Sums and products of quadratic matrices

Sums and products of quadratic matrices

Clément de Seguins Pazzis (Lycée Privé Sainte-Geneviève)

(Université de Versailles-Saint-Quentin-en-Yvelines)

UCD, October 26, 2017

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Framework

F : arbitrary field; χ( F ) its characteristic; F an algebraic closure.

Let A an F -algebra (associative, with unity 1 _A ).

a ∈ A is quadratic whenever

∃(λ, µ) ∈ F ² , a ² = λa + µ1 A

i.e.

∃p ∈ F [t] : deg(p) = 2 and p(a) = 0.

Examples:

idempotents: a ² = a involutions: a ² = 1 _A

square-zero elements: a ² = 0.

Famous historical problems: Products of idempotents

Theorem (J. Erdos (1967))

M ∈ M n ( F ) is a product of idempotents iff M singular or M = I n . Theorem (Ballantine (1967))

M ∈ M _n ( F ) is the product of p idempotents iff rk(M − I _n ) ≤ p dim(Ker M).

Proof of necessity: if M = P ₁ · · · P _p with P ₁ , . . . , P _p idempotents then

Im(M − I _n ) ⊂ X

i

Im(P i − I _n )

whence

rk(M−I n ) ≤ X

dim Im(P _i −I n ) = X

dim(Ker P _i ) ≤ p dim(Ker M).

Famous historical problems: Products of idempotents

Theorem (J. Erdos (1967))

M ∈ M n ( F ) is a product of idempotents iff M singular or M = I n . Theorem (Ballantine (1967))

M ∈ M _n ( F ) is the product of p idempotents iff rk(M − I _n ) ≤ p dim(Ker M).

Proof of necessity: if M = P ₁ · · · P _p with P ₁ , . . . , P _p idempotents then

Im(M − I _n ) ⊂ X

i

Im(P i − I _n )

whence

rk(M−I n ) ≤ X

dim Im(P _i −I n ) = X

dim(Ker P _i ) ≤ p dim(Ker M).

Famous historical problems: Products of idempotents

Theorem (J. Erdos (1967))

M ∈ M n ( F ) is a product of idempotents iff M singular or M = I n . Theorem (Ballantine (1967))

M ∈ M _n ( F ) is the product of p idempotents iff rk(M − I _n ) ≤ p dim(Ker M).

Proof of necessity: if M = P ₁ · · · P _p with P ₁ , . . . , P _p idempotents then

Im(M − I _n ) ⊂ X

i

Im(P i − I _n ) whence

rk(M−I n ) ≤ X

dim Im(P _i −I n ) = X

dim(Ker P _i ) ≤ p dim(Ker M).

Famous historical problems: Products of idempotents

Theorem (J. Erdos (1967))

M ∈ M n ( F ) is a product of idempotents iff M singular or M = I n . Theorem (Ballantine (1967))

M ∈ M _n ( F ) is the product of p idempotents iff rk(M − I _n ) ≤ p dim(Ker M).

Proof of necessity: if M = P ₁ · · · P _p with P ₁ , . . . , P _p idempotents then

Im(M − I _n ) ⊂ X

i

Im(P i − I _n )

whence

rk(M−I n ) ≤ X

dim Im(P _i −I n ) = X

dim(Ker P _i ) ≤ p dim(Ker M).

Products of involutions

Theorem (Folklore)

M ∈ M n ( F ) is a product of involutions iff det M = ±1.

Theorem (Djokovi´c, Hoffman-Paige, Wonenburger (circa 1966)) M ∈ GL n ( F ) is the product of two involutions iff M ≃ M ⁻¹ . Proof of necessity: if M = AB where A ² = B ² = I _n , then

M ⁻¹ = B ⁻¹ A ⁻¹ = BA = BMB ⁻¹ .

Products of involutions

Theorem (Folklore)

M ∈ M n ( F ) is a product of involutions iff det M = ±1.

Theorem (Djokovi´c, Hoffman-Paige, Wonenburger (circa 1966)) M ∈ GL n ( F ) is the product of two involutions iff M ≃ M ⁻¹ . Proof of necessity: if M = AB where A ² = B ² = I _n , then

M ⁻¹ = B ⁻¹ A ⁻¹ = BA = BMB ⁻¹ .

Products of involutions

Theorem (Folklore)

M ∈ M n ( F ) is a product of involutions iff det M = ±1.

