SENSE AND SIMPLICITY:
a talk on two matrices
on the occasion of Malo Hautus’
afscheidsrede
Harry Trentelman
University of Groningen, The Netherlands
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.1/25
Controllability and the Hautus test
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.2/25
Controllability
Linear dynamical system with inputs:
ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , the state, u(t) ∈ Rm , control input, A ∈ Mn×n (R),
B ∈ Mn×m (R).
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.3/25
Controllability
Linear dynamical system with inputs:
ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , the state, u(t) ∈ Rm , control input, A ∈ Mn×n (R),
B ∈ Mn×m (R).
The system is called controllable if for every pair of states x0 , x1
there exists T > 0 and an input function u on [0, T ] such that
x(0) = x0 and x(T ) = x1 .
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.3/25
Controllability
Linear dynamical system with inputs:
ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , the state, u(t) ∈ Rm , control input, A ∈ Mn×n (R),
B ∈ Mn×m (R).
The system is called controllable if for every pair of states x0 , x1
there exists T > 0 and an input function u on [0, T ] such that
x(0) = x0 and x(T ) = x1 .
We say: the pair (A, B) is controllable. Kalman, 1963
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.3/25
Test for controllability
n × nm matrix (B, AB, A2 B, . . . , An−1 B)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.4/25
Test for controllability
n × nm matrix (B, AB, A2 B, . . . , An−1 B)
Theorem (Kalman 1963): (A, B) is controllable if and only if
(B, AB, A2 B, . . . , An−1 B) has rank n.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.4/25
Test for controllability
n × nm matrix (B, AB, A2 B, . . . , An−1 B)
Theorem (Kalman 1963): (A, B) is controllable if and only if
(B, AB, A2 B, . . . , An−1 B) has rank n.
im(B, AB, A2 B, . . . , An−1 B) ⊆ Rn is equal to the the
reachable subspace
{x1 ∈ Rn | ∃ u such that x(0) = 0 and x1 = x(T )}
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.4/25
The Hautus test
Famous article by Malo Hautus, ’Controllability and observability
conditions of linear autonomous systems’, Proceedings Koninklijke
Nederlandse Academie voor Wetenschappen, series A, 1969.
Main result:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.5/25
The Hautus test
Famous article by Malo Hautus, ’Controllability and observability
conditions of linear autonomous systems’, Proceedings Koninklijke
Nederlandse Academie voor Wetenschappen, series A, 1969.
Main result:
Theorem (Hautus 1969): (A, B) is controllable if and only if the
n × (n + m) matrix (A − λI, B) has rank n for every
eigenvalue λ of A.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.5/25
The Hautus test
Famous article by Malo Hautus, ’Controllability and observability
conditions of linear autonomous systems’, Proceedings Koninklijke
Nederlandse Academie voor Wetenschappen, series A, 1969.
Main result:
Theorem (Hautus 1969): (A, B) is controllable if and only if the
n × (n + m) matrix (A − λI, B) has rank n for every
eigenvalue λ of A.
Equivalently: there is no left eigenvector of A that is a left zero
vector of B .
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.5/25
Proving the Hautus test
Standard result in every introductory course to linear systems.
(⇒)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.6/25
Proving the Hautus test
Standard result in every introductory course to linear systems.
(⇒)
rank(A − λI, B) < n =⇒
∃ η 6= 0 such that ηA = λη , ηB = 0 =⇒
∃ η 6= 0 such that η(B, AB, A2 B, . . . , An−1 B) = 0 =⇒
rank(B, AB, A2 B, . . . , An−1 B) < n =⇒
(A, B) not controllable
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.6/25
Proving the Hautus test
(⇐)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.7/25
Proving the Hautus test
(⇐)
Up to now (for students in Groningen):
assume (A, B) not controllable. Then the reachable subspace is a
proper A-invariant subspace of Rn containing im(B)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.7/25
Proving the Hautus test
(⇐)
Up to now (for students in Groningen):
assume (A, B) not controllable. Then the reachable subspace is a
proper A-invariant subspace of Rn containing im(B)
Write matrices for A and B with respect to a basis adapted to this
subspace.
This yields a contradiction with (A − λI, B) full row rank for all
λ.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.7/25
Proving the Hautus test
(⇐)
Up to now (for students in Groningen):
assume (A, B) not controllable. Then the reachable subspace is a
proper A-invariant subspace of Rn containing im(B)
Write matrices for A and B with respect to a basis adapted to this
subspace.
