dimensional analysis of matrices state-space models and...

Dimensional Analysis of MatricesState-Space Models and Dimensionless Units

HARISH J. PALANTHANDALAM-MADAPUSI, DENNIS S. BERNSTEIN, and RAVINDER VENUGOPAL

» L E C T U R E N O T E S

Physical dimensions and units, such as mass (kg),length (m), time (s), and charge (C), provide the linkbetween mathematics and the physical world. It is

well known that careful attention to physical dimensionscan provide valuable insight into relationships amongphysical 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 ofa matrix, has been extensively applied [1]–[10]. Interestinghistorical remarks on the development of dimensionalanalysis are given in [11], while detailed discussions aregiven 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) = x2 + x2 ,

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 unlabeledcoefficients 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 massm, the expression

√k/m has the dimensions of recipro-

cal time. However, when used within the context ofharmonic solutions of an oscillator, the same expressionhas the interpretation of rad/s, where the dimension-less unit “rad” is inserted to facil itate the use oftrigonometric functions. Although this insertion is adhoc, the recognition that radians are dimensionless pro-vides reasonable justification.

A publication of special note is the book [6], which takesan in-depth look at the role of dimensions including matri-ces populated with dimensioned quantities. Although thistext provides no situations in which the “usual” rules ofdimensional analysis lead to incorrect answers, the carefulreexamination in [6] of the treatment of dimensions, espe-cially for matrices, motivates the present article.

The main objective of this article is to examine thedimensional structure of the dynamics matrix A that arisesin 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 dimensionedquantities. 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 additionaldimensionless quantity for each physical dimension andeach product of dimensions.

This approach immediately clarifies the mysteriousappearance 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 � kg0 is the identity element in the group of massdimensions, and [ ]m � m0 is the identity element in thegroup of length dimensions. In fact, [ ]m is the traditionalradian, whose appearance is natural and need not beinserted with the justification that “radians are dimen-sionless.” Rather, [ ]m appears because the mathematicalstructure 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 expressionω = v/r, where ω is angular velocity, v is translational veloc-ity, and r is radius. Then [ω] = [v]/[r] = (m/s)/m =m0/s = [ ]m/s = rad/s. Again, there is no need to artificiallyinsert the dimensionless unit “rad” in order to obtain theangular velocity in the expected units. We also note that, foran angle θ in radians, the fact that [ ]αm = (m0)α = m0 = [ ]m

for all real numbers α implies that

[sin θ] =[θ − θ3

3!+ · · ·

]= [ ]m ,

which is consistent with the fact that both θ and sin θ areratios of lengths.

In real computations involving physical quantities, thatis, aside from pure theory, it is necessary to keep track ofphysical dimensions and their associated units. Elucida-tion of the physical dimension structure of state spacemodels can thus be useful for verifying the model struc-ture and ensuring that the units are consistent within thecontext of state-space computations.Digital Object Identifier 10.1109/MCS.2007.906924

100 IEEE CONTROL SYSTEMS MAGAZINE » DECEMBER 2007

ALGEBRAIC STRUCTURE OF UNITSFor simplicity, we consider the fundamental dimensionsmass (kg), length (m), and time (s) only. For convenience,we use kg, m, and s to represent the respective physicaldimension as well as the associated unit. Let R and Cdenote the real and complex numbers, respectively.Define Gkg � {kgα : α ∈ R} , Gm � {mβ : β ∈ R} , andGs � {sγ : γ ∈ R}. Note that Gkg, Gm, and Gs are Abelian(commutative) groups (see “What Is a Group?”) with theidentity elements [ ]kg, [ ]m, and [ ]s, respectively, whichare dimensionless units referred to as the massian, lengthi-an, and timian. The lengthian [ ]m in Gm, when interpretedwithin the context of a circle, is the radian. Next, definethe set G of all mixed units

G � {kgαmβsγ : α, β, γ ∈ R}. (1)

Since, for all α, β, γ ∈ R , kgαmβsγ = kgαsγ mβ =mβkgαsγ = mβsγ kgα = sγ mβkgα = sγ kgαmβ , we have thefollowing result.

Fact 1G 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 numberin fluid dynamics defined by

Re � vsLν

,

where vs is the mean fluid velocity, L is the characteristiclength of the flow, and ν is the kinematic fluid viscosity,has the units

[Re] = [ ]kg,m,s.

Similarly, the dimensionless Froude number in fluidmechanics defined by

Fr � vs

Lg,

where g is acceleration due to gravity, has the units

[Fr] = [ ]m,s.

Table 1 classifies several dimensionless quantities based ontheir units.

The set D of dimensioned scalars consists of elements ofthe form akgαmβsγ , where a ∈ C and α, β, γ ∈ R . Weallow a ∈ C to accommodate complex eigenvalues andeigenvectors. We define the units operator [ ] as

[akgαmβsγ ] � kgαmβsγ .