Theorem (Djokovi´c, Hoffman-Paige, Wonenburger (circa 1966)) M ∈ GL n ( F ) is the product of two involutions iff M ≃ M ⁻¹ . Proof of necessity: if M = AB where A ² = B ² = I _n , then

M ⁻¹ = B ⁻¹ A ⁻¹ = BA = BMB ⁻¹ .

Sums of idempotents

Theorem (Hartwig-Putcha, Wu (1990))

M ∈ M n ( F ) is a sum of idempotents iff tr M = k.1 F for some integer k ≥ rk(M).

Proof of necessity: if M = P p

i=1 P _i , then tr M = X

i

tr P _i = ( X

i

rk P _i ).1 F

and X

i

rk P _i ≥ rk M.

Sums of idempotents

Theorem (Hartwig-Putcha, Wu (1990))

M ∈ M n ( F ) is a sum of idempotents iff tr M = k.1 F for some integer k ≥ rk(M).

Proof of necessity: if M = P p

i=1 P _i , then tr M = X

i

tr P _i = ( X

i

rk P _i ).1 F

and X

i

rk P _i ≥ rk M.

Length problem for sums of idempotents

Open problem: given p ∈ N , which matrices of M n ( F ) that are sums of p idempotents?

Complete solution unknown if p ≥ 3.

Solution known for p = 2 (Hartwig-Putcha, 1990).

Length problem for sums of idempotents

Open problem: given p ∈ N , which matrices of M n ( F ) that are sums of p idempotents?

Complete solution unknown if p ≥ 3.

Solution known for p = 2 (Hartwig-Putcha, 1990).

Length problem for sums of idempotents

Open problem: given p ∈ N , which matrices of M n ( F ) that are sums of p idempotents?

Complete solution unknown if p ≥ 3.

Solution known for p = 2 (Hartwig-Putcha, 1990).

Our problem

Let p , q in F [t] with degree 2.

x ∈ A is a (p, q)-sum whenever

∃(a, b) ∈ A ² : x = a + b and p(a) = q(b) = 0.

x ∈ A is a (p, q)-product whenever

∃(a, b) ∈ A ² : x = ab and p(a) = q(b) = 0.

Note: the set of all (p, q)-sums (resp. (p, q)-products) in A is a union of conjugacy classes.

Problem: if A = M n ( F ) (or A = End(V ) where V

finite-dimensional vector space over F with dimension n),

characterize the (p, q)-sums and the (p, q)-products in terms of

Our problem

Let p , q in F [t] with degree 2.

x ∈ A is a (p, q)-sum whenever

∃(a, b) ∈ A ² : x = a + b and p(a) = q(b) = 0.

x ∈ A is a (p, q)-product whenever

∃(a, b) ∈ A ² : x = ab and p(a) = q(b) = 0.

Note: the set of all (p, q)-sums (resp. (p, q)-products) in A is a union of conjugacy classes.

Problem: if A = M n ( F ) (or A = End(V ) where V

finite-dimensional vector space over F with dimension n),

characterize the (p, q)-sums and the (p, q)-products in terms of

Our problem

Let p , q in F [t] with degree 2.

x ∈ A is a (p, q)-sum whenever

∃(a, b) ∈ A ² : x = a + b and p(a) = q(b) = 0.

x ∈ A is a (p, q)-product whenever

∃(a, b) ∈ A ² : x = ab and p(a) = q(b) = 0.

Note: the set of all (p, q)-sums (resp. (p, q)-products) in A is a union of conjugacy classes.

Problem: if A = M n ( F ) (or A = End(V ) where V

finite-dimensional vector space over F with dimension n),

characterize the (p, q)-sums and the (p, q)-products in terms of

Our problem

Let p , q in F [t] with degree 2.

x ∈ A is a (p, q)-sum whenever

∃(a, b) ∈ A ² : x = a + b and p(a) = q(b) = 0.

x ∈ A is a (p, q)-product whenever

∃(a, b) ∈ A ² : x = ab and p(a) = q(b) = 0.

Note: the set of all (p, q)-sums (resp. (p, q)-products) in A is a union of conjugacy classes.

Problem: if A = M n ( F ) (or A = End(V ) where V

finite-dimensional vector space over F with dimension n),

characterize the (p, q)-sums and the (p, q)-products in terms of

Our problem

Let p , q in F [t] with degree 2.

x ∈ A is a (p, q)-sum whenever

∃(a, b) ∈ A ² : x = a + b and p(a) = q(b) = 0.

x ∈ A is a (p, q)-product whenever

∃(a, b) ∈ A ² : x = ab and p(a) = q(b) = 0.

