Dimensional Analysis of Matrices State-Space Models and Dimensionless Units [Lecture Notes]

Abstract

Physical dimensions and units, such as mass (kg), length (m), time (s), and charge (C), provide the link between mathematics and the physical world. It is well known that careful attention to physical dimensions can provide valuable insight into relationships among physical quantities. In this regard, the Buckingham Pi theorem, which is essentially an application of the fundamental theorem of linear algebra on the sum of the rank and defect of a matrix, has been extensively applied.

Full text

Dimensional Analysis of Matrices
State-Space Models and Dimensionless Units
HARISH J. PALANTHANDALAM-MADAPUSI, DENNIS S. BERNSTEIN, and RAVINDER VENUGOPAL
»
LECTURE NOTES
P
hysical dimensions and units, such as mass (kg),
length (m), time (s), and charge (C), provide the link
between mathematics and the physical world. It is
well known that careful attention to physical dimensions
can provide valuable insight into relationships among
physical quantities. In this regard, the Buckingham Pi the-
orem (see “The Buckingham Pi Theorem in a Nutshell”),
which is essentially an application of the fundamental the-
orem of linear algebra on the sum of the rank and defect of
a matrix, has been extensively applied [1]–[10]. Interesting
historical remarks on the development of dimensional
analysis are given in [11], while detailed discussions are
given in [12, Chapter 10] and [13].
In the control literature, with its historically strong mathe-
matical influence, it is not unusual to see expressions such as
V(x,
˙
x) = x
2
+
˙
x
2
,
where
x
and
˙
x
denote position and velocity states, respec-
tively. Although this expression appears to be dimension-
ally incorrect, the reader usually assumes that unlabeled
coefficients are present to convert units from squared posi-
tion to squared velocity or vice versa.
A related issue concerns the appearance of dimen-
sionless units. For example, for a stiffness
k
and a mass
m
, the expression

k/m
has the dimensions of recipro-
cal time. However, when used within the context of
harmonic solutions of an oscillator, the same expression
has the interpretation of rad/s, where the dimension-
less unit “rad” is inserted to facilitate the use of
trigonometric functions. Although this insertion is ad
hoc, the recognition that radians are dimensionless pro-
vides reasonable justification.
A publication of special note is the book [6], which takes
an in-depth look at the role of dimensions including matri-
ces populated with dimensioned quantities. Although this
text provides no situations in which the “usual” rules of
dimensional analysis lead to incorrect answers, the careful
reexamination in [6] of the treatment of dimensions, espe-
cially for matrices, motivates the present article.
The main objective of this article is to examine the
dimensional structure of the dynamics matrix
A
that arises
in the linear state-space system
˙
x = Ax
. To do this, we
extend results of [6] and provide a self-contained treat-
ment of the dimensional structure of
A
and its exponential.
Our investigation of the physical dimensions of
A
moti-
vates us to look at the algebraic structure of dimensioned
quantities. This development forces us to define multiple,
distinct, group identity elements, which are the dimension-
less units. One such dimensionless unit is the radian. How-
ever, to complete the analysis, we introduce an additional
dimensionless quantity for each physical dimension and
each product of dimensions.
This approach immediately clarifies the mysterious
appearance of radians in the example above. Specifical-
ly,
[
√
k/m] = ([k]/[m])
1/2
= ((N/m)/kg)
1/2
= []
m
[]
kg
/s
,
where
[a]
denotes the physical dimensions of
a
,
[]
kg
 kg
0
is the identity element in the group of mass
dimensions, and
[]
m
 m
0
is the identity element in the
group of length dimensions. In fact,
[]
m
is the traditional
radian, whose appearance is natural and need not be
inserted with the justification that “radians are dimen-
sionless.” Rather,
[]
m
appears because the mathematical
structure of physical dimensions requires that it be pre-
sent. By the same reasoning, the massian
[]
kg
is also pre-
sent in
[

k/m]
.
As an additional example, consider the expression
ω = v/r,
where
ω
is angular velocity,
v
is translational veloc-
ity, and
r
is radius. Then
[ω] = [v]/[r] = (m/s)/m =
m
0
/s = []
m
/s = rad/s
. Again, there is no need to artificially
insert the dimensionless unit “
rad
” in order to obtain the
angular velocity in the expected units. We also note that, for
an angle
θ
in radians, the fact that
[]
α
m
= (m
0
)
α
= m
0
= []
m
for all real numbers
α
implies that
[sin θ] =

θ −
θ
3
3!
+···

= []
m
,
which is consistent with the fact that both
θ
and
sin θ
are
ratios of lengths.
In real computations involving physical quantities, that
is, aside from pure theory, it is necessary to keep track of
physical dimensions and their associated units. Elucida-
tion of the physical dimension structure of state space
models can thus be useful for verifying the model struc-
ture and ensuring that the units are consistent within the
context of state-space computations.
Digital Object Identifier 10.1109/MCS.2007.906924
100 IEEE CONTROL SYSTEMS MAGAZINE » DECEMBER 2007