Note that [0 kgαmβsγ ] � kgαmβsγ . Let a1kgα1 mβ1 sγ1 anda2kgα2 mβ2 sγ2 be dimensioned scalars. Then the product oftwo dimensioned scalars always exists and is defined tobe a1kgα1 mβ1 sγ1 a2kgα2 mβ2 sγ2 = a1a2kgα1+α2 mβ1+β2 sγ1+γ2 .However, the sum a1kgα1 mβ1 sγ1 + a2kgα2 mβ2 sγ2 is definedonly if α1 = α2, β1 = β2, and γ1 = γ2 , in which casea1kgα1 mβ1 sγ1 + a2kgα2 mβ2 sγ2 = (a1 + a2)kgα1 mβ1 sγ1 . Fur-thermore, although quantities such as akgα and bsγ arenot elements of D, we assume that all operations occurafter these quantities are embedded in the appropriategroup containing all of the common units. For example,(akgα)(bsγ ) � (akgα[ ]s)(bsγ [ ]kg) = abkgαsγ .

Dimensioned vectors and dimensioned matrices are denot-ed by Dn and Dn×m, respectively, all of whose entries aredimensioned scalars (see “Energy Versus Moment” for anexample of the difference between dimensioned scalarsand dimensioned vectors). Let P ∈ Dn×m and define

[P] �

⎡⎢⎣

[P1,1] · · · [P1,m]...

. . ....

[Pn,1] · · · [Pn,m]

⎤⎥⎦ ∈ Gn×m , (2)

where Pi, j is the (i, j) entry of P and Gn×m denotes the setof n × m matrices with entries in G. Note that [PT] = [P]T.If P ∈ Dn×m and Q ∈ Dm×p, then PQ exists if all additionoperations required to form the product are defined.

Fact 2Let P ∈ Dn×m and Q ∈ Dm×p. Then PQ exists if and only if,for all i = 1, . . . , n and j = 1, . . . , p,

[Pi,1][Q1, j] = [Pi,2][Q2, j] = · · · = [Pi,n][Qn, j]. (3)

Furthermore, if PQ exists, then

[PQ] = [P][Q]. (4)

Fact 3Let P ∈ Dn×n. If P2 exists, then

[P1,1] = [P2,2] = · · · = [Pn,n]. (5)

ProofSince P2 exists, it follows that, for all i, j = 1, . . . , n,

[(P2)i,i] = [Pi,1][P1,i] = [Pi,2][P2,i] = · · · = [Pi,n][Pn,i].

Now, let i, j ∈ {1, . . . , n} . Then [Pi,i][Pi,i] = [Pi, j][Pj,i] =[Pj,i][Pi, j] = [Pj, j][Pj, j]. Hence [Pi,i] = [Pj, j]. �

Fact 4Let P ∈ Dn×n. If P2 exists, then, for all positive integers k,Pk exists and [Pk] = [P]k . Furthermore, for all i = 1, . . . , nand for all positive integers k,

DECEMBER 2007 « IEEE CONTROL SYSTEMS MAGAZINE 101

[Pk] = [(Pi,i)k−1][P]. (6)

ProofSince, for all i, j = 1, . . . , n,

[(P2)i, j] = [Pi,1][P1, j] = [Pi,2][P2, j] = · · · = [Pi,n][Pn, j],

it follows that

[(P2)i, j] = [Pi,i][Pi, j].

Hence [P2] = [Pi,i][P]. Induction yields (6). �

Fact 5Let P ∈ Dn×n . Then P2 exists if and only if there existz1, z2 ∈ Gn such that zT

2 z1 exists and

[P] = z1zT2 . (7)

ProofSufficiency is immediate. To prove necessity, define

z1 �

⎡⎢⎢⎢⎣

[P1,1][P2,1]

...

[Pn,1]

⎤⎥⎥⎥⎦ , z2 �

⎡⎢⎢⎢⎣

[P1,1]/[P1,1][P1,2]/[P1,1]

...

[P1,n]/[P1,1]

⎤⎥⎥⎥⎦ .

Let u1, . . . , up be fundamental dimensions and let G�=

{∏pi=1 uαi

i : α1, . . . , αp ∈ R be the corresponding Abelian

group. Then the set D of dimensioned scalars consists of ele-

ments of the form a∏p

i=1 uαii , where a ∈ C and α1, . . . , αp ∈ 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 Q1, Q2, . . . , Qq ∈ D be dimensioned scalars such that, for

i = 1, . . . , q, Qi�= ai

∏pj=1 uαij

j , and assume that

K∑k=1

ck Qβ1k1 · · · Qβqk

q = 0, (S1)

where c1, . . . , cK ∈ R are nonzero. Let A �= [αij ]T ∈ Rp×q , and let

r�= rank A . Then there exists �

�= [γij ] ∈ Rq×(q−r) such that

rank � = q − r, A� = 0, and, for i = 1, . . . , q − r,

�i�= Qγ1i

1 · · · Qγqiq (S2)

are dimensionless.