This yields a contradiction with (A − λI, B) full row rank for all
λ.
In the original paper from 1969 one can find Malo’s original proof:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.7/25
Proving the Hautus test
(⇐)
Up to now (for students in Groningen):
assume (A, B) not controllable. Then the reachable subspace is a
proper A-invariant subspace of Rn containing im(B)
Write matrices for A and B with respect to a basis adapted to this
subspace.
This yields a contradiction with (A − λI, B) full row rank for all
λ.
In the original paper from 1969 one can find Malo’s original proof:
Sense and Simplicity!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.7/25
Hautus’ proof of the Hautus test
(⇐)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.8/25
Hautus’ proof of the Hautus test
(⇐)
(A, B) not controllable ⇒
∃ η 6= 0 such that η(B, AB, A2 B, . . . , An−1 B) = 0
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.8/25
Hautus’ proof of the Hautus test
(⇐)
(A, B) not controllable ⇒
∃ η 6= 0 such that η(B, AB, A2 B, . . . , An−1 B) = 0
Let ψ(z) minimal degree polynomial such that ηψ(A) = 0
Then deg(ψ) ≥ 1
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.8/25
Hautus’ proof of the Hautus test
(⇐)
(A, B) not controllable ⇒
∃ η 6= 0 such that η(B, AB, A2 B, . . . , An−1 B) = 0
Let ψ(z) minimal degree polynomial such that ηψ(A) = 0
Then deg(ψ) ≥ 1
Factorize ψ(z) = φ(z)(z − λ).
Then 0 = ηψ(A) = ηφ(A)(A − λI)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.8/25
Hautus’ proof of the Hautus test
(⇐)
(A, B) not controllable ⇒
∃ η 6= 0 such that η(B, AB, A2 B, . . . , An−1 B) = 0
Let ψ(z) minimal degree polynomial such that ηψ(A) = 0
Then deg(ψ) ≥ 1
Factorize ψ(z) = φ(z)(z − λ).
Then 0 = ηψ(A) = ηφ(A)(A − λI)
Define ζ := ηφ(A). Note: ζ 6= 0!
Then ζA = λζ . Also ζB = ηφ(A)B = 0. Contradiction!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.8/25
Hautus’ proof of the Hautus test
(⇐)
(A, B) not controllable ⇒
∃ η 6= 0 such that η(B, AB, A2 B, . . . , An−1 B) = 0
Let ψ(z) minimal degree polynomial such that ηψ(A) = 0
Then deg(ψ) ≥ 1
Factorize ψ(z) = φ(z)(z − λ).
Then 0 = ηψ(A) = ηφ(A)(A − λI)
Define ζ := ηφ(A). Note: ζ 6= 0!
Then ζA = λζ . Also ζB = ηφ(A)B = 0. Contradiction!
nice proof!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.8/25
The original paper
Motivated by this nice little proof I decided to take a closer look at
the original paper
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.9/25
The original paper
Motivated by this nice little proof I decided to take a closer look at
the original paper
I found several other interesting results with equally nice proofs!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.9/25
The original paper
Motivated by this nice little proof I decided to take a closer look at
the original paper
I found several other interesting results with equally nice proofs!
I will discuss one of these because it will be useful in the pole
placement problem later on.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.9/25
Reduction of the number of inputs
Again we look at the system
ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , the state space, u(t) ∈ Rm , control input,
A ∈ Mn×n (R), B ∈ Mn×m (R). Note: m is the number of
inputs.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.10/25
Reduction of the number of inputs
Again we look at the system
ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , the state space, u(t) ∈ Rm , control input,
A ∈ Mn×n (R), B ∈ Mn×m (R). Note: m is the number of
inputs.
Question: can we reduce the number of inputs so that the system
remains controllable?
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.10/25
Reduction of the number of inputs
Again we look at the system
ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , the state space, u(t) ∈ Rm , control input,
A ∈ Mn×n (R), B ∈ Mn×m (R). Note: m is the number of
inputs.
Question: can we reduce the number of inputs so that the system
remains controllable?
Does there exists C ∈ Mm×m0 (R) with m0 < m such that the
system
ẋ(t) = Ax(t) + BCv(t)
with v(t) ∈
0
m
R is still controllable?