ALGEBRAIC STRUCTURE OF UNITS
For simplicity, we consider the fundamental dimensions
mass (
kg
), length (
m
), and time (
s
) only. For convenience,
we use
kg
,
m
, and
s
to represent the respective physical
dimension as well as the associated unit. Let
R
and
C
denote the real and complex numbers, respectively.
Define
G
kg
 {kg
α
: α ∈ R}
,
G
m
 {m
β
: β ∈ R}
, and
G
s
 {s
γ
: γ ∈ R}
. Note that
G
kg
,
G
m
, and
G
s
are Abelian
(commutative) groups (see “What Is a Group?”) with the
identity elements
[]
kg
, []
m
, and
[]
s
, respectively, which
are dimensionless units referred to as the massian, lengthi-
an, and timian. The lengthian
[]
m
in
G
m
, when interpreted
within the context of a circle, is the radian. Next, define
the set
G
of all mixed units
G
 {kg
α
m
β
s
γ
: α, β, γ ∈ R}.(1)
Since, for all
α, β, γ ∈
R
,
kg
α
m
β
s
γ
= kg
α
s
γ
m
β
=
m
β
kg
α
s
γ
= m
β
s
γ
kg
α
= s
γ
m
β
kg
α
= s
γ
kg
α
m
β
, we have the
following result.
Fact 1
G
is an Abelian group with the identity element
[]
kg
[]
m
[]
s
.
The four products of the identity elements are represent-
ed by
[]
kg,m
 []
kg
[]
m
, []
kg,s
 []
kg
[]
s
, []
m,s
 []
m
[]
s
,
and
[]
kg,m,s
 []
kg
[]
m
[]
s
, of which only the last is an ele-
ment of
G
. Note that the dimensionless Reynolds number
in fluid dynamics defined by
Re

v
s
L
ν
,
where
v
s
is the mean fluid velocity,
L
is the characteristic
length of the flow, and
ν
is the kinematic fluid viscosity,
has the units
[Re] = []
kg,m,s
.
Similarly, the dimensionless Froude number in fluid
mechanics defined by
Fr

v
s
Lg
,
where
g
is acceleration due to gravity, has the units
[Fr] = []
m,s
.
Table 1 classifies several dimensionless quantities based on
their units.
The set
D
of dimensioned scalars consists of elements of
the form
akg
α
m
β
s
γ
, where
a ∈
C
and
α, β, γ ∈
R
. We
allow
a ∈ C
to accommodate complex eigenvalues and
eigenvectors. We define the units operator
[]
as
[akg
α
m
β
s
γ
]  kg
α
m
β
s
γ
.
Note that
[0 kg
α
m
β
s
γ
]  kg
α
m
β
s
γ
. Let
a
1
kg
α
1
m
β
1
s
γ
1
and
a
2
kg
α
2
m
β
2
s
γ
2
be dimensioned scalars. Then the product of
two dimensioned scalars always exists and is defined to
be
a
1
kg
α
1
m
β
1
s
γ
1
a
2
kg
α
2
m
β
2
s
γ
2
= a
1
a
2
kg
α
1
+α
2
m
β
1
+β
2
s
γ
1
+γ
2
.
However, the sum
a
1
kg
α
1
m
β
1
s
γ
1
+ a
2
kg
α
2
m
β
2
s
γ
2
is defined
only if
α
1
= α
2
,β
1
= β
2
,
and
γ
1
= γ
2
, in which case
a
1
kg
α
1
m
β
1
s
γ
1
+ a
2
kg
α
2
m
β
2
s
γ
2
= (a
1
+ a
2
)kg
α
1
m
β
1
s
γ
1
. Fur-
thermore, although quantities such as
akg
α
and
bs
γ
are
not elements of
D
, we assume that all operations occur
after these quantities are embedded in the appropriate
group containing all of the common units. For example,
(akg
α
)(bs
γ
)  (akg
α
[]
s
)(bs
γ
[]
kg
) = abkg
α
s
γ
.
Dimensioned vectors and dimensioned matrices are denot-
ed by
D
n
and
D
n×m
, respectively, all of whose entries are
dimensioned scalars (see “Energy Versus Moment” for an
example of the difference between dimensioned scalars
and dimensioned vectors). Let
P ∈ D
n×m
and define
[P]

⎡
⎢
⎣
[P
1,1
] ··· [P
1,m
]
.
.
.
.
.
.
.
.
.
[P
n,1
] ··· [P
n,m
]
⎤
⎥
⎦
∈ G
n×m
,(2)
where
P
i, j
is the
(i, j)
entry of
P
and
G
n×m
denotes the set
of
n × m
matrices with entries in
G
. Note that
[P
T
] = [P]
T
.
If
P ∈ D
n×m
and
Q ∈ D
m×p
, then
PQ
exists if all addition
operations required to form the product are defined.
Fact 2
Let
P ∈ D
n×m
and
Q ∈ D
m×p
. Then
PQ
exists if and only if,
for all
i = 1,...,n
and
j = 1,...,p,
[P
i,1
][Q
1, j
] = [P
i,2
][Q
2, j
] = ··· = [P
i,n
][Q
n, j
].(3)
Furthermore, if
PQ
exists, then
[PQ] = [P][Q].(4)
Fact 3
Let
P ∈ D
n×n
. If
P
2
exists, then
[P
1,1
] = [P
2,2
] = ··· = [P
n,n
].(5)
Proof
Since
P
2
exists, it follows that, for all
i, j = 1,...,n
,
[(P
2
)
i,i
] = [P
i,1
][P
1,i
] = [P
i,2
][P
2,i
] = ··· = [P
i,n
][P
n,i
].
Now, let
i, j ∈{1,...,n}
. Then
[P
i,i
][P
i,i
] = [P
i, j
][P
j,i
] =
[P
j,i
][P
i, j
] = [P
j, j
][P
j, j
]
. Hence
[P
i,i
] = [P
j, j
]. 
Fact 4
Let
P ∈ D
n×n
. If
P
2
exists, then, for all positive integers
k
,
P
k
exists and
[P
k
] = [P]
k
. Furthermore, for all
i = 1,...,n
and for all positive integers
k
,
DECEMBER 2007 « IEEE CONTROL SYSTEMS MAGAZINE 101

[P
k
] = [(P
i,i
)
k−1
][P].(6)
Proof
Since, for all
i, j = 1,...,n,
[(P
2
)
i, j
] = [P
i,1
][P
1, j
] = [P
i,2
][P
2, j
] = ··· = [P
i,n
][P
n, j
],
it follows that
[(P
2
)
i, j
] = [P
i,i
][P
i, j
].
Hence
[P
2
] = [P
i,i
][P]
. Induction yields (6).