PROOFIt 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

��= [γij ] ∈ 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 αkjγk i = 0, it

follows that, for all i = 1, . . . , q − r,

�i = Qγ1i

1 · · · Qγqiq

= aγ1i

1 · · · aγqiq

p∏j=1

u∑q

k=1αkj γki

j

= aγ1i

1 · · · aγqiq

p∏j=1

u0j

is dimensionless. �As an example, consider the law of conservation of momen-

tum in a collision between two rigid bodies given by

m1v−1 + m2v−

2 = m1v+1 + m2v+

2 , (S3)

where m1 and m2 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

[m1] = [m2] = kg and [v−1 ] = [v−

2 ] = [v+1 ] = [v+

2 ] = m/s. Further-

more, choosing u1 = kg, u2 = m, u3 = s, Q1 = m1, Q2 = m2 ,

Q3 = v−1 , Q4 = v−

2 , Q5 = v+1 , and Q6 = v+

2 , it follows that p = 3,

q = 6,

A =[ 1 1 0 0 0 0

0 0 1 1 1 10 0 −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

� =

⎡⎢⎢⎢⎢⎢⎣

1 0 0 0−1 0 0 00 1 1 00 −1 0 00 0 −1 10 0 0 1

⎤⎥⎥⎥⎥⎥⎦ (S5)

yields the dimensionless quantities

The Buckingham Pi Theorem in a Nutshell


Since P2 exists it follows that [(P2)1,1]/[P1,1] = zT2 z1 exists.

Furthermore, let k ∈ {1, . . . , n} and define z3 ∈ Gn by

z3 �

⎡⎢⎢⎢⎣

[P1,k][P2,k]

...

[Pn,k]

⎤⎥⎥⎥⎦ .

Then, zT2 z3 exists and thus the rows of [P] are dimensioned

scalar multiples of each other. Hence

[P] =

⎡⎢⎢⎢⎣

[P1,1] [P1,2] · · · [P1,n][P2,1] [P1,2][P2,1]/[P1,1] · · · [P1,n][P2,1]/[P1,1]

......

. . ....

[Pn,1] [P1,2][Pn,1]/[P1,1] · · · [P1,n][Pn,1]/[P1,1]

⎤⎥⎥⎥⎦

= z1zT2 . �

Fact 6Let P ∈ Dn×n. Then eP ∈ Dn×n exists if and only if P2 existsand [P] = [P2]. Furthermore, if eP exists then

[eP] = [P]. (8)

�1 = m1

m2, �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/s2, and [m] = kg, we

choose u1 = kg, u2 = m, u3 = s, Q1 = T, Q2 = l, Q3 = g, and

Q4 = m. Noting that p = 3, q = 3,

A =[ 0 0 0 1

0 1 1 01 0 −2 0

], (S6)

and r = 3, it follows that there exists q − r = 1 dimensionless

quantity given by

�1 = T√

g√l

. (S7)

Therefore,

T = �1

√lg

, (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/s2 ,

[d ] = m, [v ] = m/s, [ρ] = kg/m3, [N] = [ ]m/s, [ν] = m2/s , we

choose u1 = kg, u2 = m, u3 = s, Q1 = F, Q2 = d, Q3 = v ,

Q4 = ρ, Q5 = N, and Q6 = ν . Therefore, we have

p = 3, q = 6,

A =[ 1 0 0 1 0 0

1 1 1 −3 0 2−2 0 −1 0 −1 −1

], (S9)

and r = 3. Thus we have q − r = 6 − 3 = 3 dimensionless quan-

tities. Choosing � to be

� =

⎡⎢⎢⎢⎢⎢⎣

0 0 11 1 −21 −1 −20 0 −10 1 0

−1 0 0

⎤⎥⎥⎥⎥⎥⎦ , (S10)

it follows that

�1 = dvν

, �2 = dNv

, �3 = Fd2v2ρ

,

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.


ProofBy definition, the matrix exponential eP ∈ Dn×n is given by

eP = I + 11!

P + 12!

P2 + · · · . (9)

Necessity is immediate. To prove sufficiency, note that,since P2 exists and [P] = [P2], it follows from Fact 4 that[P] = [P2] implies that [P] = [Pk] for all positive integers k.Thus eP exists. Next, it follows from (9) that (8) holds. �

Fact 7Let P ∈ Dn×n and assume that eP exists. Then, for alli = 1, . . . , n,

[Pi,i] = [ ]kg,m,s. (10)

ProofThe result follows immediately from facts 6 and 4. �

For a real scalar q and P ∈ Dn×m , the Schur powerP{q} ∈ Dn×m is defined by

(P{q})i, j � (Pi, j)q, (11)

assuming the right hand side exists. The notation[P]C ∈ Cn×m denotes the numerical part of the dimen-sioned matrix P ∈ Dn×m. Note that

P = [P]C ◦ [P], (12)

where ◦ is the Schur (entry-wise) product. We write [P]Cas [P]R if [P]C is real. Let IR denote the identity matrix inRn×n . Furthermore, let Q ∈ Dm×p and assume that PQexists. 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 8Let P ∈ Dn×m, and let y ∈ Dn and u ∈ Dm be such that

y = Pu. (14)

Then

[P] = [y][uT]{−1}. (15)

ProofThe ith component equation of (15) is

[Pi,1][u1] + [Pi,2][u2] + · · · + [Pi,m][um] = [yi].