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.10/25
Reduction of the number of inputs
Let λ be an eigenvalue of A
ω(λ) := the maximal number of eigenvectors with eigenvalue λ
ω(λ) = n − rank(A − λI)
ω(A) := maxλ ω(λ)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.11/25
Reduction of the number of inputs
Let λ be an eigenvalue of A
ω(λ) := the maximal number of eigenvectors with eigenvalue λ
ω(λ) = n − rank(A − λI)
ω(A) := maxλ ω(λ)
Theorem: If (A, B) is controllable then there exists a m × ω(A)
matrix C such that (A, BC) is controllable.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.11/25
Reduction of the number of inputs
Let λ be an eigenvalue of A
ω(λ) := the maximal number of eigenvectors with eigenvalue λ
ω(λ) = n − rank(A − λI)
ω(A) := maxλ ω(λ)
Theorem: If (A, B) is controllable then there exists a m × ω(A)
matrix C such that (A, BC) is controllable.
The number of inputs can be reduced to ω(A)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.11/25
Reduction of the number of inputs
Let λ be an eigenvalue of A
ω(λ) := the maximal number of eigenvectors with eigenvalue λ
ω(λ) = n − rank(A − λI)
ω(A) := maxλ ω(λ)
Theorem: If (A, B) is controllable then there exists a m × ω(A)
matrix C such that (A, BC) is controllable.
The number of inputs can be reduced to ω(A)
Example: if A has n distinct eigenvalues then ω(A) = 1.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.11/25
Reduction of the number of inputs
How to find such m × ω(A) matrix C ? Use the Hautus test!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.12/25
Reduction of the number of inputs
How to find such m × ω(A) matrix C ? Use the Hautus test!
Let λ1 , λ2 , . . . , λν be the distinct eigenvalues of A.
rank(A − λk I, B) = n
rank(A − λk I) = n − ω(λk )
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.12/25
Reduction of the number of inputs
How to find such m × ω(A) matrix C ? Use the Hautus test!
Let λ1 , λ2 , . . . , λν be the distinct eigenvalues of A.
rank(A − λk I, B) = n
rank(A − λk I) = n − ω(λk )
There must be ω(λk ) columns of B that together with n − ω(λk )
columns of A − λk I form a basis of Rn .
Hence: ∃ a m × ω(A) matrix Ck (only 1’s and 0’s) such that
rank(A − λk I, BCk ) = n
(k = 1, 2 . . . ν)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.12/25
Reduction of the number of inputs
How to find such m × ω(A) matrix C ? Use the Hautus test!
Let λ1 , λ2 , . . . , λν be the distinct eigenvalues of A.
rank(A − λk I, B) = n
rank(A − λk I) = n − ω(λk )
There must be ω(λk ) columns of B that together with n − ω(λk )
columns of A − λk I form a basis of Rn .
Hence: ∃ a m × ω(A) matrix Ck (only 1’s and 0’s) such that
rank(A − λk I, BCk ) = n
(k = 1, 2 . . . ν)
Parameter vector α = (α1 , α2 , . . . , αν )
Define C(α) = α1 C1 + α2 C2 + . . . + αν Cν
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.12/25
Reduction of the number of inputs
How to find such m × ω(A) matrix C ? Use the Hautus test!
Let λ1 , λ2 , . . . , λν be the distinct eigenvalues of A.
rank(A − λk I, B) = n
rank(A − λk I) = n − ω(λk )
There must be ω(λk ) columns of B that together with n − ω(λk )
columns of A − λk I form a basis of Rn .
