FUNDAMENTAL THEOREM of ALGEBRA
through Linear Algebra
Anant R. Shastri
I. 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
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Teachers’ Enrichment Programme in
Linear Algebra at IITB, Aug. 2012
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Teachers’ Enrichment Programme in
Linear Algebra at IITB, Aug. 2012
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.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Based on an article in
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Based on an article in
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Amer. Math. Monthly- 110,(2003), pp.
620-623. by Derksen, a German
Mathematician.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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
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Ex. 1 Show that every odd degree polynomial
P(t) ∈ R[t] has a real root.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 1 Show that every odd degree polynomial
P(t) ∈ R[t] has a real root.
This is where Intermediate Value Theorem is
used.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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 + a1 t n−1 + · · · and
see that as t → +∞, p(t) → +∞, and
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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 + a1 t n−1 + · · · and
see that as t → +∞, p(t) → +∞, and
As t → −∞, P(t) → −∞. It follows that there
exist t0 ∈ R such that P(t0 ) = 0.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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 + a1 t n−1 + · · · and
see that as t → +∞, p(t) → +∞, and
As t → −∞, P(t) → −∞. It follows that there
exist t0 ∈ R such that P(t0 ) = 0.
From now onwards we only use linear algebra.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Companion Matrix Let
P(t) = t n + a1 t n−1 + · · · + an be a monic
polynominal of degree n. Its companian matrix
CP is defined to be the n × n matrix


0
1
0 ··· 0
0
 0
0
1 ··· 0
0 


 ..

.
.
.
.
.
.
CP =  .
.
.
. .


 0
···
··· 0
1 
−an −an−1
· · · · · · −a1
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Companion Matrix Let
P(t) = t n + a1 t n−1 + · · · + an be a monic
polynominal of degree n. Its companian matrix
CP is defined to be the n × n matrix


0
1
0 ··· 0
0
 0
0
1 ··· 0
0 


 ..

.
.
.
.
.
.
CP =  .
.
.
. .