Therefore,

[Pi,1][u1] = [Pi,2][u2] = · · · = [Pi,m][um] = [yi],

and thus [Pi, j] = [yi]/[uj]. Hence (15) holds. �

Agroup (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 = {kgαmβsγ : 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 ofratios of the basic physical dimensions.

Dimensionless Unit Name Examples[ ]kg Massian Air-fuel ratio

Stoichiometric mass ratio

[ ]m Lengthian RadianStrainPoisson's ratioFresnel numberAspect ratio

[ ]s Timian Courant-Friedrichs-Lewy (CFL) number

Damkohler numbers

[ ]kg,m Densian Density ratioMoment-of-inertia ratio

[ ]kg,s Flowian Mass-flow ratioStiffness ratio

[ ]m,s Velocian Froude numberFourier numberMach numberStokes number

[ ]kg,m,s Forcian Reynolds numberWeber number Coefficient of frictionLift coefficientDrag coefficient


Next, let P ∈ Dn×n. Then, the determinant det P of P isdefined to be

det P =∑

p∈Pn

σ(p)P1,p1 P2,p2 · · · Pn,pn, (16)

where Pn is the set of all permutations p = (p1, . . . , pn) of(1, 2, . . . , n), and σ(p) is the signature of the permutationp, which is 1 if p is achieved by applying an even numberof transpositions to (1, 2, . . . , n) and −1 if p is reached byapplying an odd number of transpositions to (1, 2, . . . , n).Note that if P ∈ Dn×n then det P exists if and only if[P1,p1 P2,p2 · · · Pn,pn ] is the same for all p ∈ Pn. Hence, if detP exists, we have

[det P] = [P1,p1 P2,p2 · · · Pn,pn ] (17)

for all p ∈ Pn. Note that

det [P]C = [det P]C (18)

anddet P = (det [P]C)[det P]. (19)

The following result presents necessary and sufficient con-ditions for the existence of det P.

Fact 9Let P ∈ Dn×n. Then det P exists if and only if there existz1, z2 ∈ Gn such that

[P] = z1zT2 . (20)

ProofSufficiency is immediate. To prove necessity, first let n = 2.Then, since det P exists, it follows that

[P1,1][P1,2]

= [P2,1][P2,2]

. (21)

Thus the columns of [P] are dimensioned scalar multiples ofeach other. Next, let n = 3 and assume that det P exists.Then it follows from the cofactor expansion of det P that thedeterminant of every 2 × 2 submatrix of P exists. Hence (21)holds. Next, it follows that [P1,1P2,3P3,2] = [P1,2P2,3P3,1]and hence

[P1,1][P1,2]

= [P3,1][P3,2]

. (22)

Furthermore, using [P1,2P2,3P3,1] = [P1,3P2,2P3,1] and[P1,2P2,1P3,2] = [P1,3P2,1P3,2], it follows that [P1,2]/[P1,3] =[P2,2]/[P2,3] and [P1,2]/[P1,3] = [P3,2]/[P3,3] . Thus thecolumns of [P] are dimensioned scalar multiples of each

other. Likewise, for all n ≥ 1, it can be seen that, since det Pexists, the columns of [P] are dimensioned scalar multiplesof each other. Thus, defining

z1 �

⎡⎢⎢⎢⎣

[P1,1][P2,1]

...

[Pn,1]

⎤⎥⎥⎥⎦ , z2 �

⎡⎢⎢⎢⎣

[P1,1]/[P1,1][P1,2]/[P1,1]

...

[P1,n]/[P1,1]

⎤⎥⎥⎥⎦ ,

it follows that (20) holds. �Note that if P2 exists then det P exists. However, the

following example shows that the converse does not hold.

Example 1Let P ∈ D2×2 be such that

[P] =[

m m2

s ms

]. (23)

Then det P exists, but P2 does not exist.Let P ∈ Dn×n. Then λ ∈ D and v ∈ Dn are an eigenvalue-

eigenvector pair of P if [v]C is not zero and λ and v satisfy

Pv = λv. (24)

Fact 10Let P ∈ Dn×n. Then P has an eigenvalue-eigenvector pairλ ∈ D, v ∈ Dn if and only if det P exists and, for alli = 1, . . . , n and j = 1, . . . , n,

[Pi,i] = [Pj, j]. (25)

Since energy is force times displacement, it follows that the

units of energy are J = Nm = kgm2/s2. On the other hand,

since moment times angular displacement is energy, it follows

that the units of moment are J/rad = Nm/rad = kgm2/s2rad.

Furthermore, since rad = [ ]m = m0 , it follows that

J/rad = kgm2/s2rad = kgm2/s2 = 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 D3. 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


In this case,

[P] = [λv][vT]{−1} (26)

and, for all i = 1, . . . , n,

[Pi,i] = [λ]. (27)

ProofTo prove necessity, note that it follows from Fact 8 that(24) implies (26). It thus follows from Fact 9 that det Pexists. Furthermore, it follows from (24) that, for alli = 1, . . . , n,

[Pi,i][vi] = [λ][vi].

Thus

[Pi,i] = [λ].

Hence, for i = 1, . . . , n, j = 1, . . . , n, it follows that[Pi,i] = [Pj, j].

To prove sufficiency, from (20) and (25) it follows that

[(z1)1(z2)1] = [(z1)2(z2)2] = · · · = [(z1)n(z2)n], (28)

where (z1)i denotes the ith component of z1 . Thus,λG � zT

2 z1 exists. Note that λGz1 = z1zT2 z1 = [P]z1 . Next, let

λC ∈ C and vC ∈ Cn be such that

[P]CvC = λCvC. (29)

Then defining λ ∈ D and v ∈ Dn by λ � λCλG andv � vC ◦ z1 it follows that

Pv = ([P]C ◦ [P])(vC ◦ [v])

= ([P]CvC) ◦ z1zT2 z1

= (λCvC) ◦ λGz1

= λCλG(vC ◦ z1)

= λv. �

Next, let P ∈ Dn×n. Then, if det [P]C �= 0, we define theinverse P−1 of P by

P−1 � 1det P

PA, (30)

where the adjugate PA is defined by (PA)i, j �(−1)i+ jdet P[ j,i] , where P[ j,i] denotes the (n − 1) × (n − 1)

cofactor of Pi,i. Hence

[P−1] = 1[det P]

[PA] (31)

and

[P−1]C = 1[det P]C

[PA]C. (32)

The following example shows that, for P ∈ Dn×n suchthat P−1 exists, in general [P−1][P] �= [P][P−1].

Example 2Let P ∈ Dn× be such that

[P] = [ ]m,s

[m 1/s

ms2 s

]

and assume that P−1 exists. Then

[P−1] = [ ]m,s

[1/m 1/ms2

s 1/s

],

[P][P−1] = [ ]m,s

[1 1/s2

s2 1

],

and

[P−1][P] = [ ]m,s

[1 1/ms

ms 1

]

Thus [P−1][P] �= [P][P−1].

DIMENSIONS OF MATRICES IN STATE-SPACE MODELSConsider the system

x(t) = Ax(t) + Bu(t), (33)

y(t) = Cx(t) + Du(t), (34)

where [t] = s, x(t) ∈ Dn, y(t) ∈ D l, u(t) ∈ Dm , A ∈ Dn×n,

B ∈ Dn×m, C ∈ D l×n , and D ∈ D l×m . Every component ofx(t), y(t), u(t) , and thus every entry of A, B, C, D , is adimensioned 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] = 1s

[x(t)][xT(t)]{−1}, (37)

[B] = 1s

[x(t)][uT(t)]{−1}, (38)

[C] = [y(t)][xT(t)]{−1}, (39)

and

[D] = [y(t)][uT(t)]{−1}. (40)


ProofThe 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(sIs − A)−1B + D, (41)

where s ∈ D is the Laplace variable, [s] = 1/s, andIs � IR ◦ s[A].

Fact 12

[H(s)] = [y(t)][uT(t)]{−1}. (42)

ProofNote that

[C(sI − A)−1B] = [y(t)][xT(t)]{−1}[x(t)][xT(t)]{−1}

× [x(t)][uT(t)]{−1},= [y(t)][uT(t)]{−1}

= [D]. �

Fact 13For all i = 1, . . . , n,

[Ai,i] = [ ]kg,ms−1. (43)

Furthermore, det A exists and satisfies

[det A] = [ ]kg,ms−n. (44)

Proof. It follows from (37) that

[Ai,i] = 1s

[xi(t)][xi(t)]

= [ ]kg,m

s.

Next, note that

[Ai,pi] = 1s

[xi(t)][xpi(t)]

.

Thus, for all p ∈ Pn,

[A1,p1 A2,p2 · · · An,pn ] = 1sn

[x1(t)][x2(t)] · · · [xn(t)][xp1(t)][xp2(t)] · · · [xpn(t)]

= [ ]kg,m

sn .

Since [A1,p1 A2,p2 · · · An,pn ] is the same for all p ∈ Pn, det Aexists. Finally, since [det A] = ∏n

i=1[A1,pi] for all p ∈ Pn, itfollows that

[det A] = [A1,p1 A2,p2 · · · An,pn ] = [ ]kg,m

sn . �

Fact 14Let t ∈ D be such that [t] = s. Then

det [At] = [ ]kg,m,s. (45)

MATRIX EXPONENTIAL

Lemma 1Let t ∈ D be such that [t] = s. Then the following state-ments hold:

i) For all positive integers k, Ak exists.ii) For all k ≥ 1, [Ak] = (1/sk−1)[A].iii) For all k ≥ 1, [Ak] = (1/s)[Ak−1].iv) For all k ≥ 1, [Aktk] = [At].v) [A]{−1} = (1/s2)[A].