Fact 5
Let
P ∈ D
n×n
. Then
P
2
exists if and only if there exist
z
1
, z
2
∈ G
n
such that
z
T
2
z
1
exists and
[P] = z
1
z
T
2
.(7)
Proof
Sufficiency is immediate. To prove necessity, define
z
1

⎡
⎢
⎢
⎢
⎣
[P
1,1
]
[P
2,1
]
.
.
.
[P
n,1
]
⎤
⎥
⎥
⎥
⎦
, z
2

⎡
⎢
⎢
⎢
⎣
[P
1,1
]/[P
1,1
]
[P
1,2
]/[P
1,1
]
.
.
.
[P
1,n
]/[P
1,1
]
⎤
⎥
⎥
⎥
⎦
.
L
et
u
1
, ..., u
p
be fundamental dimensions and let
G

=
{

p
i=1
u
α
i
i
: α
1
, ...,α
p
∈ R
be the corresponding Abelian
group. Then the set
D
of dimensioned scalars consists of ele-
ments of the form
a

p
i=1
u
α
i
i
, where
a ∈ C
and
α
1
, ..., α
p
∈ R
.
The following theorem, called the Buckingham Pi theorem [S1],
shows that a relationship between
q
dimensioned quantities
induces a collection of dimensionless quantities.
THEOREM S1
Let
Q
1
, Q
2
, ..., Q
q
∈ D
be dimensioned scalars such that, for
i = 1, ..., q
,
Q
i

= a
i

p
j=1
u
α
ij
j
, and assume that
K

k=1
c
k
Q
β
1k
1
···Q
β
qk
q
= 0,(S1)
where
c
1
, ..., c
K
∈ R
are nonzero. Let
A

= [α
ij
]
T
∈
R
p×q
, and let
r

= rank A
. Then there exists


= [γ
ij
] ∈
R
q ×(q −r )
such that
rank  = q − r, A  = 0
, and, for
i = 1,...,q − r,

i

= Q
γ
1i
1
···Q
γ
qi
q
(S2)
are dimensionless.
PROOF
It follows from the fundamental theorem of linear algebra [S2, p.
33] that
rank A + def A = q
and thus
def A = q − r,
where
def A
is the dimension of the nullspace of
A
. Next, let


= [γ
ij
] ∈
R
q×(q−r )
be such that the columns of

form a basis for
the nullspace of
A
. Then it follows that
rank  = q − r
and
A  = 0
. Next, since the
(j, i)
entry of
A 
is

q
k=1
α
kj
γ
ki
= 0
, it
follows that, for all
i = 1, ..., q − r,

i
= Q
γ
1i
1
···Q
γ
qi
q
= a
γ
1i
1
···a
γ
qi
q
p

j=1
u

q
k=1
α
kj
γ
ki
j
= a
γ
1i
1
···a
γ
qi
q
p

j=1
u
0
j
is dimensionless.

As an example, consider the law of conservation of momen-
tum in a collision between two rigid bodies given by
m
1
v
−
1
+ m
2
v
−
2
= m
1
v
+
1
+ m
2
v
+
2
,(S3)
where
m
1
and
m
2
are the masses of the bodies,
v
−
1
and
v
−
2
are the
velocities of the bodies before collision, and
v
+
1
and
v
+
2
are the veloci-
ties of the bodies after collision, respectively. Note that
[m
1
] = [m
2
] = kg
and
[v
−
1
] = [v
−
2
] = [v
+
1
] = [v
+
2
] = m/s
. Further-
more, choosing
u
1
= kg
,
u
2
= m
,
u
3
= s
,
Q
1
= m
1
, Q
2
= m
2
,
Q
3
= v
−
1
,
Q
4
= v
−
2
,
Q
5
= v
+
1
, and
Q
6
= v
+
2
, it follows that
p = 3
,
q = 6
,
A =

110000
001111
00−1 −1 −1 −1

,(S4)
and
r = 2
. Therefore, in accordance with Theorem S1, there exist
q − r = 4
dimensionless quantities. These dimensionless quan-
tities can be computed by determining a basis for the null space
of
A
. For example, choosing
 =
⎡
⎢
⎢
⎢
⎢
⎢
⎣
1000
−10 00
0110
0 −100
00−11
0001
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(S5)
yields the dimensionless quantities
The Buckingham Pi Theorem in a Nutshell
102 IEEE CONTROL SYSTEMS MAGAZINE » DECEMBER 2007

Since
P
2
exists it follows that
[(P
2
)
1,1
]/[P
1,1
] = z
T
2
z
1
exists.
Furthermore, let
k ∈{1,...,n}
and define
z
3
∈ G
n
by
z
3

⎡
⎢
⎢
⎢
⎣
[P
1,k
]
[P
2,k
]
.
.
.
[P
n,k
]
⎤
⎥
⎥
⎥
⎦
.
Then,
z
T
2
z
3
exists and thus the rows of
[P]
are dimensioned
scalar multiples of each other. Hence
[P] =
⎡
⎢
⎢
⎢
⎣
[P
1,1
][P
1,2
] ··· [P
1,n
]
[P
2,1
][P
1,2
][P
2,1
]/[P
1,1
] ··· [P
1,n
][P
2,1
]/[P
1,1
]
.
.
.
.
.
.
.
.
.
.
.
.
[P
n,1
][P
1,2
][P
n,1
]/[P
1,1
] ··· [P
1,n
][P
n,1
]/[P
1,1
]
⎤
⎥
⎥
⎥
⎦
= z
1
z
T
2
. 
Fact 6
Let
P ∈ D
n×n
. Then
e
P
∈ D
n×n
exists if and only if
P
2
exists
and
[P] = [P
2
]
. Furthermore, if
e
P
exists then
[e
P
] = [P].(8)