Hence: ∃ a m × ω(A) matrix Ck (only 1’s and 0’s) such that
rank(A − λk I, BCk ) = n
(k = 1, 2 . . . ν)
Parameter vector α = (α1 , α2 , . . . , αν )
Define C(α) = α1 C1 + α2 C2 + . . . + αν Cν
The set of α’s for which (A, BC(α)) is not controllable is the
union of n proper algebraic varieties.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.12/25
Hautus and the pole placement problem
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.13/25
The stabilization problem
Linear system: ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , u(t) ∈ Rm , A ∈ Mn×n (R), B ∈ Mn×m (R)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.14/25
The stabilization problem
Linear system: ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , u(t) ∈ Rm , A ∈ Mn×n (R), B ∈ Mn×m (R)
State feedback u(t) = F x(t), F ∈ Mm×n (R)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.14/25
The stabilization problem
Linear system: ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , u(t) ∈ Rm , A ∈ Mn×n (R), B ∈ Mn×m (R)
State feedback u(t) = F x(t), F ∈ Mm×n (R)
Controlled system ẋ(t) = (A + BF )x(t)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.14/25
The stabilization problem
Linear system: ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , u(t) ∈ Rm , A ∈ Mn×n (R), B ∈ Mn×m (R)
State feedback u(t) = F x(t), F ∈ Mm×n (R)
Controlled system ẋ(t) = (A + BF )x(t)
Stabilization problem: find F such that every eigenvalue λ of
A + BF satisfies Re(λ) < 0
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.14/25
The stabilization problem
Linear system: ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , u(t) ∈ Rm , A ∈ Mn×n (R), B ∈ Mn×m (R)
State feedback u(t) = F x(t), F ∈ Mm×n (R)
Controlled system ẋ(t) = (A + BF )x(t)
Stabilization problem: find F such that every eigenvalue λ of
A + BF satisfies Re(λ) < 0
If such F exists then (A, B) is called stabilizable
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.14/25
The stabilization problem
Linear system: ẋ(t) = Ax(t) + Bu(t)
x(t) ∈ Rn , u(t) ∈ Rm , A ∈ Mn×n (R), B ∈ Mn×m (R)
State feedback u(t) = F x(t), F ∈ Mm×n (R)
Controlled system ẋ(t) = (A + BF )x(t)
Stabilization problem: find F such that every eigenvalue λ of
A + BF satisfies Re(λ) < 0
If such F exists then (A, B) is called stabilizable
Theorem (Hautus test for stabilizability): (A, B) is stabilizable if
and only if (A − λI, B) has rank n for every eigenvalue λ of A
with Re(λ) ≥ 0
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.14/25
The pole placement problem
System ẋ(t) = Ax(t) + Bu(t), state feedback u(t) = F x(t)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.15/25
The pole placement problem
System ẋ(t) = Ax(t) + Bu(t), state feedback u(t) = F x(t)
Eigenvalues of A + BF are the roots of the characteristic
polynomial χA+BF of A + BF .
Question: what are conditions on A and B such that for every
monic, real polynomial p of degree n there exists F ∈ Mm×n (R)
such that χA+BF = p?
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.15/25
The pole placement problem
System ẋ(t) = Ax(t) + Bu(t), state feedback u(t) = F x(t)
Eigenvalues of A + BF are the roots of the characteristic
polynomial χA+BF of A + BF .
Question: what are conditions on A and B such that for every
monic, real polynomial p of degree n there exists F ∈ Mm×n (R)
such that χA+BF = p?
Famous result by W.M. Wonham (1967): the pole placement
theorem (‘On pole assignment in multi-input controllable linear
systems’, IEEE Transactions on Automatic Control, 1967.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.15/25
The pole placement problem
System ẋ(t) = Ax(t) + Bu(t), state feedback u(t) = F x(t)
Eigenvalues of A + BF are the roots of the characteristic
polynomial χA+BF of A + BF .
Question: what are conditions on A and B such that for every
monic, real polynomial p of degree n there exists F ∈ Mm×n (R)
such that χA+BF = p?
Famous result by W.M. Wonham (1967): the pole placement
theorem (‘On pole assignment in multi-input controllable linear
systems’, IEEE Transactions on Automatic Control, 1967.
Theorem: For every polynomial
p(z) = z n + pn−1 z n−1 + . . . + p1 z + p0 with real
coefficients there exists F ∈ Mm×n (R) such that χA+BF = p
if and only if (A, B) is controllable.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.15/25
Proving the pole placement theorem
In this audience, many people teach or have tought this to their
students. The most common proof is as follows:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.16/25
Proving the pole placement theorem
In this audience, many people teach or have tought this to their
students. The most common proof is as follows:
(⇒)
(A, B) not controllable ⇒
∃ λ and η 6= 0 such that ηA = λη and ηB = 0 ⇒
For every F we have η(A + BF ) = λη ⇒
For every F , λ is an eigenvalue of A + BF ⇒
there exists a polynomial p with χA+BF 6= p for all F (take any
polynomial such that p(λ) 6= 0)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.16/25
Proving the pole placement theorem
(⇐)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.17/25
Proving the pole placement theorem
(⇐)
For the converse implication we need to construct for every monic,
real polynomial p of degree n a F ∈ Mm×n (R) such that
χA+BF = p
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.17/25
Proving the pole placement theorem
(⇐)
For the converse implication we need to construct for every monic,
real polynomial p of degree n a F ∈ Mm×n (R) such that
χA+BF = p
This is usually done using Heymann’s lemma, (M. Heymann,
‘Comments to: Pole assignment in multi-input controllable linear
systems’, IEEE Trans. Aut. Control, 1968) (followed by a comment
of W.M. Wonham)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.17/25
Proving the pole placement theorem
(⇐)
For the converse implication we need to construct for every monic,
real polynomial p of degree n a F ∈ Mm×n (R) such that
χA+BF = p
This is usually done using Heymann’s lemma, (M. Heymann,
‘Comments to: Pole assignment in multi-input controllable linear
systems’, IEEE Trans. Aut. Control, 1968) (followed by a comment
of W.M. Wonham)
Heymann’s
lemma:
Let
(A, B) be controllable and let
b1 , b2 , . . . bm be the column vectors of B .