 0
···
··· 0
1 
−an −an−1
· · · · · · −a1
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 polynomial
P(t) ∈ K[t] of degree n has a root in K iff every
linear map Kn → Kn has an eigen value in K.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 4 Show that every R-linear map
f : R2n+1 → R2n+1 has real eigen value.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 5 Show that the space HERMn (C) of all
complex Hermitian n × n matrices (i.e., all A
such that A∗ = A) is a R-vector space of
dimension n2 .
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 5 Show that the space HERMn (C) of all
complex Hermitian n × n matrices (i.e., all A
such that A∗ = A) is a R-vector space of
dimension n2 .
Ex. 6 Given A ∈ Mn (C), the mappings
1
1
αA (B) = (AB +BA∗ ); βA (B) = (AB −BA∗ )
2
2ı
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
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Answer:
αA ◦ βA (B) =
=
=
=
1
∗
4ı αA (AB − BA )
1
∗
4ı [A(AB − BA ) +
1
∗
4ı [A(AB + BA ) −
1
4ı βA ◦ αA (B).
Anant R. Shastri I. I. T. Bombay
(AB − BA∗ )A∗ ]
(AB + BA∗ )A∗ ]
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 7 If αA and βA have a common eigen vector
then A has an eigen value in C.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 7 If αA and βA have a common eigen vector
then A has an eigen value in C.
Answer: If αA (B) = λB and βA (B) = µB then
consider
AB = (αA + ıβA )(B) = (λ + ıµ)B.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 7 If αA and βA have a common eigen vector
then A has an eigen value in C.
Answer: If αA (B) = λB and βA (B) = µB then
consider
AB = (αA + ıβA )(B) = (λ + ıµ)B.
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Since B is an eigen vector, at least one of the
column vectors , say u 6= 0.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 7 If αA and βA have a common eigen vector
then A has an eigen value in C.
Answer: If αA (B) = λB and βA (B) = µB then
consider
AB = (αA + ıβA )(B) = (λ + ıµ)B.
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Since B is an eigen vector, at least one of the
column vectors , say u 6= 0.
It follows that Au = (λ + ıµ)u and hence λ + ıµ
is an eigen value for A.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 8 Show that any two commuting linear
maps α, β : R2n+1 → R2n+1 have a common
eigen vector.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 8 Show that any two commuting linear
maps α, β : R2n+1 → R2n+1 have a common
eigen vector.
Answer: Use induction and subspaces kernel and
image 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
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Put Ker α − λIn = V , and
Range(α − λIn ) = W .
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Put Ker α − λIn = V , and
Range(α − λIn ) = W .
Then β(V ) ⊂ V and β(W ) ⊂ W . Also
α(V ) ⊂ V ; α(W ) ⊂ W .
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Put Ker α − λIn = V , and
Range(α − λIn ) = 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.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Put Ker α − λIn = V , and
Range(α − λIn ) = 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 .
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 9 Every C-linear map A : C2n+1 → C2n+1 has
an eigen value.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 9 Every C-linear map A : C2n+1 → C2n+1 has
an eigen value.
Solution: By Ex. 5 and Ex. 8,
αA , βA : Herm2n+1 → Herm2n+1 have a common
eigenvector.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 9 Every C-linear map A : C2n+1 → C2n+1 has
an eigen value.
Solution: By Ex. 5 and Ex. 8,
αA , βA : Herm2n+1 → Herm2n+1 have a common
eigenvector.
By Ex. 7, A has an eigenvalue.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 10 Show that the space Symn (K) of
symmetric n × n matrices forms a subspace of
dimension n(n + 1)/2 of Mn (K).
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 10 Show that the space Symn (K) of
symmetric n × n matrices forms a subspace of
dimension n(n + 1)/2 of Mn (K).
Ex. 11 Given A ∈ Mn (K), show that
1
φA : B 7→ (AB + BAt ); ψA : B 7→ ABAt
2
define two commuting endomorphisms of
Symn (K). Show that if B is a common eigen
vector of φA , ψA then (A2 + aA + bId)B = 0 for
some a, b ∈ K. Further if K = C, conclude that
A has an eigen value.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Answer: The first part is easy.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Answer: The first part is easy.
To see the second part, suppose
φA (B) = AB + BAt = λB
and
ψA (B) = ABAt = µB
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Answer: The first part is easy.
To see the second part, suppose
φA (B) = AB + BAt = λB
and
ψA (B) = ABAt = µB
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Multiply first relation by A on the left and use
the second to obtain
A2 B + 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
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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
(A2 + aA + b)v = 0.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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
(A2 + aA + b)v = 0.
Now write
A2 + aA + bId = (A − λ1 Id)(A − λ2 Id). If
(A − λ2 Id)v = 0 then λ2 is an eigen value of A.
and we are through.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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
(A2 + aA + b)v = 0.
Now write
A2 + aA + bId = (A − λ1 Id)(A − λ2 Id). If
(A − λ2 Id)v = 0 then λ2 is an eigen value of A.
and we are through.
Otherwise u = (A − λ2 Id) 6= 0 and
(A − λ1 Id)u = 0 and hence λ1 is an eigen value
of A.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Let S1 (K, r ) denote the following statement:
Any endomorphism A : Kn → Kn has an eigen
value for all n not divisible by 2r . Let S2 (K, r )
denote the statement: Any two commuting
endomorphisms A1 , A2 : Kn → Kn have a
common eigen vector for all n not divisible by 2r .
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Let S1 (K, r ) denote the following statement:
Any endomorphism A : Kn → Kn has an eigen
value for all n not divisible by 2r . Let S2 (K, r )
denote the statement: Any two commuting
endomorphisms A1 , A2 : Kn → Kn have a
common eigen vector for all n not divisible by 2r .
Ex. 12 Prove that S1 (K, r ) =⇒ S2 (K, r ).
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Let S1 (K, r ) denote the following statement:
Any endomorphism A : Kn → Kn has an eigen
value for all n not divisible by 2r . Let S2 (K, r )
denote the statement: Any two commuting
endomorphisms A1 , A2 : Kn → Kn have a
common eigen vector for all n not divisible by 2r .
Ex. 12 Prove that S1 (K, r ) =⇒ S2 (K, r ).
Answer: Exactly as in Ex. 8.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 13 Prove that S1 (C, r ) =⇒ S1 (C, r + 1).
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 13 Prove that S1 (C, r ) =⇒ S1 (C, r + 1).
Answer:
Let n = 2k m where m is odd and k ≤ r . Let
A ∈ Mn (C). Then φA , ψA are two mutually
commuting operators on Symn (C) which is of
dimension n(n + 1)/2 = 2k−1 m(2k m + 1) which
is not divisible by 2r .
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 13 Prove that S1 (C, r ) =⇒ S1 (C, r + 1).
Answer:
Let n = 2k m where m is odd and k ≤ r . Let
A ∈ Mn (C). Then φA , ψA are two mutually
commuting operators on Symn (C) which is of
dimension n(n + 1)/2 = 2k−1 m(2k m + 1) which
is not divisible by 2r .
Therefore S1 (C, r ) together with Ex. 12 implies
that φA , ψA have a common eigen vector.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 13 Prove that S1 (C, r ) =⇒ S1 (C, r + 1).
Answer:
Let n = 2k m where m is odd and k ≤ r . Let
A ∈ Mn (C). Then φA , ψA are two mutually
commuting operators on Symn (C) which is of
dimension n(n + 1)/2 = 2k−1 m(2k m + 1) which
is not divisible by 2r .
Therefore S1 (C, r ) together with Ex. 12 implies
that φA , ψA have a common eigen vector.
By Ex. 11, this implies A has an eigen value in
C.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 14 Conclude that every non constant
polynomial over complex numbers has a root.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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Ex. 14 Conclude that every non constant
polynomial over complex numbers has a root.
Answer:
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
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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
S1 (C, 1). By Ex. 13 applied repeatedly, we get
S1 (C, r ) for all r . Write n = 2r m where m is
odd. Then S1 (C, r + 1) implies every
endomorphism A : Cn → Cn has an eigenvalue.
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA
THANK YOU
FOR YOUR ATTENTION
Anant R. Shastri I. I. T. Bombay
Derksen’s Proof of FUNDAMENTAL THEOREM of ALGEBRA