Appendix A: Linear Algebra
In this appendix, we summarize some important results from vector and matrix algebra that are useful in our development in this book. Most of the vector and matrix properties presented in the following sections are elementary and can be found in standard texts on linear algebra.
A.1 Matrices
An m n matrix A is an ordered rectangular array that has m n elements. The matrix A can be written in the form
A ¼ a
ij¼
a
11a
12a
1na
21a
22a
2n⋮ ⋮ ⋱ ⋮
a
m1a
m2a
mn2 6 6 6 6 6 4
3 7 7 7 7
7 5 ðA:1Þ
The matrix A is called an m n matrix since it has m rows and n columns. The scalar element a
ijlies in the ith row and jth column of the matrix A. Therefore, the index i, which takes the values 1, 2, . . ., m, denotes the row number, while the index j, which takes the values 1, 2, . . ., n, denotes the column number.
A matrix A is said to be square if m ¼ n. An example of a square matrix is
A ¼
3 :0 2:0 0:95 6 :3 0 :0 10 :0 9 :0 3 :5 1 :25 2
6 4
3 7 5
In this example m ¼ n ¼ 3, consequently, A is a 3 3 matrix.
The transpose of an m n matrix A is an n m matrix denoted as A
Tand de fined as
# Springer Nature Switzerland AG 2019
A. Shabana, Vibration of Discrete and Continuous Systems,
Mechanical Engineering Series,https://doi.org/10.1007/978-3-030-04348-3
379
A
T¼ a
ji¼
a
11a
21a
m1a
12a
22a
m2⋮ ⋮ ⋱ ⋮
a
1na
2na
mn2 6 6 4
3 7 7
5 ðA:2Þ
For example, let A be the matrix
A ¼ 2 :0 4:0 7:5 23:5 0 :0 8 :5 10 :0 0 :0
The transpose of A is given by
A ¼
2 :0 0:0
4:0 8:5
7:5 10:0 23 :5 0:0 2
6 6 6 4
3 7 7 7 5
That is, the transpose of the matrix A is obtained by interchanging the rows and columns.
Definitions A square matrix A is said to be symmetric if a
ij¼ a
ji, that is, if the elements on the upper right half can be obtained by flipping the matrix about the diagonal. For example,
A ¼
3 :0 2:0 1:5
2:0 0 :0 2:3 1 :5 2 :3 1:5 2
6 4
3 7 5
is a symmetric matrix. Note that if A is symmetric, then A is the same as its transpose, that is, A ¼ A
T.
A square matrix is said to be an upper triangular matrix if a
ij¼ 0 for i > j. That is, every element below each diagonal element of an upper triangular matrix is zero. An example of an upper triangular matrix is
A ¼
6 :0 2:5 10:2 11:0 0 :0 8:0 5:5 6 :0 0 :0 0:0 3:2 4:0 0 :0 0:0 0:0 2:2 2
6 6 6 4
3 7 7 7 5
A square matrix is said to be a lower triangular matrix if a
ij¼ 0 for i < j. That is,
every element above the diagonal elements of a lower triangular matrix is zero. An
example of a lower triangular matrix is
A ¼
6 :0 0 :0 0 :0 0 :0 2 :5 8 :0 0 :0 0 :0 10 :2 5 :5 3 :2 0 :0
11:0 6:0 4:0 2:2 2
6 6 6 4
3 7 7 7 5
The diagonal matrix is a square matrix such that a
ij¼ 0 if i 6¼ j; that is, a diagonal matrix has element a
ijalong the diagonal with all other elements equal to zero. For example,
A ¼
5 :0 0:0 0:0 0 :0 1:0 0:0 0 :0 0:0 7:0 2
6 4
3 7 5
is a diagonal matrix.
The null matrix or the zero matrix is de fined to be the matrix in which all the elements are equal to zero. The unit matrix or the identity matrix is a diagonal matrix whose diagonal elements are nonzero and equal to one.
A skew-symmetric matrix is a matrix such that a
ij¼ a
ji. Note that since a
ij¼ a
jifor all i and j values, the diagonal elements should be equal to zero. An example of a skew-symmetric matrix ~ Ais
~A ¼
0 :0 3:0 5:0 3 :0 0 :0 2 :5 5 :0 2:5 0 :0 2
6 4
3 7 5
Observe that ~ A
T¼ ~A.