1
=
m
1
m
2
,
2
=
v
−
1
v
−
2
,

3
=
v
−
1
v
+
1
,
4
=
v
+
1
v
+
2
.
Note that these dimensionless quantities are not unique.
An application of the Buckingham Pi Theorem is to derive
physical relationships between dimensioned quantities. For
example, consider the problem of deriving an expression for the
time period of oscillations of a pendulum. We expect the time
period
T
to depend on the length
l
of the pendulum, the acceler-
ation
g
due to gravity, and perhaps the mass
m
of the pendu-
lum. Since
[T ] = s, [l] = m, [g] = m/s
2
,
and
[m] = kg
, we
choose
u
1
= kg, u
2
= m, u
3
= s
,
Q
1
= T, Q
2
= l, Q
3
= g,
and
Q
4
= m
. Noting that
p = 3
,
q = 3
,
A =

00 0 1
01 1 0
10−20

,(S6)
and
r = 3
, it follows that there exists
q − r = 1
dimensionless
quantity given by

1
=
T
√
g
√
l
.(S7)
Therefore,
T = 
1

l
g
,(S8)
where the dimensionless constant

1
can be determined experi-
mentally to be
2π
. Note that the time period does not depend on
the mass of the pendulum, a result due to Galilieo.
As a final example, consider the force generated by a pro-
peller on an aircraft. Presumably, the force
F
depends on the
diameter
d
of the propeller, the velocity
v
of the airplane, the
density
ρ
of the air, the rotational speed
N
of the propeller, and
the dynamic viscosity
ν
of the air. Noting that
[F ] = kgm/s
2
,
[d] = m, [v] = m/s, [ρ] = kg/m
3
, [N] = []
m
/s, [ν] = m
2
/s
, we
choose
u
1
= kg, u
2
= m, u
3
= s, Q
1
= F, Q
2
= d, Q
3
= v
,
Q
4
= ρ, Q
5
= N,
and
Q
6
= ν
. Therefore, we have
p = 3, q = 6,
A =

100100
111−30 2
−20−10−1 −1

,(S9)
and
r = 3
. Thus we have
q − r = 6 − 3 = 3
dimensionless quan-
tities. Choosing

to be
 =
⎡
⎢
⎢
⎢
⎢
⎢
⎣
001
11−2
1 −1 −2
00−1
010
−10 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
,(S10)
it follows that

1
=
dv
ν
,
2
=
dN
v
,
3
=
F
d
2
v
2
ρ
,
where

1
is the Reynolds number,

2
is the top-speed ratio, and

3
is the dynamic-force ratio.
REFERENCES
[S1] E. Buckingham, “On physically similar systems: Illustration of
the use of dimensional equations,” Phys. Rev., vol. 4, no. 4, pp.
345–376, 1914.
[S2] D. S. Bernstein, Matrix Mathematics: Theory, Facts, and For-
mulas with Application to Linear Systems Theory. Princeton, NJ:
Princeton University Press, 2005.
DECEMBER 2007 « IEEE CONTROL SYSTEMS MAGAZINE 103

Proof
By definition, the matrix exponential
e
P
∈ D
n×n
is given by
e
P
= I +
1
1!
P +
1
2!
P
2
+··· .(9)
Necessity is immediate. To prove sufficiency, note that,
since
P
2
exists and
[P] = [P
2
]
, it follows from Fact 4 that
[P] = [P
2
]
implies that
[P] = [P
k
]
for all positive integers
k
.
Thus
e
P
exists. Next, it follows from (9) that (8) holds.

Fact 7
Let
P ∈ D
n×n
and assume that
e
P
exists. Then, for all
i = 1,...,n,
[P
i,i
] = []
kg,m,s
.(10)
Proof
The result follows immediately from facts 6 and 4.

For a real scalar
q
and
P ∈ D
n×m
, the Schur power
P
{q}
∈ D
n×m
is defined by
(P
{q}
)
i, j
 (P
i, j
)
q
,(11)
assuming the right hand side exists. The notation
[P]
C
∈ C
n×m
denotes the numerical part of the dimen-
sioned matrix
P ∈ D
n×m
. Note that
P = [P]
C
◦ [P],(12)
where
◦
is the Schur (entry-wise) product. We write
[P]
C
as
[P]
R
if
[P]
C
is real. Let
I
R
denote the identity matrix in
R
n×n
. Furthermore, let
Q ∈ D
m×p
and assume that
PQ
exists. Then
[PQ]
C
= [P]
C
[Q]
C
and
PQ = ([P]
C
◦ [P])([Q]
C
◦ [Q])
= ([P]
C
[Q]
C
) ◦ ([P][Q]) = [PQ]
C
◦ [PQ]. (13)
Fact 8
Let
P ∈ D
n×m
, and let
y ∈ D
n
and
u ∈ D
m
be such that
y = Pu.(14)
Then
[P] = [y][u
T
]
{−1}
.(15)
Proof
The
i
th component equation of (15) is
[P
i,1
][u
1
] + [P
i,2
][u
2
] +···+[P
i,m
][u
m
] = [y
i
].
Therefore,
[P
i,1
][u
1
] = [P
i,2
][u
2
] = ··· = [P
i,m
][u
m
] = [y
i
],
and thus
[P
i, j
] = [y
i
]/[u
j
]
. Hence (15) holds.

A
group
(G, ∗)
is a set
G
with a binary operation
∗ : G × G → G
that satisfies the following axioms:
A1) For all
a, b ∈ G
,
a ∗ b ∈ G
.
A2) For all
a, b, c ∈ G
,
(a ∗ b) ∗ c = a ∗ (b ∗ c)
.
A3) There exists an identity element
e ∈ G
such that, for
all
a ∈ G
,
e ∗ a = a ∗ e = a
.
A4) For all
a ∈ G
, there exists
b ∈ G
such that
a ∗ b = b ∗ a = e
, where
e
is the identity element in
G
.
Note that, A1–A4 do not imply that, for all
a
, b ∈ G
,
a ∗ b = b ∗ a
. However, if, for all
a, b ∈ G
,
a ∗ b = b ∗ a
,
then the group
G
is an Abelian group.
The set of real numbers with
e = 0
and the binary oper-
ation of addition is a group. However, the set of real num-
bers with
e = 1
and the binary operation of multiplication is
not a group since A4 is not satisfied for
a = 0
. Furthermore,
since addition is commutative, the set of real numbers with
the addition operation is an Abelian group.
The set of units
G ={kg
α
m
β
s
γ
: α, β, γ ∈ R}
with
e = []
kg,m,s
and the binary operation of multiplication is an
Abelian group.
What Is a Group?
TABLE 1 Classification of dimensionless units and examples.
These seven dimensionless units are defined in terms of
ratios of the basic physical dimensions.
Dimensionless Unit Name Examples
[]
kg
Massian Air-fuel ratio
Stoichiometric mass ratio
[]
m
Lengthian Radian
Strain
Poisson's ratio
Fresnel number
Aspect ratio
[]
s
Timian Courant-Friedrichs-Lewy
(CFL) number
Damkohler numbers
[]
kg,m
Densian Density ratio
Moment-of-inertia ratio
[]
kg,s
Flowian Mass-flow ratio
Stiffness ratio
[]
m,s
Velocian Froude number
Fourier number
Mach number
Stokes number
[]
kg,m,s
Forcian Reynolds number
Weber number
Coefficient of friction
Lift coefficient
Drag coefficient
104 IEEE CONTROL SYSTEMS MAGAZINE » DECEMBER 2007

Next, let
P ∈ D
n×n
. Then, the determinant det
P
of
P
is
defined to be
det P =

p∈ P
n
σ(p)P
1,p
1
P
2,p
2
···P
n,p
n
,(16)
where
P
n
is the set of all permutations
p = (p
1
,...,p
n
)
of
(1, 2, ..., n)
, and
σ(p)
is the signature of the permutation
p
, which is 1 if
p
is achieved by applying an even number
of transpositions to
(1, 2, ..., n)
and
−1
if
p
is reached by
applying an odd number of transpositions to
(1, 2, ..., n)
.
Note that if
P ∈ D
n×n
then det
P
exists if and only if
[P
1,p
1
P
2,p
2
···P
n,p
n
]
is the same for all
p ∈ P
n
. Hence, if det
P
exists, we have
[det P] = [P
1,p
1
P
2,p
2
···P
n,p
n
] (17)
for all
p ∈ P
n
. Note that
det [P]
C
= [det P]
C
(18)
and
det P = (det [P]
C
)[det P].(19)
The following result presents necessary and sufficient con-
ditions for the existence of det
P
.
Fact 9
Let
P ∈ D
n×n
. Then det
P
exists if and only if there exist
z
1
, z
2
∈ G
n
such that
[P] = z
1
z
T
2
.(20)
Proof
Sufficiency is immediate. To prove necessity, first let
n = 2
.
Then, since det
P
exists, it follows that
[P
1,1
]
[P
1,2
]
=
[P
2,1
]
[P
2,2
]
.(21)
Thus the columns of
[P]
are dimensioned scalar multiples of
each other. Next, let
n = 3
and assume that det
P
exists.
Then it follows from the cofactor expansion of det
P
that the
determinant of every
2 × 2
submatrix of
P
exists. Hence (21)
holds. Next, it follows that
[P
1,1
P
2,3
P
3,2
] = [P
1,2
P
2,3
P
3,1
]
and hence
[P
1,1
]
[P
1,2
]
=
[P
3,1
]
[P
3,2
]
.(22)
Furthermore, using
[P
1,2
P
2,3
P
3,1
] = [P
1,3
P
2,2
P
3,1
]
and
[P
1,2
P
2,1
P
3,2
] = [P
1,3
P
2,1
P
3,2
]
, it follows that
[P
1,2
]/[P
1,3
] =
[P
2,2
]/[P
2,3
]
and
[P
1,2
]/[P
1,3
] = [P
3,2
]/[P
3,3
]
. Thus the
columns of
[P]
are dimensioned scalar multiples of each
other. Likewise, for all
n ≥ 1
, it can be seen that, since det
P
exists, the columns of
[P]
are dimensioned scalar multiples
of each other. Thus, defining
z
1

⎡
⎢
⎢
⎢
⎣
[P
1,1
]
[P
2,1
]
.
.
.
[P
n,1
]
⎤
⎥
⎥
⎥
⎦
, z
2

⎡
⎢
⎢
⎢
⎣
[P
1,1
]/[P
1,1
]
[P
1,2
]/[P
1,1
]
.
.
.
[P
1,n
]/[P
1,1
]
⎤
⎥
⎥
⎥
⎦
,
it follows that (20) holds.

Note that if
P
2
exists then det
P
exists. However, the
following example shows that the converse does not hold.
Example 1
Let
P ∈ D
2×2
be such that
[P] =

mm
2
sms

.(23)
Then det
P
exists, but
P
2
does not exist.
Let
P ∈ D
n×n
. Then
λ ∈ D
and
v ∈ D
n
are an eigenvalue-
eigenvector pair of
P
if
[v]
C
is not zero and
λ
and
v
satisfy
Pv = λv.(24)
Fact 10
Let
P ∈ D
n×n
. Then
P
has an eigenvalue-eigenvector pair
λ ∈ D, v ∈ D
n
if and only if det
P
exists and, for all
i = 1,...,n
and
j = 1,...,n,
[P
i,i
] = [P
j, j
].(25)
S
ince energy is force times displacement, it follows that the
units of energy are
J = Nm = kgm
2
/s
2
. On the other hand,
since moment times angular displacement is energy, it follows
that the units of moment are
J/rad = Nm/rad = kgm
2
/s
2
rad
.
Furthermore, since
rad = []
m
= m
0
, it follows that
J/rad = kgm
2
/s
2
rad = kgm
2
/s
2
= J
, and hence moment has
the same units as energy.
Although the above analysis suggests that energy and
moment are indistinguishable, we know intuitively that they
are different. This apparent contradiction is resolved by the
fact that energy is a dimensioned scalar in
D
, while moment
is a dimensioned vector in
D
3
. In fact, the work done by a
moment through an angle is the dot product of the moment
and a dimensionless angle vector, which is a dimensionless
vector perpendicular to the plane containing the angle. The
direction of the angle vector is determined by the right-hand
rule, and its dimensionless magnitude is given by the radian
measure of the angle.
Energy Versus Moment
DECEMBER 2007 « IEEE CONTROL SYSTEMS MAGAZINE 105