Then for any
i = 1, 2, . . . , m there exists Fi ∈ Mm×n (R) such that
(A + BFi , bi ) is controllable
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.17/25
Proving the pole placement theorem
Heymann’s lemma reduces the problem to the controllable case
with one input. This is an easy problem:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.18/25
Proving the pole placement theorem
Heymann’s lemma reduces the problem to the controllable case
with one input. This is an easy problem:
With respect to a suitable basis, A + BFi is in ‘companion form’
and bi = (0, 0, . . . , 1)T .
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.18/25
Proving the pole placement theorem
Heymann’s lemma reduces the problem to the controllable case
with one input. This is an easy problem:
With respect to a suitable basis, A + BFi is in ‘companion form’
and bi = (0, 0, . . . , 1)T .
Less known: in the paper by Malo Hautus, ‘Stabilization,
controllablity and observability of linear autonomous systems’,
Proceedings Koninklijke Nederlandse Academie voor
Wetenschappen, series A, 1970, there is a completely different
proof, that does not use Heymann’s lemma!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.18/25
Proving the pole placement theorem
Heymann’s lemma reduces the problem to the controllable case
with one input. This is an easy problem:
With respect to a suitable basis, A + BFi is in ‘companion form’
and bi = (0, 0, . . . , 1)T .
Less known: in the paper by Malo Hautus, ‘Stabilization,
controllablity and observability of linear autonomous systems’,
Proceedings Koninklijke Nederlandse Academie voor
Wetenschappen, series A, 1970, there is a completely different
proof, that does not use Heymann’s lemma!
I would like to take a look at that proof here: history of Systems
Theory!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.18/25
A theorem of R. Rado
Hautus’s proof uses a combinatorial theorem by R. Rado, ‘A
theorem on independence relations’, Quart. J. Math. (Oxford), 1942.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.19/25
A theorem of R. Rado
Hautus’s proof uses a combinatorial theorem by R. Rado, ‘A
theorem on independence relations’, Quart. J. Math. (Oxford), 1942.
Theorem: If Σ = {S1 , S2 , . . . , Sn } is a collection of subsets of a
vector space V such that for k = 1, 2, . . . , n the union of each
k-tuple of sets in Σ contains at least k independent vectors, then
there exists a set of independent vectors {x1 , x2 , . . . , xn } in V
such that xk ∈ Sk (k = 1, 2 . . . , n).
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.19/25
A theorem of R. Rado
Hautus’s proof uses a combinatorial theorem by R. Rado, ‘A
theorem on independence relations’, Quart. J. Math. (Oxford), 1942.
Theorem: If Σ = {S1 , S2 , . . . , Sn } is a collection of subsets of a
vector space V such that for k = 1, 2, . . . , n the union of each
k-tuple of sets in Σ contains at least k independent vectors, then
there exists a set of independent vectors {x1 , x2 , . . . , xn } in V
such that xk ∈ Sk (k = 1, 2 . . . , n).
Corollary (finite dimensional case): If {M1 , M2 , . . . , Mn } is a
collection of matrices with n rows such that
rank(Mi1 , Mi2 , . . . , Mik ) ≥ k for each choice of mutually
distinct i1 , i2 , . . . , ik , then there exists a basis {x1 , x2 , . . . , xn }
of Rn with xi ∈ im(Mi ) (i = 1, 2, . . . , n).
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.19/25
A theorem of R. Rado
Hautus’s proof uses a combinatorial theorem by R. Rado, ‘A
theorem on independence relations’, Quart. J. Math. (Oxford), 1942.