Note: the set of all (p, q)-sums (resp. (p, q)-products) in A is a union of conjugacy classes.

Problem: if A = M n ( F ) (or A = End(V ) where V

finite-dimensional vector space over F with dimension n),

characterize the (p, q)-sums and the (p, q)-products in terms of

Some reduction tricks

We can assume that p and q are monic.

(p, q)-sums are (q, p)-sums.

In M n ( F ), (p, q)-products are (q, p)-products

(if M = AB with p(A) = q(B) = 0, then M ^T = B ^T A ^T and q(B ^T ) = p(A ^T ) = 0; one concludes by noting that M ≃ M ^T ).

Translation trick for sums: let λ ∈ F . Set p ₁ := p(t − λ).

x = a + b with p(a) = q(b) = 0 iff x + λ1 A = (a + λ1 A ) + B with p ₁ (a + λ1 _A ) = q(b) = 0.

Multiplication trick for sums: let λ ∈ F ^∗ . Set p ₁ := p(t/λ) and q ₁ := q(t/λ). Then, x = a + b with p(a) = q(b) = 0 iff

λx = λa + λb with p ₁ (λa) = q ₁ (λb) = 0.

Some reduction tricks

We can assume that p and q are monic.

(p, q)-sums are (q, p)-sums.

In M n ( F ), (p, q)-products are (q, p)-products

(if M = AB with p(A) = q(B) = 0, then M ^T = B ^T A ^T and q(B ^T ) = p(A ^T ) = 0; one concludes by noting that M ≃ M ^T ).

Translation trick for sums: let λ ∈ F . Set p ₁ := p(t − λ).

x = a + b with p(a) = q(b) = 0 iff x + λ1 A = (a + λ1 A ) + B with p ₁ (a + λ1 _A ) = q(b) = 0.

Multiplication trick for sums: let λ ∈ F ^∗ . Set p ₁ := p(t/λ) and q ₁ := q(t/λ). Then, x = a + b with p(a) = q(b) = 0 iff

λx = λa + λb with p ₁ (λa) = q ₁ (λb) = 0.

Some reduction tricks

We can assume that p and q are monic.

(p, q)-sums are (q, p)-sums.

In M n ( F ), (p, q)-products are (q, p)-products

(if M = AB with p(A) = q(B) = 0, then M ^T = B ^T A ^T and q(B ^T ) = p(A ^T ) = 0; one concludes by noting that M ≃ M ^T ).

Translation trick for sums: let λ ∈ F . Set p ₁ := p(t − λ).

x = a + b with p(a) = q(b) = 0 iff x + λ1 A = (a + λ1 A ) + B with p ₁ (a + λ1 _A ) = q(b) = 0.

Multiplication trick for sums: let λ ∈ F ^∗ . Set p ₁ := p(t/λ) and q ₁ := q(t/λ). Then, x = a + b with p(a) = q(b) = 0 iff

λx = λa + λb with p ₁ (λa) = q ₁ (λb) = 0.

Some reduction tricks

We can assume that p and q are monic.

(p, q)-sums are (q, p)-sums.

In M n ( F ), (p, q)-products are (q, p)-products

(if M = AB with p(A) = q(B) = 0, then M ^T = B ^T A ^T and q(B ^T ) = p(A ^T ) = 0; one concludes by noting that M ≃ M ^T ).

Translation trick for sums: let λ ∈ F . Set p ₁ := p(t − λ).

x = a + b with p(a) = q(b) = 0 iff x + λ1 A = (a + λ1 A ) + B with p ₁ (a + λ1 _A ) = q(b) = 0.

Multiplication trick for sums: let λ ∈ F ^∗ . Set p ₁ := p(t/λ) and q ₁ := q(t/λ). Then, x = a + b with p(a) = q(b) = 0 iff

λx = λa + λb with p ₁ (λa) = q ₁ (λb) = 0.

Some reduction tricks

We can assume that p and q are monic.

(p, q)-sums are (q, p)-sums.

In M n ( F ), (p, q)-products are (q, p)-products

(if M = AB with p(A) = q(B) = 0, then M ^T = B ^T A ^T and q(B ^T ) = p(A ^T ) = 0; one concludes by noting that M ≃ M ^T ).

Translation trick for sums: let λ ∈ F . Set p ₁ := p(t − λ).

x = a + b with p(a) = q(b) = 0 iff x + λ1 A = (a + λ1 A ) + B with p ₁ (a + λ1 _A ) = q(b) = 0.