In this case,
[P] = [λv][v
T
]
{−1}
(26)
and, for all
i = 1,...,n,
[P
i,i
] = [λ].(27)
Proof
To prove necessity, note that it follows from Fact 8 that
(24) implies (26). It thus follows from Fact 9 that det
P
exists. Furthermore, it follows from (24) that, for all
i = 1,...,n,
[P
i,i
][v
i
] = [λ][v
i
].
Thus
[P
i,i
] = [λ].
Hence, for
i = 1,...,n, j = 1,...,n,
it follows that
[P
i,i
] = [P
j, j
]
.
To prove sufficiency, from (20) and (25) it follows that
[(z
1
)
1
(z
2
)
1
] = [(z
1
)
2
(z
2
)
2
] = ··· = [(z
1
)
n
(z
2
)
n
],(28)
where
(z
1
)
i
denotes the
i
th component of
z
1
. Thus,
λ
G
 z
T
2
z
1
exists. Note that
λ
G
z
1
= z
1
z
T
2
z
1
= [P]z
1
. Next, let
λ
C
∈ C
and
v
C
∈ C
n
be such that
[P]
C
v
C
= λ
C
v
C
.(29)
Then defining
λ ∈ D
and
v ∈ D
n
by
λ
 λ
C
λ
G
and
v
 v
C
◦ z
1
it follows that
Pv = ([P]
C
◦ [P])(v
C
◦ [v])
= ([P]
C
v
C
) ◦ z
1
z
T
2
z
1
= (λ
C
v
C
) ◦ λ
G
z
1
= λ
C
λ
G
(v
C
◦ z
1
)
= λv.

Next, let
P ∈ D
n×n
. Then, if det
[P]
C
= 0
, we define the
inverse
P
−1
of
P
by
P
−1

1
det P
P
A
,(30)
where the adjugate
P
A
is defined by
(P
A
)
i, j

(−1)
i+ j
det P
[ j,i]
, where
P
[ j,i]
denotes the
(n − 1) × (n − 1)
cofactor of
P
i,i
. Hence
[P
−1
] =
1
[det P]
[P
A
] (31)
and
[P
−1
]
C
=
1
[det P]
C
[P
A
]
C
.(32)
The following example shows that, for
P ∈ D
n×n
such
that
P
−1
exists, in general
[P
−1
][P] = [P][P
−1
]
.
Example 2
Let
P ∈ D
n×
be such that
[P] = []
m,s

m1/s
ms
2
s

and assume that
P
−1
exists. Then
[P
−1
] = []
m,s

1/m1/ms
2
s1/s

,
[P][P
−1
] = []
m,s

11/s
2
s
2
1

,
and
[P
−1
][P] = []
m,s

11/ms
ms 1

Thus
[P
−1
][P] = [P][P
−1
]
.
DIMENSIONS OF MATRICES
IN STATE-SPACE MODELS
Consider the system
˙
x(t) = Ax(t) + Bu(t), (33)
y(t) = Cx(t) + Du(t), (34)
where
[t] = s
,
x(t) ∈ D
n
, y(t) ∈ D
l
, u(t) ∈ D
m
,
A ∈ D
n×n
,
B ∈ D
n×m
, C ∈ D
l×n
, and
D ∈ D
l×m
. Every component of
x(t), y(t), u(t)
, and thus every entry of
A, B, C, D
, is a
dimensioned scalar. Taking units on both sides of (33) yields
[
˙
x(t)] = [A][x(t)] = [B][u(t)], (35)
[y(t)] = [C][x(t)] = [D][u(t)]. (36)
The following result is given on page 150 of [6].
Fact 11
[A] =
1
s
[x(t)][x
T
(t)]
{−1}
, (37)
[B] =
1
s
[x(t)][u
T
(t)]
{−1}
, (38)
[C] = [y(t)][x
T
(t)]
{−1}
, (39)
and
[D] = [y(t)][u
T
(t)]
{−1}
.(40)
106 IEEE CONTROL SYSTEMS MAGAZINE » DECEMBER 2007

Proof
The result follows from
[
˙
x(t)] = (1/s)[x(t)]
and Fact 8.

Next, define the transfer function matrix
H(s) ∈ D
l×m
by
H(s)
 C(sI
s
− A)
−1
B + D,(41)
where
s ∈ D
is the Laplace variable,
[s] = 1/s
, and
I
s
 I
R
◦ s[A]
.
Fact 12
[H(s)] = [y(t)][u
T
(t)]
{−1}
.(42)
Proof
Note that
[C(sI − A)
−1
B] = [y(t)][x
T
(t)]
{−1}
[x(t)][x
T
(t)]
{−1}
× [x(t)][u
T
(t)]
{−1}
,
= [y(t)][u
T
(t)]
{−1}
= [D]. 
Fact 13
For all
i = 1,...,n,
[A
i,i
] = []
kg,m
s
−1
.(43)
Furthermore, det
A
exists and satisfies
[det A] = []
kg,m
s
−n
.(44)
Proof. It follows from (37) that
[A
i,i
] =
1
s
[x
i
(t)]
[x
i
(t)]
=
[]
kg,m
s
.
Next, note that
[A
i,p
i
] =
1
s
[x
i
(t)]
[x
p
i
(t)]
.
Thus, for all
p ∈ P
n
,
[A
1,p
1
A
2,p
2
···A
n,p
n
] =
1
s
n
[x
1
(t)][x
2
(t)] ···[x
n
(t)]
[x
p
1
(t)][x
p
2
(t)] ···[x
p
n
(t)]
=
[]
kg,m
s
n
.
Since
[A
1,p
1
A
2,p
2
···A
n,p
n
]
is the same for all
p ∈ P
n
, det
A
exists. Finally, since [det
A] =

