. 8.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Let S1(K, r) denote the following statement:Any endomorphism A : Kn → Kn has an eigenvalue for all n not divisible by 2r . Let S2(K, r)denote the statement: Any two commutingendomorphisms A1,A2 : Kn → Kn have acommon eigen vector for all n not divisible by 2r .
I Ex. 12 Prove that S1(K, r) =⇒ S2(K, r).
I Answer: Exactly as in Ex. 8.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Let S1(K, r) denote the following statement:Any endomorphism A : Kn → Kn has an eigenvalue for all n not divisible by 2r . Let S2(K, r)denote the statement: Any two commutingendomorphisms A1,A2 : Kn → Kn have acommon eigen vector for all n not divisible by 2r .
I Ex. 12 Prove that S1(K, r) =⇒ S2(K, r).
I Answer: Exactly as in Ex. 8.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 13 Prove that S1(C, r) =⇒ S1(C, r + 1).
I Answer:Let n = 2km where m is odd and k ≤ r . LetA ∈ Mn(C). Then φA, ψA are two mutuallycommuting operators on Symn(C) which is ofdimension n(n + 1)/2 = 2k−1m(2km + 1) whichis not divisible by 2r .
I Therefore S1(C, r) together with Ex. 12 impliesthat φA, ψA have a common eigen vector.
I By Ex. 11, this implies A has an eigen value inC.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 13 Prove that S1(C, r) =⇒ S1(C, r + 1).
I Answer:Let n = 2km where m is odd and k ≤ r . LetA ∈ Mn(C). Then φA, ψA are two mutuallycommuting operators on Symn(C) which is ofdimension n(n + 1)/2 = 2k−1m(2km + 1) whichis not divisible by 2r .
I Therefore S1(C, r) together with Ex. 12 impliesthat φA, ψA have a common eigen vector.
I By Ex. 11, this implies A has an eigen value inC.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 13 Prove that S1(C, r) =⇒ S1(C, r + 1).
I Answer:Let n = 2km where m is odd and k ≤ r . LetA ∈ Mn(C). Then φA, ψA are two mutuallycommuting operators on Symn(C) which is ofdimension n(n + 1)/2 = 2k−1m(2km + 1) whichis not divisible by 2r .
I Therefore S1(C, r) together with Ex. 12 impliesthat φA, ψA have a common eigen vector.
I By Ex. 11, this implies A has an eigen value inC.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 13 Prove that S1(C, r) =⇒ S1(C, r + 1).
I Answer:Let n = 2km where m is odd and k ≤ r . LetA ∈ Mn(C). Then φA, ψA are two mutuallycommuting operators on Symn(C) which is ofdimension n(n + 1)/2 = 2k−1m(2km + 1) whichis not divisible by 2r .
I Therefore S1(C, r) together with Ex. 12 impliesthat φA, ψA have a common eigen vector.
I By Ex. 11, this implies A has an eigen value inC.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 14 Conclude that every non constantpolynomial over complex numbers has a root.
I Answer:
I By Ex.3, it is enough to show that every n × ncomplex matrix has an eigen value.
I By Ex.9 this holds for odd n. This impliesS1(C, 1). By Ex. 13 applied repeatedly, we getS1(C, r) for all r . Write n = 2rm where m isodd. Then S1(C, r + 1) implies everyendomorphism A : Cn → Cn has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 14 Conclude that every non constantpolynomial over complex numbers has a root.
I Answer:
I By Ex.3, it is enough to show that every n × ncomplex matrix has an eigen value.
I By Ex.9 this holds for odd n. This impliesS1(C, 1). By Ex. 13 applied repeatedly, we getS1(C, r) for all r . Write n = 2rm where m isodd. Then S1(C, r + 1) implies everyendomorphism A : Cn → Cn has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 14 Conclude that every non constantpolynomial over complex numbers has a root.
I Answer:
I By Ex.3, it is enough to show that every n × ncomplex matrix has an eigen value.
I By Ex.9 this holds for odd n. This impliesS1(C, 1). By Ex. 13 applied repeatedly, we getS1(C, r) for all r . Write n = 2rm where m isodd. Then S1(C, r + 1) implies everyendomorphism A : Cn → Cn has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
I Ex. 14 Conclude that every non constantpolynomial over complex numbers has a root.
I Answer:
I By Ex.3, it is enough to show that every n × ncomplex matrix has an eigen value.
I By Ex.9 this holds for odd n. This impliesS1(C, 1). By Ex. 13 applied repeatedly, we getS1(C, r) for all r . Write n = 2rm where m isodd. Then S1(C, r + 1) implies everyendomorphism A : Cn → Cn has an eigenvalue.
Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Anant R. Shastri I. I. T. Bombay Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA