A Proof of Fundamental Theorem of Algebra Through Linear ...ars/MA106/ma106-fta.pdfI Amer. Math. Monthly- 110,(2003), pp. 620-623. Anant R. Shastri Derksen’s Proof of FTA I Based

A Proof of Fundamental Theorem of AlgebraThrough Linear Algebra due to Derksen

Anant R. Shastri

February 13, 2011

Anant R. Shastri Derksen’s Proof of FTA

I MA106-Linear Algebra 2011

I We present a proof ofthe Fundamental Theorem of Algebra(FTA)Every non constant polynomial in one variablewith coefficients in C has a root in C.

I through a sequence of easily do-able exercises.

The proof uses elementary linear algebraexcept that we begin withthe intermediate value theorem in provingthat every odd degee real polynomial has areal root.

I through a sequence of easily do-able exercises.The proof uses elementary linear algebraexcept that we begin with

the intermediate value theorem in provingthat every odd degee real polynomial has areal root.

I through a sequence of easily do-able exercises.The proof uses elementary linear algebraexcept that we begin withthe intermediate value theorem in provingthat every odd degee real polynomial has areal root.

I Based on an article in

I Amer. Math. Monthly- 110,(2003), pp.620-623.

I Based on an article in

I Amer. Math. Monthly- 110,(2003), pp.620-623.

Sketch of the Proof

I In what follows K will denote any field.However, we need worry about only two caseshere: K is either R or C.

I We begin with a basic result in real analysis:

Intermediate Value Theoremwhich immediately yields:

I Every odd degree polynomial p(t) ∈ R[t] has areal root.

I It should be noted that there is no purelyalgebraic proof of FTA which does not use IVT.

I Indeed, all proofs of FTA use IVT explicitly orimplicitly. The simple reason is that IVT isequivalent to any other axiom that is used in theconstruction of real numbers.

I From now onwards we only use linear algebra.

I We begin with a basic result in real analysis:Intermediate Value Theorem

which immediately yields:

I We begin with a basic result in real analysis:Intermediate Value Theoremwhich immediately yields:

Sketch of the Proof

I Companion Matrix Letp(t) = tn + a1t

n−1 + · · ·+ an be a monicpolynomial of degree n. Its companion matrix Cpis 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.1. Show that the characteristic polynomial ofCp is p, i.e., det(Cp − tIn) = (−1)np(t).

Sketch of the Proof

I Companion Matrix Letp(t) = tn + a1t

n−1 + · · ·+ an be a monicpolynomial of degree n. Its companion matrix Cpis 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.1. Show that the characteristic polynomial of

Cp is p, i.e., det(Cp − tIn) = (−1)np(t).Anant R. Shastri Derksen’s Proof of FTA

I Ex. 2. Show that every non constant polynomialp(t) ∈ K[t] of degree n has a root in K iff everyendomorphism Kn → Kn has an eigenvalue in K

I iff every n × n matrix over K has an eigenvaluein K.

I Recall that by an endomorphism of a vectorspace V we mean a linear transformationV → V .

I Ex. 3 Show that every R-linear mapf : R2n+1 → R2n+1 has a real eigenvalue.

Sketch of the Proof

For positive integers r , let S1(K, r) denote thefollowing statement:Every endomorphism A : Kn → Kn has aneigenvector for all n not divisible by 2r .Also let S2(K, r) denote the statementAny two commuting endomorphisms Kn → Kn havea common eigenvector for all n not divisible by 2r .

We shall prove one result over any field K and twospecial results for K = R and C.

Sketch of the Proof

For positive integers r , let S1(K, r) denote thefollowing statement:Every endomorphism A : Kn → Kn has aneigenvector for all n not divisible by 2r .Also let S2(K, r) denote the statementAny two commuting endomorphisms Kn → Kn havea common eigenvector for all n not divisible by 2r .We shall prove one result over any field K and twospecial results for K = R and C.

Sketch of the Proof

For positive integers r , let S1(K, r) denote thefollowing statement:Every endomorphism A : Kn → Kn has aneigenvector for all n not divisible by 2r .Also let S2(K, r) denote the statementAny two commuting endomorphisms Kn → Kn havea common eigenvector for all n not divisible by 2r .We shall prove one result over any field K and twospecial results for K = R and C.

Sketch of Proof

I (a) S1(K, r) =⇒ S2(K, r), r ≥ 1.

I (b) S2(R, 1) =⇒ S1(C, 1).I (c) S1(C, r) =⇒ S1(C, r + 1), r ≥ 1.I Begin with IVT which is the same as S1(R, 1).

Putting K = R in (a) we get S2(R, 1).

