FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebraars/TEP-fta.pdf · Linear Algebra at IITB, Aug. 2012 I We present a proof of Fundamental Theorem of Algebra through a sequence

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

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).


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).


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) =

1

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) =

1

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 .






AB = (αA + ıβA)(B) = (λ + ıµ)B .






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.



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FUNDAMENTAL THEOREM of ALGEBRA through Linear Algebraars/TEP-fta.pdf · Linear Algebra at IITB, Aug. 2012 I We present a proof of Fundamental Theorem of Algebra through a sequence