FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebraars/TEP-fta.pdf · Linear Algebra at IITB,...

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

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.

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.

I Based on an article in

I Amer. Math. Monthly- 110,(2003), pp.620-623. by Derksen, a GermanMathematician.

I Based on an article in

I Amer. Math. Monthly- 110,(2003), pp.620-623. by Derksen, a GermanMathematician.

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

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.

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

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).

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

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).

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.

I Ex. 4 Show that every R-linear mapf : R2n+1 → R2n+1 has real eigen value.

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) =

2ı(AB−BA∗)

define R-linear maps HERMn(C)→ HERMn(C).Show that αA, βA commute with each other.

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) =

2ı(AB−BA∗)

define R-linear maps HERMn(C)→ HERMn(C).Show that αA, βA commute with each other.

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).

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.

AB = (αA + ıβA)(B) = (λ + ıµ)B .

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.

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.

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 .

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.

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.

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.

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.

I Answer: The first part is easy.I To see the second part, suppose

A2B + µB − λAB = 0

(A2 − λA + µId)B = 0.

I Answer: The first part is easy.I To see the second part, suppose

A2B + µB − λAB = 0

(A2 − λA + µId)B = 0.

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.

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.

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.

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.

I Answer:

THANK YOU

FOR YOUR ATTENTION

FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebraars/TEP-fta.pdf · Linear Algebra at IITB,...

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