If, in addition, A−1 exists, thenvi) [A−1] = [AT]{−1} .

vii) [A−1] = s2[A]T.

ProofStatements 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] = [A1,1] · · · [Ai−1,i−1] [Ai+1,i+1] · · ·[An,n]/([A1,1] · · · [An,n]) = 1/[Ai,i] . Thus, the diagonalentries of [A][A−1] satisfy

([A][A−1])i, i = [ ]kg,m,s, i = 1, . . . , n.

Therefore,

([A][A−1])i,i = [Ai,1][A−11,i ] + [Ai,2][A−1

2,i ]

+ · · · + [Ai,n][A−1n,i ][An,n]

= [ ]kg,m,s,

which implies that

[(A−1)i, j] = [Aj,i]−1. (46)

Thus, vi) is satisfied.To prove vii), note that

([A]T)i, j = 1s

[xj(t)]

[xi(t)]. (47)

Next, from (46) it follows that

[(A−1)i, j] = [Aj,i]−1 = s

[xj(t)]

[xi(t)]. (48)

Thus from (47) and (48), it follows that

[A]T = 1s2 [A−1]. (49)

To prove v), using vi) in (49), we have

[A]T = 1s2 [AT]{−1}. (50)

Taking transposes yields v). �


Fact 15[A−1] = s[x(t)][xT(t)]{−1}. (51)

Furthermore,

[A−1][A] = [A][A−1]. (52)

ProofNote that

[A−1] = [AT]{−1} = s[x(t)][xT(t)]{−1}.

Hence [ A−1][ A] = [ A][ A−1] = [ x(t)][ xT(t)]{−1}[ x(t)][xT(t)]{−1} . �

Fact 16Let t ∈ D be such that [t] = s. Then

[eAt] = [At] = [x(t)][xT(t)]{−1}. (53)

EIGENVALUES AND EIGENVECTORS OF A

Fact 17Let λ ∈ D be an eigenvalue of A, and let v ∈ Dn be an asso-ciated eigenvector. Then, for all i = 1, . . . , n,

[λ] = [Ai,i] (54)

and

[v] = [xT(t)]{−1}[v][x(t)]. (55)

ProofSince Av = λv , it follows that, for all i = 1, . . . , n,

[Ai,i][vi] = [λ][vi] , and thus [λ] = [Ai,i] . Next, sinceAv = λv, it follows that

1s

[x(t)][xT(t)]{−1}[v] = 1s

[v],

which implies (55). �

DC MOTOR EXAMPLEConsider a dc motor with constant armature current Ia.Defining the state vector to be x � [ if ω ]T, where if is thefield current and ω is the motor angular velocity, we have

A =[− Rf

Lf0

BIaJ − c

J

], (56)

where Rf and Lf are the field resistance and inductance,respectively, B is the electromagnetic constant of themotor, J is the inertia of the motor shaft and externalload, and c is the angular damping coefficient. The unitsof Rf, Lf, Ia, B, J, and c are m2kg/C2s, m2kg/C2 , C/s,kgm2/C2, kgm2, and kgm2/s, respectively.

Taking units yields

[x(t)] =[

C/s[ ]m/s

]. (57)

Thus

[A] = 1s

[x(t)][xT(t)]{−1} =[

[ ]C/s [ ]mC/s[ ]m/Cs [ ]m/s

], (58)

where [ ]C denotes the Coulombian. Hence[det A] = [ ]m,C/s2. Furthermore,

[det A]C = det [A]R =[

cRf

Lf J

]R

. (59)

Thus,

det A =[

cRf

JLf

]R

[ ]m,C

s2 . (60)

Next, if [cRf]R �= 0 then det [A]R �= 0 and [A−1] is given by

[A−1] =[

[ ]Cs [ ]mCs[ ]ms/C [ ]ms

]. (61)

Finally,

[eAt] =[

[ ]C,s [ ]m,sC[ ]m,s/C [ ]m,s

]. (62)

SPRING-DAMPER SYSTEM EXAMPLEConsider the spring-mass system shown in Figure 1. Bydefining the state x(t) � [ q1 q1 q2 q2 ]T , where qi andqi are the displacement and velocity of the ith mass, respec-tively, we have

A �

⎡⎢⎢⎣

0 1 0 0− (k1+k2)

m1− (c1+c2)

m1

k2m1

c2m1

0 0 0 1k1m2

c1m2

− k2m2

− c2m2

⎤⎥⎥⎦ . (63)

Taking units yields

[x(t)] =

⎡⎢⎢⎣

mm/sm

m/s

⎤⎥⎥⎦ . (64)

FIGURE 1 Two-mass spring-damper system.

k2

m1 m2

c1

k1

c2


Thus,

[A] = 1s

[x(t)][xT(t)]{−1}

= [ ]m

⎡⎢⎢⎣

1/s 1 1/s 11/s2 1/s 1/s2 1/s1/s 1 1/s 11/s2 1/s 1/s2 1/s

⎤⎥⎥⎦ . (65)

Hence [det A] = [ ]m/s4. Furthermore,

[det A]C = det [A]R =[

k2

m1m2

]R

. (66)