Multiplication trick for sums: let λ ∈ F ^∗ . Set p ₁ := p(t/λ) and q ₁ := q(t/λ). Then, x = a + b with p(a) = q(b) = 0 iff

λx = λa + λb with p ₁ (λa) = q ₁ (λb) = 0.

Some reduction tricks

We can assume that p and q are monic.

(p, q)-sums are (q, p)-sums.

In M n ( F ), (p, q)-products are (q, p)-products

(if M = AB with p(A) = q(B) = 0, then M ^T = B ^T A ^T and q(B ^T ) = p(A ^T ) = 0; one concludes by noting that M ≃ M ^T ).

Translation trick for sums: let λ ∈ F . Set p ₁ := p(t − λ).

x = a + b with p(a) = q(b) = 0 iff x + λ1 A = (a + λ1 A ) + B with p ₁ (a + λ1 _A ) = q(b) = 0.

Multiplication trick for sums: let λ ∈ F ^∗ . Set p ₁ := p(t/λ) and q ₁ := q(t/λ). Then, x = a + b with p(a) = q(b) = 0 iff

λx = λa + λb with p ₁ (λa) = q ₁ (λb) = 0.

Reduced situation for split polynomials

If p and q split, only three cases need to be considered:

p = t ² − αt, q = t ² − t with α nonzero (matrices of the form αP + Q with P, Q idempotents);

Hartwig-Putcha (1990) α = ±1 and F = C ;

de Seguins Pazzis (2010) arbitrary α, arbitrary field.

p = q = t ² (sums of two square-zero matrices);

Wang-Wu (1991) F = C ; Botha (2012) arbitrary field.

p = t ² − t and q = t ² (idempotent plus square-zero) Wang (1995) F = C ;

de Seguins Pazzis (2012) arbitrary field.

Reduced situation for split polynomials

If p and q split, only three cases need to be considered:

p = t ² − αt, q = t ² − t with α nonzero (matrices of the form αP + Q with P, Q idempotents);

Hartwig-Putcha (1990) α = ±1 and F = C ;

de Seguins Pazzis (2010) arbitrary α, arbitrary field.

p = q = t ² (sums of two square-zero matrices);

Wang-Wu (1991) F = C ; Botha (2012) arbitrary field.

p = t ² − t and q = t ² (idempotent plus square-zero) Wang (1995) F = C ;

de Seguins Pazzis (2012) arbitrary field.

Reduced situation for split polynomials

If p and q split, only three cases need to be considered:

p = t ² − αt, q = t ² − t with α nonzero (matrices of the form αP + Q with P, Q idempotents);

Hartwig-Putcha (1990) α = ±1 and F = C ;

de Seguins Pazzis (2010) arbitrary α, arbitrary field.

p = q = t ² (sums of two square-zero matrices);

Wang-Wu (1991) F = C ; Botha (2012) arbitrary field.

p = t ² − t and q = t ² (idempotent plus square-zero) Wang (1995) F = C ;

de Seguins Pazzis (2012) arbitrary field.

Some additional notation

Jordan cell of size n with respect to λ ∈ F :

J _n (λ) :=



 

 



λ 1 (0)

λ . ..

. .. 1

(0) λ



 

 



∈ M n ( F ).

Direct sum of two square matrices:

A ⊕ B :=

A 0 0 B

.

Some additional notation

Jordan cell of size n with respect to λ ∈ F :

J _n (λ) :=



 

 



λ 1 (0)

λ . ..

. .. 1

(0) λ



 

 



∈ M n ( F ).

Direct sum of two square matrices:

A ⊕ B :=

A 0 0 B

.

Statement of some known results

Let p , q in F [t] split and monic with degree 2.

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ).

Set

σ := x ₁ + x ₂ + y ₁ + y ₂ = tr(p) + tr(q) and

Root(p) + Root(q) =

x _i + y _j | (i, j) ∈ {1, 2} ² . (set of all (p, q)-sums in F !).

Roughly, the (p, q)-sums are the matrices with spectrum

“symmetric around" σ

2 ·

Statement of some known results

Let p , q in F [t] split and monic with degree 2.

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ).

Set

σ := x ₁ + x ₂ + y ₁ + y ₂ = tr(p) + tr(q) and

Root(p) + Root(q) =

x _i + y _j | (i, j) ∈ {1, 2} ² . (set of all (p, q)-sums in F !).