n
i=1
[A
1,p
i
]
for all
p ∈ P
n
, it
follows that
[det A] = [A
1,p
1
A
2,p
2
···A
n,p
n
] =
[]
kg,m
s
n
. 
Fact 14
Let
t ∈ D
be such that
[t] = s
. Then
det [At] = []
kg,m,s
.(45)
MATRIX EXPONENTIAL
Lemma 1
Let
t ∈ D
be such that
[t] = s
. Then the following state-
ments hold:
i) For all positive integers
k
,
A
k
exists.
ii) For all
k ≥ 1
,
[A
k
] = (1/s
k−1
)[A]
.
iii) For all
k ≥ 1
,
[A
k
] = (1/s)[A
k−1
]
.
iv) For all
k ≥ 1
,
[A
k
t
k
] = [At].
v)
[A]
{−1}
= (1/s
2
)[A]
.
If, in addition,
A
−1
exists, then
vi)
[A
−1
] = [A
T
]
{−1}
.
vii)
[A
−1
] = s
2
[A]
T
.
Proof
Statements i)–iv) follow from Fact 4. Next, we prove vi).
Since
(A
−1
)
i,i
= det A
[i,i]
/det A
, it follows that
[(A
−1
)
i,i
] =
= det [A
[i,i]
]/det [A] = [A
1,1
] ··· [A
i−1,i−1
][A
i+1,i+1
] ···
[A
n,n
]/([A
1,1
] ···[A
n,n
]) = 1/[A
i,i
]
. Thus, the diagonal
entries of
[A][A
−1
]
satisfy
([A][A
−1
])
i, i
= []
kg,m,s
, i = 1,...,n.
Therefore,
([A][A
−1
])
i,i
= [A
i,1
][A
−1
1,i
] + [A
i,2
][A
−1
2,i
]
+···+[A
i,n
][A
−1
n,i
][A
n,n
]
= []
kg,m,s
,
which implies that
[(A
−1
)
i, j
] = [A
j,i
]
−1
.(46)
Thus, vi) is satisfied.
To prove vii), note that
([A]
T
)
i, j
=
1
s
[x
j
(t)]
[x
i
(t)]
.(47)
Next, from (46) it follows that
[(A
−1
)
i, j
] = [A
j,i
]
−1
= s
[x
j
(t)]
[x
i
(t)]
.(48)
Thus from (47) and (48), it follows that
[A]
T
=
1
s
2
[A
−1
].(49)
To prove
v)
, using vi) in (49), we have
[A]
T
=
1
s
2
[A
T
]
{−1}
.(50)
Taking transposes yields v).

DECEMBER 2007 « IEEE CONTROL SYSTEMS MAGAZINE 107

Fact 15
[A
−1
] = s[x(t)][x
T
(t)]
{−1}
.(51)
Furthermore,
[A
−1
][A] = [A][A
−1
].(52)
Proof
Note that
[A
−1
] = [A
T
]
{−1}
= s[x(t)][x
T
(t)]
{−1}
.
Hence
[ A
−1
][ A] = [ A][ A
−1
] = [ x(t)][ x
T
(t )]
{−1}
[ x(t)]
[x
T
(t)]
{−1}
.

Fact 16
Let
t ∈ D
be such that
[t] = s
. Then
[e
At
] = [At] = [x(t)][x
T
(t)]
{−1}
.(53)
EIGENVALUES AND EIGENVECTORS OF
A
Fact 17
Let
λ ∈ D
be an eigenvalue of
A
, and let
v ∈ D
n
be an asso-
ciated eigenvector. Then, for all
i = 1,...,n,
[λ] = [A
i,i
] (54)
and
[v] = [x
T
(t)]
{−1}
[v][x(t)].(55)
Proof
Since
Av = λv
, it follows that, for all
i = 1,...,n,
[A
i,i
][v
i
] = [λ][v
i
]
, and thus
[λ] = [A
i,i
]
. Next, since
Av = λv
, it follows that
1
s
[x(t)][x
T
(t)]
{−1}
[v] =
1
s
[v],
which implies (55).

DC MOTOR EXAMPLE
Consider a dc motor with constant armature current
I
a
.
Defining the state vector to be
x
 [
i
f
ω
]
T
, where
i
f
is the
field current and
ω
is the motor angular velocity, we have
A =

−
R
f
L
f
0
BI
a
J
−
c
J

,(56)
where
R
f
and
L
f
are the field resistance and inductance,
respectively,
B
is the electromagnetic constant of the
motor,
J
is the inertia of the motor shaft and external
load, and
c
is the angular damping coefficient. The units
of
R
f
, L
f
, I
a
, B, J,
and
c
are
m
2
kg/C
2
s
,
m
2
kg/C
2
,
C/s
,
kgm
2
/C
2
, kgm
2
, and
kgm
2
/s
, respectively.
Taking units yields
[x(t)] =