Thus,

det A =[

k2

m1m2

]R

[ ]m

s4 . (67)

Next, if [k2]R �= 0 then det [A]R �= 0 and [A−1] is given by

[A−1] = [ ]m

⎡⎢⎢⎣

s s2 s s2

1 s 1 ss s2 s s2

1 s 1 s

⎤⎥⎥⎦ . (68)

Finally,

[eAt] = [ ]m

⎡⎢⎢⎣

[ ]s s [ ]s s1/s [ ]s 1/s [ ]s

[ ]s s [ ]s s1/s [ ]s 1/s [ ]s

⎤⎥⎥⎦ . (69)

CONCLUSIONSPhysical dimensions are the link between mathematicalmodels and the real world. In this article we extendedresults of [6] by determining the dimensional structureof a matrix under which standard operations involvingthe inverse, powers, exponential, and eigenvalues arevalid. These results were applied to state space models.We also distinguished between different types ofdimensionless units, namely, the massian, lengthian,timian, densian, flowian, velocian, and forcian. Thesedimensionless units arise naturally from the structureof the groups of units, and appear throughout scienceand engineering.

ACKNOWLEDGMENTSWe would like to thank Jan Willems for helpful sugges-tions and comments.

REFERENCES[1] G. Birkhoff, Hydrodynamics—A Study in Logic, Fact, and Similitude. NewYork: Dover, 1955.

[2] P.W. Bridgman, Dimensional Analysis. New Haven, CT: Yale Univ. Press, 1963.

[3] H.E. Huntley, Dimensional Analysis. New York: Dover, 1967.

[4] C.C. Lin and L.A. Segel, Mathematics Applied to Deterministic Problems inthe Natural Sciences. New York: Macmillan, 1974.

[5] W.D. Curtis, J.D. Logan, and W.A. Parker, “Dimensional analysis and thepi theorem,” Lin. Alg. Appl., vol. 47, pp. 117–126, 1982.

[6] G.W. Hart, Multidimensional Analysis: Algebras and Systems for Science andEngineering. 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 techniquefor 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-matica, vol. 14, no. 1, pp. 84–99, 1953.

[11] E.O. Macagno, “Historico-critical review of dimensional analysis,” J.Franklin Inst., vol. 292, no. 6, pp. 391–402, 1971.

[12] D.H. Krantz, R.D. Luce, P. Suppes, and A. Tversky, Foundations of Mea-surement. New York: Dover, 2007.

[13] H. Whitney, “The mathematics of physical quantities: Part II: Quantitystructures and dimensional analysis,” Amer. Math. Monthly, vol. 75, no. 3, pp.227–256, 1971.

AUTHOR INFORMATIONHarish J. Palanthandalam-Madapusi ([email protected])received the B.E. degree from the University of Mumbai inmechanical engineering in 2001. In 2001 and 2002 he was aresearch 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 ofMechanical 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 iseditor-in-chief of the IEEE Control Systems Magazine, and heis the author of Matrix Mathematics: Theory, Facts, and Formu-las with Applications to Linear Systems published by PrincetonUniversity Press in 2005. His interests are in system identifi-cation and adaptive control for aerospace applications.

Ravinder Venugopal received the B.Tech. degree inaerospace 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 ofMichigan. From 1997 to 1999 he was a postdoctoralresearch fellow at the Aerospace Engineering Departmentof the University of Michigan. He is the founder and CEOof Sysendes, Inc. His research interests include discrete-time adaptive control, active noise and vibration control,and hydraulic control for industrial applications.


» F E E D B A C K

MISSING LINKSI am contacting you concerning thearticle [1], which appeared in theDecember 2007 issue of IEEE ControlSystems Magazine, coauthored by usand R. Venugopal. In this regard, Imust bring to your attention the factthat I have discovered two additionalarticles on the subject. These articlespresent further insights into theproperties of dimensions. The pur-pose of this letter is to brieflydescribe these insights.

The articles I am referring to are[2] and [3], which build on the book[4]. These works develop orientationanalysis, which augments standarddimensional analysis in verifyingcandidate relationships and uncover-ing new relationships. Orientationanalysis keeps track of the orientationof each quantity and ensures that, ineach physical relationship, the quan-tities on both sides have the same ori-entation. Although this idea isintuitively clear, it has been widelyignored in textbooks.

From an orientation point of view,quantities such as length, velocity,acceleration, force, and moment havea direction in space, whereas quanti-ties such as mass, time, energy, andpower do not. Not surprisingly, the

orientation of a position, velocity, oracceleration vector along the x-direc-tion is different from that of a corre-sponding vector along the y-direction.From this point of view, two forcevectors acting along different lines ofaction, although representing thesame physical quantity, are funda-mentally different because they havedifferent directions. Although closelyrelated, orientation is not the same asvector direction. For instance, area hasorientation but is not a vector.