Theorem: If Σ = {S1 , S2 , . . . , Sn } is a collection of subsets of a
vector space V such that for k = 1, 2, . . . , n the union of each
k-tuple of sets in Σ contains at least k independent vectors, then
there exists a set of independent vectors {x1 , x2 , . . . , xn } in V
such that xk ∈ Sk (k = 1, 2 . . . , n).
Corollary (finite dimensional case): If {M1 , M2 , . . . , Mn } is a
collection of matrices with n rows such that
rank(Mi1 , Mi2 , . . . , Mik ) ≥ k for each choice of mutually
distinct i1 , i2 , . . . , ik , then there exists a basis {x1 , x2 , . . . , xn }
of Rn with xi ∈ im(Mi ) (i = 1, 2, . . . , n).
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.19/25
Hautus’ proof of the pole placement theorem
Lemma (Hautus): Let (A, B) be controllable, and let
µ1 , µ2 , . . . , µk be k distinct numbers (1 ≤ k ≤ n) such that
none of the µi is an eigenvalue of A. Define Mi ∈ Mn×m (R) by
Mi = (A − µi I)−1 B
(i = 1, 2 . . . k)
Then rank(M1 , M2 , . . . , Mk ) ≥ k.
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.20/25
Hautus’ proof of the pole placement theorem
Lemma (Hautus): Let (A, B) be controllable, and let
µ1 , µ2 , . . . , µk be k distinct numbers (1 ≤ k ≤ n) such that
none of the µi is an eigenvalue of A. Define Mi ∈ Mn×m (R) by
Mi = (A − µi I)−1 B
(i = 1, 2 . . . k)
Then rank(M1 , M2 , . . . , Mk ) ≥ k.
Proof: The proof uses the fact that if (A, B) is controllable then for
each k with 1 ≤ k ≤ n:
rank(B, AB, . . . , Ak−1 B) ≥ k
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.20/25
Hautus’ proof of the pole placement theorem
Lemma (Hautus): Let (A, B) be controllable, and let
µ1 , µ2 , . . . , µk be k distinct numbers (1 ≤ k ≤ n) such that
none of the µi is an eigenvalue of A. Define Mi ∈ Mn×m (R) by
Mi = (A − µi I)−1 B
(i = 1, 2 . . . k)
Then rank(M1 , M2 , . . . , Mk ) ≥ k.
Proof: The proof uses the fact that if (A, B) is controllable then for
each k with 1 ≤ k ≤ n:
rank(B, AB, . . . , Ak−1 B) ≥ k
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.20/25
Hautus’ proof of the pole placement theorem
Lemma (Hautus): Let (A, B) be controllable, and let
µ1 , µ2 , . . . , µk be k distinct numbers (1 ≤ k ≤ n) such that
none of the µi is an eigenvalue of A. Define Mi ∈ Mn×m (R) by
Mi = (A − µi I)−1 B
(i = 1, 2 . . . k)
Then rank(M1 , M2 , . . . , Mk ) ≥ k.
Proof: The proof uses the fact that if (A, B) is controllable then for
each k with 1 ≤ k ≤ n:
rank(B, AB, . . . , Ak−1 B) ≥ k
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.20/25
Hautus’ proof of the pole placement theorem
Now let λ1 , λ2 , . . . , λn be n distinct real numbers such that none
of the λi is an eigenvalue of A
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.21/25
Hautus’ proof of the pole placement theorem
Now let λ1 , λ2 , . . . , λn be n distinct real numbers such that none
of the λi is an eigenvalue of A
The collection of matrices {M1 , M2 , . . . , Mn } with
Mk = (A − λk I)−1 B satisfies the corollary of Rado’s theorem
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.21/25
Hautus’ proof of the pole placement theorem
Now let λ1 , λ2 , . . . , λn be n distinct real numbers such that none
of the λi is an eigenvalue of A
The collection of matrices {M1 , M2 , . . . , Mn } with
Mk = (A − λk I)−1 B satisfies the corollary of Rado’s theorem
Hence there is a basis {x1 , x2 , . . . , xn } of Rn such that
xi ∈ im(Mi ), i.e. xi = (A − λi I)−1 Bui for some ui
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.21/25
Hautus’ proof of the pole placement theorem
Now let λ1 , λ2 , . . . , λn be n distinct real numbers such that none
of the λi is an eigenvalue of A
The collection of matrices {M1 , M2 , . . . , Mn } with
Mk = (A − λk I)−1 B satisfies the corollary of Rado’s theorem
Hence there is a basis {x1 , x2 , . . . , xn } of Rn such that
xi ∈ im(Mi ), i.e. xi = (A − λi I)−1 Bui for some ui