Roughly, the (p, q)-sums are the matrices with spectrum

“symmetric around" σ

2 ·

Theorem

Let M ∈ M n ( F ). Then M is a (p, q)-sum iff M similar over F to M ₁ ⊕ · · · ⊕ M _k , each M _i being of one of the following types:

J _n (λ) ⊕ J _n (σ − λ),

(λ ∈ F such that λ 6= σ − λ, n > 0);

J _n + 1 (λ) ⊕ J _n (σ − λ),

(n ≥ 0, λ ∈ Root(p) + Root(q));

J _n + 2 (λ) ⊕ J n (σ − λ),

( n ≥ 0, λ ∈ Root(p) + Root(q), exactly one of p and q has a double root);

J _2n (λ),

(n > 0, λ ∈ F such that λ = σ − λ).

J _2n + 1 (λ),

(n ≥ 0, λ ∈ Root(p) + Root(q) such that λ = σ − λ).

Theorem

Let M ∈ M n ( F ). Then M is a (p, q)-sum iff M similar over F to M ₁ ⊕ · · · ⊕ M _k , each M _i being of one of the following types:

J _n (λ) ⊕ J _n (σ − λ),

(λ ∈ F such that λ 6= σ − λ, n > 0);

J _n + 1 (λ) ⊕ J _n (σ − λ),

(n ≥ 0, λ ∈ Root(p) + Root(q));

J _n + 2 (λ) ⊕ J n (σ − λ),

( n ≥ 0, λ ∈ Root(p) + Root(q), exactly one of p and q has a double root);

J _2n (λ),

(n > 0, λ ∈ F such that λ = σ − λ).

J _2n + 1 (λ),

(n ≥ 0, λ ∈ Root(p) + Root(q) such that λ = σ − λ).

Theorem

Let M ∈ M n ( F ). Then M is a (p, q)-sum iff M similar over F to M ₁ ⊕ · · · ⊕ M _k , each M _i being of one of the following types:

J _n (λ) ⊕ J _n (σ − λ),

(λ ∈ F such that λ 6= σ − λ, n > 0);

J _n + 1 (λ) ⊕ J _n (σ − λ),

(n ≥ 0, λ ∈ Root(p) + Root(q));

J _n + 2 (λ) ⊕ J n (σ − λ),

( n ≥ 0, λ ∈ Root(p) + Root(q), exactly one of p and q has a double root);

J _2n (λ),

(n > 0, λ ∈ F such that λ = σ − λ).

J _2n + 1 (λ),

(n ≥ 0, λ ∈ Root(p) + Root(q) such that λ = σ − λ).

Theorem

Let M ∈ M n ( F ). Then M is a (p, q)-sum iff M similar over F to M ₁ ⊕ · · · ⊕ M _k , each M _i being of one of the following types:

J _n (λ) ⊕ J _n (σ − λ),

(λ ∈ F such that λ 6= σ − λ, n > 0);

J _n + 1 (λ) ⊕ J _n (σ − λ),

(n ≥ 0, λ ∈ Root(p) + Root(q));

J _n + 2 (λ) ⊕ J n (σ − λ),

( n ≥ 0, λ ∈ Root(p) + Root(q), exactly one of p and q has a double root);

J _2n (λ),

(n > 0, λ ∈ F such that λ = σ − λ).

J _2n + 1 (λ),

(n ≥ 0, λ ∈ Root(p) + Root(q) such that λ = σ − λ).

Theorem

Let M ∈ M n ( F ). Then M is a (p, q)-sum iff M similar over F to M ₁ ⊕ · · · ⊕ M _k , each M _i being of one of the following types:

J _n (λ) ⊕ J _n (σ − λ),

(λ ∈ F such that λ 6= σ − λ, n > 0);

J _n + 1 (λ) ⊕ J _n (σ − λ),

(n ≥ 0, λ ∈ Root(p) + Root(q));

J _n + 2 (λ) ⊕ J n (σ − λ),

( n ≥ 0, λ ∈ Root(p) + Root(q), exactly one of p and q has a double root);

J _2n (λ),

(n > 0, λ ∈ F such that λ = σ − λ).

J _2n + 1 (λ),

(n ≥ 0, λ ∈ Root(p) + Root(q) such that λ = σ − λ).

Theorem

Let M ∈ M n ( F ). Then M is a (p, q)-sum iff M similar over F to M ₁ ⊕ · · · ⊕ M _k , each M _i being of one of the following types:

J _n (λ) ⊕ J _n (σ − λ),

(λ ∈ F such that λ 6= σ − λ, n > 0);

J _n + 1 (λ) ⊕ J _n (σ − λ),

(n ≥ 0, λ ∈ Root(p) + Root(q));

J _n + 2 (λ) ⊕ J n (σ − λ),

( n ≥ 0, λ ∈ Root(p) + Root(q), exactly one of p and q has a double root);

J _2n (λ),

(n > 0, λ ∈ F such that λ = σ − λ).

J _2n + 1 (λ),

(n ≥ 0, λ ∈ Root(p) + Root(q) such that λ = σ − λ).

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Decomposability

Let V vector space over F , u ∈ End(V )

V = V ₁ ⊕ V ₂ with V _i invariant under u (and dim V _i > 0);

Assume that u _|V

(p, q)-sum for all i.

Then u _|V

= a _i + b _i with p(a _i ) = q(b _i ) = 0.

Set a ∈ End(V ) and b ∈ End(V ) s.t. a _|V

= a _i and b _|V

= b _i . p(a) = 0, q(b) = 0, u = a + b, whence u is a (p, q)-sum.

In that case u is decomposable.

Problem: understand the indecomposable (p, q)-sums.

Warning: if u is a (p, q)-sum and V = V ₁ ⊕ V ₂ with V ₁ , V ₂ invariant under u, u _|V

and u _|V

might not be (p, q)-sums!

Example: p = q = t ²

u ↔

1 0

.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

Regularity

Definition

Let u ∈ End(V ) (V finite-dimensional vector space over F ).

u is (p, q)-regular iff no eigenvalue of u in F is a (p, q)-sum u is (p, q)-exceptional iff all eigenvalues of u in F are (p, q)-sums.

Theorem

In End(V ), any indecomposable (p, q)-sum is either regular or exceptional.

Thus, one needs to study only:

the indecomposable regular (p, q)-sums;

(p, q)-sums.

The commutation lemma

Let a ∈ A and p = t ² − tr(p) t + p(0) ∈ F [t] such that p(a) = 0.

Set

a ^⋆ := tr(p)1 _A − a (p-conjugate of a) so that

aa ^⋆ = a ^⋆ a = p(0)1 A .

Similarly, take q ∈ F [t] monic with degree 2 and b ∈ A s.t.

q(b) = 0. Set b ^⋆ the q-conjugate of b.

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba) = a ^⋆ b + b ^⋆ a.

Lemma (Commutation lemma)

a and b commute with ab ^⋆ + ba ^⋆ .

The commutation lemma

Let a ∈ A and p = t ² − tr(p) t + p(0) ∈ F [t] such that p(a) = 0.

Set

a ^⋆ := tr(p)1 _A − a (p-conjugate of a) so that

aa ^⋆ = a ^⋆ a = p(0)1 A .

Similarly, take q ∈ F [t] monic with degree 2 and b ∈ A s.t.

q(b) = 0. Set b ^⋆ the q-conjugate of b.

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba) = a ^⋆ b + b ^⋆ a.

Lemma (Commutation lemma)

a and b commute with ab ^⋆ + ba ^⋆ .

The commutation lemma

Let a ∈ A and p = t ² − tr(p) t + p(0) ∈ F [t] such that p(a) = 0.

Set

a ^⋆ := tr(p)1 _A − a (p-conjugate of a) so that

aa ^⋆ = a ^⋆ a = p(0)1 A .

Similarly, take q ∈ F [t] monic with degree 2 and b ∈ A s.t.

q(b) = 0. Set b ^⋆ the q-conjugate of b.

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba) = a ^⋆ b + b ^⋆ a.

Lemma (Commutation lemma)

a and b commute with ab ^⋆ + ba ^⋆ .

The commutation lemma

Let a ∈ A and p = t ² − tr(p) t + p(0) ∈ F [t] such that p(a) = 0.

Set

a ^⋆ := tr(p)1 _A − a (p-conjugate of a) so that

aa ^⋆ = a ^⋆ a = p(0)1 A .

Similarly, take q ∈ F [t] monic with degree 2 and b ∈ A s.t.

q(b) = 0. Set b ^⋆ the q-conjugate of b.

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba) = a ^⋆ b + b ^⋆ a.

Lemma (Commutation lemma)

a and b commute with ab ^⋆ + ba ^⋆ .

The commutation lemma

Let a ∈ A and p = t ² − tr(p) t + p(0) ∈ F [t] such that p(a) = 0.

Set

a ^⋆ := tr(p)1 _A − a (p-conjugate of a) so that

aa ^⋆ = a ^⋆ a = p(0)1 A .

Similarly, take q ∈ F [t] monic with degree 2 and b ∈ A s.t.

q(b) = 0. Set b ^⋆ the q-conjugate of b.

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba) = a ^⋆ b + b ^⋆ a.

Lemma (Commutation lemma)

a and b commute with ab ^⋆ + ba ^⋆ .

The commutation lemma

Let a ∈ A and p = t ² − tr(p) t + p(0) ∈ F [t] such that p(a) = 0.

Set

a ^⋆ := tr(p)1 _A − a (p-conjugate of a) so that

aa ^⋆ = a ^⋆ a = p(0)1 A .

Similarly, take q ∈ F [t] monic with degree 2 and b ∈ A s.t.

q(b) = 0. Set b ^⋆ the q-conjugate of b.

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba) = a ^⋆ b + b ^⋆ a.

Lemma (Commutation lemma)

a and b commute with ab ^⋆ + ba ^⋆ .

Set u := a + b. Then,

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba)

= tr(q)a + tr(p)b − (a + b) ² + a ² + b ²

= σu − u ² − (p(0) + q(0))1 _A .

Consequence: v := u ² − σu commutes with a and b.

Now, assume that A = End(V ) and split

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ) (roots in F ) and set

F _p,q := Y

i,j

(t − x _i − y _j ) ∈ F [t].

u regular (resp. exceptional) iff F (u) invertible (resp.

Set u := a + b. Then,

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba)

= tr(q)a + tr(p)b − (a + b) ² + a ² + b ²

= σu − u ² − (p(0) + q(0))1 _A .

Consequence: v := u ² − σu commutes with a and b.

Now, assume that A = End(V ) and split

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ) (roots in F ) and set

F _p,q := Y

i,j

(t − x _i − y _j ) ∈ F [t].

u regular (resp. exceptional) iff F (u) invertible (resp.

Set u := a + b. Then,

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba)

= tr(q)a + tr(p)b − (a + b) ² + a ² + b ²

= σu − u ² − (p(0) + q(0))1 _A .

Consequence: v := u ² − σu commutes with a and b.

Now, assume that A = End(V ) and split

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ) (roots in F ) and set

F _p,q := Y

i,j

(t − x _i − y _j ) ∈ F [t].

u regular (resp. exceptional) iff F (u) invertible (resp.

Set u := a + b. Then,

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba)

= tr(q)a + tr(p)b − (a + b) ² + a ² + b ²

= σu − u ² − (p(0) + q(0))1 _A .

Consequence: v := u ² − σu commutes with a and b.

Now, assume that A = End(V ) and split

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ) (roots in F ) and set

F _p,q := Y

i,j

(t − x _i − y _j ) ∈ F [t].

u regular (resp. exceptional) iff F (u) invertible (resp.

Set u := a + b. Then,

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba)

= tr(q)a + tr(p)b − (a + b) ² + a ² + b ²

= σu − u ² − (p(0) + q(0))1 _A .

Consequence: v := u ² − σu commutes with a and b.

Now, assume that A = End(V ) and split

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ) (roots in F ) and set

F _p,q := Y

i,j

(t − x _i − y _j ) ∈ F [t].

u regular (resp. exceptional) iff F (u) invertible (resp.

Set u := a + b. Then,

ab ^⋆ + ba ^⋆ = tr(q)a + tr(p)b − (ab + ba)

= tr(q)a + tr(p)b − (a + b) ² + a ² + b ²

= σu − u ² − (p(0) + q(0))1 _A .

Consequence: v := u ² − σu commutes with a and b.

Now, assume that A = End(V ) and split

p(t) = (t − x ₁ )(t − x ₂ ) and q(t) = (t − y ₁ )(t − y ₂ ) (roots in F ) and set

F _p,q := Y

i,j

(t − x _i − y _j ) ∈ F [t].

u regular (resp. exceptional) iff F (u) invertible (resp.

F _p,q = t ² − σt + (x 1 + y ₁ )(x 2 + y ₂ )

t ² − σt + (x 1 + y ₂ )(x 2 + y ₁ ) so

F _p,q (u) = Λ(v) for

Λ := t + (x 1 + y ₁ )(x 2 + y ₂ )

t + (x 1 + y ₂ )(x 2 + y ₁ )

∈ F [t].

For large enough N

V = Ker F _p,q (u) ^N

| {z }

V

⊕ Im F _p,q (u) ^N

| {z }

V

.

V _i invariant under a, b, u, whence u _|V

is a (p, q)-sum!!

F _p,q = t ² − σt + (x 1 + y ₁ )(x 2 + y ₂ )

t ² − σt + (x 1 + y ₂ )(x 2 + y ₁ ) so

F _p,q (u) = Λ(v) for

Λ := t + (x 1 + y ₁ )(x 2 + y ₂ )

t + (x 1 + y ₂ )(x 2 + y ₁ )

∈ F [t].

For large enough N

V = Ker F _p,q (u) ^N

| {z }

V

⊕ Im F _p,q (u) ^N

| {z }

V

.

V _i invariant under a, b, u, whence u _|V

is a (p, q)-sum!!

F _p,q = t ² − σt + (x 1 + y ₁ )(x 2 + y ₂ )

t ² − σt + (x 1 + y ₂ )(x 2 + y ₁ ) so

F _p,q (u) = Λ(v) for

Λ := t + (x 1 + y ₁ )(x 2 + y ₂ )

t + (x 1 + y ₂ )(x 2 + y ₁ )

∈ F [t].

For large enough N

V = Ker F _p,q (u) ^N

| {z }

V

⊕ Im F _p,q (u) ^N

| {z }

V

.

V _i invariant under a, b, u, whence u _|V

is a (p, q)-sum!!

F _p,q = t ² − σt + (x 1 + y ₁ )(x 2 + y ₂ )

t ² − σt + (x 1 + y ₂ )(x 2 + y ₁ ) so

F _p,q (u) = Λ(v) for

Λ := t + (x 1 + y ₁ )(x 2 + y ₂ )

t + (x 1 + y ₂ )(x 2 + y ₁ )

∈ F [t].

For large enough N

V = Ker F _p,q (u) ^N

| {z }

V

⊕ Im F _p,q (u) ^N

| {z }

V

.

V _i invariant under a, b, u, whence u _|V

is a (p, q)-sum!!

F _p,q = t ² − σt + (x 1 + y ₁ )(x 2 + y ₂ )

t ² − σt + (x 1 + y ₂ )(x 2 + y ₁ ) so

F _p,q (u) = Λ(v) for

Λ := t + (x 1 + y ₁ )(x 2 + y ₂ )

t + (x 1 + y ₂ )(x 2 + y ₁ )

∈ F [t].

For large enough N

V = Ker F _p,q (u) ^N

| {z }

V

⊕ Im F _p,q (u) ^N

| {z }

V

.

V _i invariant under a, b, u, whence u _|V

is a (p, q)-sum!!

Quick reminder on the Frobenius canonical form

Companion matrix associated with a monic polynomial r = t ^k − a _k−1 t ^k−1 − · · · − a ₀ :

C(r ) :=



 

 

 



0 (0) a ₀

1 0 a ₁

0 . .. ... .. . .. . . .. 0 a _k−2 (0) · · · 0 1 a _k−1



 

 

 



∈ M _k ( F ).

Theorem (Frobenius canonical form theorem)

Every matrix M is similar to C(r 1 ) ⊕ · · · ⊕ C(r k ) for a unique list

(r 1 , . . . , r _k ) of non-constant monic polynomials s.t. r _i+1 divides r _i

for all i. The r _i ’s are the invariant factors of M.

Analyzing regular (p, q)-sums

Reformulation of earlier results:

Theorem

Assume p, q split over F . Let M ∈ M n ( F ) be (p, q)-regular.

Then, M is a (p, q)-sum iff each invariant factor of M equals r (t ² − σt) for some r ∈ F [t].

Extending the ground field, one easily deduces:

Theorem

If M ∈ M n ( F ) is a regular (p, q)-sum then its invariants factors are polynomials in t ² − σt.

Problem: is the converse still true in the non-split case?

Analyzing regular (p, q)-sums

Reformulation of earlier results:

Theorem

Assume p, q split over F . Let M ∈ M n ( F ) be (p, q)-regular.

Then, M is a (p, q)-sum iff each invariant factor of M equals r (t ² − σt) for some r ∈ F [t].

Extending the ground field, one easily deduces:

Theorem

If M ∈ M n ( F ) is a regular (p, q)-sum then its invariants factors are polynomials in t ² − σt.

Problem: is the converse still true in the non-split case?