As an application of orientationanalysis, consider the problem dis-cussed in [2]–[4] of deriving an expres-sion for the range R of a projectile firedhorizontally from a height h, with avelocity v0 . Since [h] = m,

[v0] = m/s, [R] = m, and [g] = m/s2 ,the problem involves four quantitiesand two dimensions. Therefore, it fol-lows from the Buckingham Pi theoremgiven in [1] that there exist two dimen-sionless quantities, namely, �1 = R/hand �2 = v0/

√hg. This result is not

useful, however, for characterizing Rin terms of v0, h, and g. However, ifwe recognize that lengths along thevertical and horizontal directions havedifferent orientations, and thus repre-sent different physical dimensions,namely, mx and my, we have[h] = my, [v0] = mx/s, [R] = mx, and[g] = my/s2. We then have four quan-tities and three dimensions, whichyields the dimensionless quantity

�1 = v0

R

√hg,

and thus the useful expressionR = �1v0

√h/g.

In addition to providing a tool forderiving and checking physical rela-

tionships, orientation analysis can beused to explain the nature of anglesand dimensionless units such as radi-ans. For example, an angle can beviewed as the ratio of two lengthsalong perpendicular directions (thatis, radial and circumferential withinthe context of a circle), and thus anglehas orientation, namely, along thedirection perpendicular to the planecontaining the two lengths, as notedin [2]. This point is consistent with thetreatment of angle as a vector point-ing out of the plane that contains theangle; this vector can be dotted with amoment vector to determine the workdone by the moment vector whenmoving through an angle. Note that aratio of identical dimensions, such aslength/length, is dimensionless butcan possess orientation.

Orientation analysis helps explainwhy radians are different from strain,which is also a ratio of lengths butdoes not have orientation. This dis-tinction is not due to the fact that radi-ans have a natural circle scale,namely, 2π, but rather the fact thatstrain is the ratio of lengths along thesame direction. On the other hand,Poisson’s ratio, which measures theratio of the resulting bulge or shrink-age due to the strain in an orthogonaldirection, is appropriately measuredin radians. As another example, thedistinguishing feature betweenmoment and energy—which weexplain in our article as being due tothe distinction between a scalar and avector—is that moment has orienta-tion, whereas energy does not.

The rules and consequences oforientation analysis, which areexplained in detail in [2] and [3], canbe used to compute the orientationsDigital Object Identifier 10.1109/MCS.2008.920947

I EEE Control Systems Magazine

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16 IEEE CONTROL SYSTEMS MAGAZINE » JUNE 2008

of various derived quantities. Forexample, as noted above, area hasorientation, although volume is ori-entationless. Even more surprising isthe consequence that cos θ is orienta-tionless, but sin θ has orientation.Since the square of every orientationis orientationless, the identitycos2 θ+ sin2 θ = 1 is orientationless.Along the same lines, ejωt = cos(ωt)+j sin(ωt) (where j � √−1) is orienta-tionless, which in turn implies that jhas orientation. Since j represents arotation about the z-axis, this obser-vation seems reasonable. Althoughthese observations are surprising,their consistency strikes me asbeyond coincidental. Somethingdeep and fundamental seems to begoing on here.

Although orientation analysisoffers insight into the nature of dimen-sionless units such as radians, thereremain unanswered questions in theworld of dimensions and units. For

example, it occurs to me that the anglebetween a force vector and a displace-ment vector, whose dot product arisesin computing work (that is, energytransferred), should not be measuredin radians but rather in forcians. Inother words, it seems that not allangles are created equal. Anothermystery is the fact that orientationanalysis does not offer a satisfactoryexplanation for the appearance ofradians in the expression for naturalfrequency ω = √

k/m, where m is massand k is the ratio of force to displace-ment. While we usually explain radi-ans in terms of a ratio of lengths, theforce and displacement in the simpleharmonic oscillator are parallel, andthus orientation analysis suggests thatthe correct unit for natural frequencyis not radian in the sense of orientationalong the z-axis but rather the orienta-tionless lengthian.

I suspect we haven't heard the lastword on this subject. It often seems

that definitively resolving fundamen-tal issues is not a simple task.

Harish J. Palanthandalam-Madapusi

Syracuse University

References[1] H.J. Palanthandalam-Madapusi, D.S. Bern-stein, and R. Venugopal, “Dimensional analysisof matrices: State-space models and dimension-less units,” IEEE Contr. Sys. Mag., vol. 27, no. 6,pp. 100–109, Dec. 2007.[2] D. Siano, “Orientational analysis—A supple-ment to dimensional analysis I,” J. Franklin Inst.,vol. 320, pp. 267–283, 1985.[3] D. Siano, “Orientational analysis, tensor analysisand the group properties of the SI SupplementaryUnits II,” J. Franklin Inst., vol. 320, pp. 285–302, 1985.[4] H.E. Huntley, Dimensional Analysis. London:McDonald, 1952; reprinted by Dover, 1967.

Editor’s Note: Perhaps a rationale fororientational analysis can be found ingeometric algebra, where area is treatedas a bivector. See C. Doran and A.Lasenby, Geometric Algebra for Physicists,Cambridge University Press, 2005.

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