Now (b)gives S1(C, 1) and repeated application of (c)gives S1(C, k) for all k ≥ 1.

I Given n, write n = 2k` where ` is odd. ThenS1(C, k + 1) implies that every polynomial ofdegree n has a root. And that completes theproof of FTA.

Sketch of Proof

I (a) S1(K, r) =⇒ S2(K, r), r ≥ 1.I (b) S2(R, 1) =⇒ S1(C, 1).

I (c) S1(C, r) =⇒ S1(C, r + 1), r ≥ 1.I Begin with IVT which is the same as S1(R, 1).

Putting K = R in (a) we get S2(R, 1).Now (b)gives S1(C, 1)

and repeated application of (c)gives S1(C, k) for all k ≥ 1.

Sketch of Proof

I (a) S1(K, r) =⇒ S2(K, r), r ≥ 1.I (b) S2(R, 1) =⇒ S1(C, 1).I (c) S1(C, r) =⇒ S1(C, r + 1), r ≥ 1.

I Begin with IVT which is the same as S1(R, 1).Putting K = R in (a) we get S2(R, 1).Now (b)gives S1(C, 1) and repeated application of (c)gives S1(C, k) for all k ≥ 1.

Sketch of Proof

I (a) S1(K, r) =⇒ S2(K, r), r ≥ 1.I (b) S2(R, 1) =⇒ S1(C, 1).I (c) S1(C, r) =⇒ S1(C, r + 1), r ≥ 1.I Begin with IVT which is the same as S1(R, 1).

Putting K = R in (a) we get S2(R, 1).Now (b)gives S1(C, 1) and repeated application of (c)gives S1(C, k) for all k ≥ 1.

Sketch of Proof

I (a) S1(K, r) =⇒ S2(K, r), r ≥ 1.I (b) S2(R, 1) =⇒ S1(C, 1).I (c) S1(C, r) =⇒ S1(C, r + 1), r ≥ 1.I Begin with IVT which is the same as S1(R, 1).

Putting K = R in (a) we get S2(R, 1).Now (b)gives S1(C, 1) and repeated application of (c)gives S1(C, k) for all k ≥ 1.

I Ex. 4. Prove statement (a):S1(K, r) =⇒ S2(K, r).

I We shall prove this by induction on thedimension n of the vector space.

I For n = 1 this is clear, since any non zero vectoris an eigenvector for all endomorphisms.

I So assume that the statement is true for smallervalues of n and let A,B be two commuting n× nmatrices over K, and n is not divisible by 2r .

S1(K, r) implies S2(K, r)

I Let λ be an eigenvalue of A and putV1 = N (A− λ I ),V2 = R(A− λ I ). Since Bcommutes with A, it follows that B commutes withA− λI ) also. Therefore B restricts toendomorphisms β : V1 → V1 and β : V2 → V2.(Refer: Tut. sheet 9.18.)

I By rank-nullity theorem,

dim V1 + dim V2 = n

and hence dim V1 or dim V2 is not divisible by 2r .

I If dim V1 is not divisible by 2r then β : V1 → V1

has an eigenvector which is also the eigenvector forA.

I If not, then dim V2 < n and not divisible by 2r .

Now we consider α, β : V2 → V2 which arerestrictions of A and B and hence are commutingendomorphisms.By induction we are through there is a commoneigen vector v ∈ V2 ⊂ V for α and β which is thena common eigen vector for A and B as well.

dim V1 + dim V2 = n

I Ex. 5 Show that the space HERMn(C) of allcomplex Hermitian n × n matrices is a R vectorspace of dimension n2.

(Refer to Exercise 8.5 and a question in theQuiz.)Recall that a matrix A is hermitian if it is equalto its transpose conjugate A∗.and the next one is from 8.6.

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 is a R vectorspace of dimension n2.(Refer to Exercise 8.5 and a question in theQuiz.)

Recall that a matrix A is hermitian if it is equalto its transpose conjugate A∗.and the next one is from 8.6.

αA(B) =1

2(AB +BA∗); βA(B) =

1

2ı(AB−BA∗)

I Ex. 5 Show that the space HERMn(C) of allcomplex Hermitian n × n matrices is a R vectorspace of dimension n2.(Refer to Exercise 8.5 and a question in theQuiz.)Recall that a matrix A is hermitian if it is equalto its transpose conjugate A∗.and the next one is from 8.6.

αA(B) =1

2(AB +BA∗); βA(B) =

1

2ı(AB−BA∗)

I Answer:

4ıαA ◦ βA(B)= 2αA(AB − BA∗)= [A(AB − BA∗) + (AB − BA∗)A∗]= [A(AB + BA∗)− (AB + BA∗)A∗]= 2ıβA(AB + BA

∗)= 4ıβA ◦ αA(B).