C/s
[]
m
/s

.(57)
Thus
[A] =
1
s
[x(t)][x
T
(t)]
{−1}
=

[]
C
/s[]
m
C/s
[]
m
/Cs [ ]
m
/s

, (58)
where
[]
C
denotes the Coulombian. Hence
[det A] = []
m,C
/s
2
.
Furthermore,
[det A]
C
= det [A]
R
=

cR
f
L
f
J

R
.(59)
Thus,
det A =

cR
f
JL
f

R
[]
m,C
s
2
.(60)
Next, if
[cR
f
]
R
= 0
then det
[A]
R
= 0
and
[A
−1
]
is given by
[A
−1
] =

[]
C
s[]
m
Cs
[]
m
s/C[]
m
s

.(61)
Finally,
[e
At
] =

[]
C,s
[]
m,s
C
[]
m,s
/C[]
m,s

.(62)
SPRING-DAMPER SYSTEM EXAMPLE
Consider the spring-mass system shown in Figure 1. By
defining the state
x(t)
 [
q
1
˙
q
1
q
2
˙
q
2
]
T
, where
q
i
and
˙
q
i
are the displacement and velocity of the
i
th
mass, respec-
tively, we have
A

⎡
⎢
⎢
⎣
0100
−
(k
1
+k
2
)
m
1
−
(c
1
+c
2
)
m
1
k
2
m
1
c
2
m
1
0001
k
1
m
2
c
1
m
2
−
k
2
m
2
−
c
2
m
2
⎤
⎥
⎥
⎦
.(63)
Taking units yields
[x(t)] =
⎡
⎢
⎢
⎣
m
m/s
m
m/s
⎤
⎥
⎥
⎦
.(64)
FIGURE 1 Two-mass spring-damper system.
k
2
m
1
m
2
c
1
k
1
c
2
108 IEEE CONTROL SYSTEMS MAGAZINE » DECEMBER 2007

Thus,
[A] =
1
s
[x(t)][x
T
(t)]
{−1}
= []
m
⎡
⎢
⎢
⎣
1/s11/s1
1/s
2
1/s1/s
2
1/s
1/s11/s1
1/s
2
1/s1/s
2
1/s
⎤
⎥
⎥
⎦
. (65)
Hence
[det A] = []
m
/s
4
.
Furthermore,
[det A]
C
= det [A]
R
=

k
2
m
1
m
2

R
.(66)
Thus,
det A =

k
2
m
1
m
2

R
[]
m
s
4
.(67)
Next, if
[k
2
]
R
= 0
then det
[A]
R
= 0
and
[A
−1
]
is given by
[A
−1
] = []
m
⎡
⎢
⎢
⎣
ss
2
ss
2
1s1s
ss
2
ss
2
1s1s
⎤
⎥
⎥
⎦
.(68)
Finally,
[e
At
] = []
m
⎡
⎢
⎢
⎣
[]
s
s[]
s
s
1/s[]
s
1/s[]
s
[]
s
s[]
s
s
1/s[]
s
1/s[]
s
⎤
⎥
⎥
⎦
.(69)
CONCLUSIONS
Physical dimensions are the link between mathematical
models and the real world. In this article we extended
results of [6] by determining the dimensional structure
of a matrix under which standard operations involving
the inverse, powers, exponential, and eigenvalues are
valid. These results were applied to state space models.
We also distinguished between different types of
dimensionless units, namely, the massian, lengthian,
timian, densian, flowian, velocian, and forcian. These
dimensionless units arise naturally from the structure
of the groups of units, and appear throughout science
and engineering.
ACKNOWLEDGMENTS
We would like to thank Jan Willems for helpful sugges-
tions and comments.
REFERENCES
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[5] W.D. Curtis, J.D. Logan, and W.A. Parker, “Dimensional analysis and the
pi theorem,” Lin. Alg. Appl., vol. 47, pp. 117–126, 1982.
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Engineering. New York: Springer, 1995.
[7] T. Szirtes, Applied Dimensional Analysis and Modeling. New York: McGraw-
Hill, 1998.
[8] T.H. Fay and S.V. Joubert, “Dimensional analysis: An elegant technique
for facilitating the teaching of mathematical modeling,” Int. J. Math. Educa-
tion Science Tech., vol. 33, pp. 280–293, 2002.
[9] M. Rybaczuk, B. Lysik, and W. Kasprzak, Dimensional Analysis in the Iden-
tification of Mathematical Models. Singapore: World Scientific, 1990.
[10] S. Drobot, “On the foundations of dimensional analysis,” Studia Mathe-
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AUTHOR INFORMATION
Harish J. Palanthandalam-Madapusi (hpalanth@umich.edu)
received the B.E. degree from the University of Mumbai in
mechanical engineering in 2001. In 2001 and 2002 he was a
research engineer at the Indian Institute of Technology, Bom-
bay. He received the Ph.D. degree from the Aerospace Engi-
neering Department at the University of Michigan in 2007.
He is currently an assistant professor in the Department of
Mechanical and Aerospace Engineering at Syracuse Univer-
sity. His interests are in the areas of system identification,
data assimilation, and estimation.
Dennis S. Bernstein is a professor in the Aerospace Engi-
neering Department at the University of Michigan. He is
editor-in-chief of the IEEE Control Systems Magazine, and he
is the author of Matrix Mathematics: Theory, Facts, and Formu-
las with Applications to Linear Systems published by Princeton
University Press in 2005. His interests are in system identifi-
cation and adaptive control for aerospace applications.
Ravinder Venugopal received the B.Tech. degree in
aerospace engineering from the Indian Institute of Tech-
nology, Madras, India, the M.S. degree in aerospace engi-
neering from Texas A & M University, and the Ph.D.
degree in aerospace engineering from the University of
Michigan. From 1997 to 1999 he was a postdoctoral
research fellow at the Aerospace Engineering Department
of the University of Michigan. He is the founder and CEO
of Sysendes, Inc. His research interests include discrete-
time adaptive control, active noise and vibration control,
and hydraulic control for industrial applications.
DECEMBER 2007 « IEEE CONTROL SYSTEMS MAGAZINE 109