Define F0 : Rn → Rm by: F0 xi := −ui (i = 1, 2, . . . , n)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.21/25
Hautus’ proof of the pole placement theorem
Now let λ1 , λ2 , . . . , λn be n distinct real numbers such that none
of the λi is an eigenvalue of A
The collection of matrices {M1 , M2 , . . . , Mn } with
Mk = (A − λk I)−1 B satisfies the corollary of Rado’s theorem
Hence there is a basis {x1 , x2 , . . . , xn } of Rn such that
xi ∈ im(Mi ), i.e. xi = (A − λi I)−1 Bui for some ui
Define F0 : Rn → Rm by: F0 xi := −ui (i = 1, 2, . . . , n)
Then (A + BF0 )xi = λi xi (i = 1, 2, . . . , n) so
λ1 , λ2 , . . . , λn are the eigenvalues of A + BF0 (distinct
eigenvalues!)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.21/25
Hautus’ proof of the pole placement theorem
Almost there...... We have now proven:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.22/25
Hautus’ proof of the pole placement theorem
Almost there...... We have now proven:
Lemma: Let (A, B) be controllable. Let λ1 , λ2 , . . . , λn be
distinct real numbers such that none of the λi is an eigenvalue of
A. Then there exists F0 ∈ Mm×n (R) such that λ1 , λ2 , . . . , λn
are the eigenvalues of A + BF0 .
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.22/25
Hautus’ proof of the pole placement theorem
Almost there...... We have now proven:
Lemma: Let (A, B) be controllable. Let λ1 , λ2 , . . . , λn be
distinct real numbers such that none of the λi is an eigenvalue of
A. Then there exists F0 ∈ Mm×n (R) such that λ1 , λ2 , . . . , λn
are the eigenvalues of A + BF0 .
Note: already very close to pole placement: every polynomial p of
the form p(z) = (z − λ1 )(z − λ2 ) · · · (z − λn ) with λi 6= λj
real and λi not an eigenvalue of A can be assigned
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.22/25
Hautus’ proof of the pole placement theorem
Almost there...... We have now proven:
Lemma: Let (A, B) be controllable. Let λ1 , λ2 , . . . , λn be
distinct real numbers such that none of the λi is an eigenvalue of
A. Then there exists F0 ∈ Mm×n (R) such that λ1 , λ2 , . . . , λn
are the eigenvalues of A + BF0 .
Note: already very close to pole placement: every polynomial p of
the form p(z) = (z − λ1 )(z − λ2 ) · · · (z − λn ) with λi 6= λj
real and λi not an eigenvalue of A can be assigned
And now? The coupe de grâce:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.22/25
Hautus’ proof of the pole placement theorem
Almost there...... We have now proven:
Lemma: Let (A, B) be controllable. Let λ1 , λ2 , . . . , λn be
distinct real numbers such that none of the λi is an eigenvalue of
A. Then there exists F0 ∈ Mm×n (R) such that λ1 , λ2 , . . . , λn
are the eigenvalues of A + BF0 .
Note: already very close to pole placement: every polynomial p of
the form p(z) = (z − λ1 )(z − λ2 ) · · · (z − λn ) with λi 6= λj
real and λi not an eigenvalue of A can be assigned
And now? The coupe de grâce:
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.22/25
Coupe de grâce
Recall Hautus’s result from 1969: if (A, B) is controllable then
there exists a m × ω(A) matrix C such that (A, BC) is
controllable
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.23/25
Coupe de grâce
Recall Hautus’s result from 1969: if (A, B) is controllable then
there exists a m × ω(A) matrix C such that (A, BC) is
controllable
Now, start with a controllable pair (A, B)
Let F0 be such that the eigenvalues of A + BF0 are distinct
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.23/25
Coupe de grâce
Recall Hautus’s result from 1969: if (A, B) is controllable then
there exists a m × ω(A) matrix C such that (A, BC) is
controllable
Now, start with a controllable pair (A, B)
Let F0 be such that the eigenvalues of A + BF0 are distinct
Note that therefore ω(A + BF0 ) = 1
Also (A + BF0 , B) is still controllable
Hence there exists a m × 1 matrix c such that (A + BF0 , Bc) is
controllable: back in the controllable case with one input!