I Ex. 7. If αA and βA have a common eigenvectorthen A has an eigenvalue 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 eigenvector, at least one of thecolumn vectors of B , say u 6= 0.

I It follows that Au = (λ+ ıµ)u and hence λ+ ıµis an eigenvalue for A.

AB = (αA + ıβA)(B) = (λ + ıµ)B .

I Ex.8. Prove statement (b):S2(R, 1) =⇒ S1(C, 1). (Exercise 11.4)

I Solution: By Ex.6, given A ∈ Mn(C),αA, βA : Hermn(C)→ Hermn(C) are twocommuting endomorphisms of the real vectorspace Hermn which is of dimension n

2, by ex. 5.

I If n is odd, so is n2.

I By Ex.4, it follows that αA, βA have a commoneigenvector.

I By Ex. 7, A has an eigenvalue.

2, by ex. 5.

I Ex. 9. Show that the space Symn(K) ofsymmetric n × n matrices forms a subspace ofdimension n(n + 1)/2 of Mn(K). (Exercise 8.5)

The next one is Ex. 11.5.

I Ex. 10 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 commoneigenvector of φA, ψA, then(A2 + aA + b In)B = 0 for some a, b ∈ K; furtherif K = C, conclude that A has an eigenvalue.

I Ex. 9. Show that the space Symn(K) ofsymmetric n × n matrices forms a subspace ofdimension n(n + 1)/2 of Mn(K). (Exercise 8.5)The next one is Ex. 11.5.

φA : B 7→1

I Ex. 9. Show that the space Symn(K) ofsymmetric n × n matrices forms a subspace ofdimension n(n + 1)/2 of Mn(K). (Exercise 8.5)The next one is Ex. 11.5.

φA : B 7→1

I Answer: The first part is easy.

I To see the second part, suppose

φA(B) = AB + BAt = λB

andψA(B) = ABA

t = 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 + µ In)B = 0.

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

andψA(B) = ABA

t = µB

A2B + µB − λAB = 0

(A2 − λA + µ In)B = 0.

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

andψA(B) = ABA

t = µB

A2B + µB − λAB = 0

(A2 − λA + µ In)B = 0.

I For the last part, first observe that since B is aneigenvector there is at least one column vector vwhich is non zero. Therefore(A2 + aA + b In)v = 0.

I Now write A2 + aA + bIn = (A−λ1In)(A−λ2In).If (A− λ2In)v = 0, then λ2 is an eigenvalue of Aand we are through.

I Otherwise put u = (A− λ2In)v 6= 0. Then(A− λ1In)u = 0 and hence λ1 is an eigenvalueof A.

I Ex. 11 Prove statement (c)S1(C, r) =⇒ S1(C, r + 1). Hence conclude thatS1(C, r) is true for all r ≥ 1.

I Answer:Let n = 2k` where 1 ≤ k ≤ r and ` is odd. LetA ∈ Mn(C). Then φA, ψA are two mutuallycommuting operators on Symn(C) which is ofdimension n(n + 1)/2 = 2k−1`(2k` + 1) which isnot divisible by 2r .

I By Ex. 10, with K = C along with the inductionhypothesis, it follows that A has an eigenvalue.

I We summerise what we did.

I In order to prove that every non constantpolynomial over complex numbers has a root wenote that

I by Ex.2, it is enough to show that every n × ncomplex matrix has an eigenvalue.

I By Ex.8, this holds for odd n. This implies thatS1(C, 1).

By Ex. 11 applied repeatedly, we getS1(C, r) for all r .Given any polynomial of degree n, write n = 2r`,where ` is odd. Then S1(C, r + 1) implies thatevery polynomial of degree n has a root.

I By Ex.8, this holds for odd n. This implies thatS1(C, 1). By Ex. 11 applied repeatedly, we getS1(C, r) for all r .

Given any polynomial of degree n, write n = 2r`,where ` is odd. Then S1(C, r + 1) implies thatevery polynomial of degree n has a root.

I By Ex.8, this holds for odd n. This implies thatS1(C, 1). By Ex. 11 applied repeatedly, we getS1(C, r) for all r .Given any polynomial of degree n, write n = 2r`,where ` is odd.

Then S1(C, r + 1) implies thatevery polynomial of degree n has a root.

I By Ex.8, this holds for odd n. This implies thatS1(C, 1). By Ex. 11 applied repeatedly, we getS1(C, r) for all r .Given any polynomial of degree n, write n = 2r`,where ` is odd. Then S1(C, r + 1) implies thatevery polynomial of degree n has a root.

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