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.23/25
Coupe de grâce
Recall Hautus’s result from 1969: if (A, B) is controllable then
there exists a m × ω(A) matrix C such that (A, BC) is
controllable
Now, start with a controllable pair (A, B)
Let F0 be such that the eigenvalues of A + BF0 are distinct
Note that therefore ω(A + BF0 ) = 1
Also (A + BF0 , B) is still controllable
Hence there exists a m × 1 matrix c such that (A + BF0 , Bc) is
controllable: back in the controllable case with one input!
The remainder of the proof is unchanged (companion form with one
input)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.23/25
Coupe de grâce
Recall Hautus’s result from 1969: if (A, B) is controllable then
there exists a m × ω(A) matrix C such that (A, BC) is
controllable
Now, start with a controllable pair (A, B)
Let F0 be such that the eigenvalues of A + BF0 are distinct
Note that therefore ω(A + BF0 ) = 1
Also (A + BF0 , B) is still controllable
Hence there exists a m × 1 matrix c such that (A + BF0 , Bc) is
controllable: back in the controllable case with one input!
The remainder of the proof is unchanged (companion form with one
input)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.23/25
Hautus’ proof of Heymann’s lemma
In most texts on linear systems a proof using Heymann’s lemma is
preferred..... (including the famous book ‘Linear Multivariable
Control, a Geometric Approach’ by W.M. Wonham)
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.24/25
Hautus’ proof of Heymann’s lemma
In most texts on linear systems a proof using Heymann’s lemma is
preferred..... (including the famous book ‘Linear Multivariable
Control, a Geometric Approach’ by W.M. Wonham)
Apparently, Malo Hautus himself also prefers this proof: in the
well-known paper ‘A simple proof of Heymann’s lemma’, IEEE
Transactions on Automatic Control’, 1977, he presents a simple
proof of Heymann’s lemma, self contained, only 21 lines
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.24/25
Hautus’ proof of Heymann’s lemma
In most texts on linear systems a proof using Heymann’s lemma is
preferred..... (including the famous book ‘Linear Multivariable
Control, a Geometric Approach’ by W.M. Wonham)
Apparently, Malo Hautus himself also prefers this proof: in the
well-known paper ‘A simple proof of Heymann’s lemma’, IEEE
Transactions on Automatic Control’, 1977, he presents a simple
proof of Heymann’s lemma, self contained, only 21 lines
More evidence for this is the book ‘Control Theory for Linear
Systems’ by H.L. Trentelman, A.A. Stoorvogel and M.L.J. Hautus
unfortunately sold out....
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.24/25
Sense and simplicity in Hautus’ work
I hope to have shared with you my enthusiasm about the work of
Malo Hautus....
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.25/25
Sense and simplicity in Hautus’ work
I hope to have shared with you my enthusiasm about the work of
Malo Hautus....
I could continue with Malo’s beautiful work on the regulator
problem, for which he developed new theory on the universal and
individual solvability of linear matrix equations......
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.25/25
Sense and simplicity in Hautus’ work
I hope to have shared with you my enthusiasm about the work of
Malo Hautus....
I could continue with Malo’s beautiful work on the regulator
problem, for which he developed new theory on the universal and
individual solvability of linear matrix equations......
or his work within the ‘Geometrix Approach’ to linear systems and
the idea of (ξ, ω) representations......
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.25/25
Sense and simplicity in Hautus’ work
I hope to have shared with you my enthusiasm about the work of
Malo Hautus....
I could continue with Malo’s beautiful work on the regulator
problem, for which he developed new theory on the universal and
individual solvability of linear matrix equations......
or his work within the ‘Geometrix Approach’ to linear systems and
the idea of (ξ, ω) representations......
or his work on the Linear Quadratic Control problem.......
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.25/25
Sense and simplicity in Hautus’ work
I hope to have shared with you my enthusiasm about the work of
Malo Hautus....
I could continue with Malo’s beautiful work on the regulator
problem, for which he developed new theory on the universal and
individual solvability of linear matrix equations......
or his work within the ‘Geometrix Approach’ to linear systems and
the idea of (ξ, ω) representations......
or his work on the Linear Quadratic Control problem.......
but I will have to stop here.....
SENSE AND SIMPLICITY: a talk on two matrices on the occasion of Malo Hautus’ afscheidsrede – p.25/25