The trace of an n n square matrix A, denoted by tr A is the sum of the diagonal elements of A. The trace of A can thus be written as
tr A ¼ X
ni¼1
a
iiðA:3Þ
Note that the trace of an n n identity matrix is n, while the trace of a skew- symmetric matrix is zero.
A.2 Matrix Operations
In this section, we discuss some of the basic matrix operations which are used throughout this book.
Matrix Addition The sum of two matrices A and B, denoted by AþB is given by A þ B ¼ a
ijþ b
ijðA:4Þ
A.2 Matrix Operations 381
where b
ijare the elements of B. In order to add two matrices A and B, it is necessary that A and B have the same dimensions, that is, the same number of rows and the same number of columns. It is clear from Eq.
A.4that matrix addition is commuta- tive, that is,
A þ B ¼ B þ A ðA:5Þ
Example A.1
The two matrices A and B are defined as
A ¼ 3 :0 1:0 5:0 2 :0 0:0 2 :0
, B ¼ 2 :0 3:0 6 :0
3:0 0:0 5:0
The sum AþB is given by
A þ B ¼ 3 :0 1:0 5:0 2 :0 0:0 2 :0
þ 2 :0 3:0 6 :0
3:0 0:0 5:0
¼ 5 :0 4:0 1 :0
1:0 0:0 3:0
while A B is given by
A B ¼ 3 :0 1:0 5:0 2 :0 0:0 2 :0
2 :0 3:0 6 :0
3:0 0:0 5:0
¼ 1 :0 2:0 11:0 5 :0 0 :0 7 :0
Matrix Multiplication The product of two matrices A and B is another matrix C defined as
C ¼ AB ðA:6Þ
The element c
ijof the matrix C is defined by multiplying the elements of the ith row in A by the elements of the jth column in B according to the rule
c
ij¼ a
i1b
1 jþ a
i2b
2 jþ þ a
inb
nj¼ X
k
a
ikb
kjðA:7Þ
Therefore, the number of columns in A must be equal to the number of rows in B.
Observe that if A is an m n matrix and B is an n p matrix, then C is an m p matrix. Observe also that, in general, AB 6¼ BA. That is, matrix multiplication is not commutative. Matrix multiplication, however, is distributive, that is, if A and B are m p matrices and C is a p n matrix, then
A þ B
ð ÞC ¼ AC þ BC ðA:8Þ
Example A.2
Let
A ¼
0 4 1
2 1 1
3 2 1
2 6 4
3 7 5, B ¼
0 1 0 0 5 2 2 6 4
3 7 5
Then
AB ¼
0 4 1
2 1 1
3 2 1
2 6 4
3 7 5
0 1 0 0 5 2 2 6 4
3 7 5 ¼
5 2 5 4 5 5 2 6 4
3 7 5
Observe that the product BA is not defined in this example since the number of columns in B is not equal to the number of rows in A.
The associative law is valid for matrix multiplications. That is, if A is an m p matrix, B is a p q matrix, and C is a q n matrix, then (AB)C ¼ A(BC) ¼ ABC.
Determinant The determinant of an n n square matrix A, denoted as jAj, is a scalar de fined as
A j j ¼
a
11a
12a
1na
21a
22a
2n⋮ ⋮ ⋱ ⋮
a
n1a
n2a
nn2 6 6 6 6 6 4
3 7 7 7 7
7 5 ðA:9Þ
In order to be able to evaluate the unique value of the determinant of A, some basic de finitions have to be made first. The minor M
ijcorresponding to the element a
ijis the determinant formed by deleting the ith row and jth column from the original determinant jAj. The cofactor C
ijof the element a
ijis de fined as
C
ij¼ 1 ð Þ
iþjM
ijðA:10Þ Using this de finition of the cofactors C
ij, which are determinants of order n 1, the value of the determinant in Eq.
A.9can be obtained in terms of the cofactors of the elements of an arbitrary row i as follows:
A j j ¼ X
nj¼1
a
ijC
ijðA:11Þ
A.2 Matrix Operations 383
If A is a 2 2 matrix defined as
A ¼ a
11a
12a
21a
22,
the cofactors C
ijassociated with the elements of the first row are
C
11¼ 1 ð Þ
2a
22¼ a
22, C
12¼ 1 ð Þ
3a
21¼ a
21ðA:12Þ According to the de finition of Eq.
A.11, the determinant of the 22 matrix A can be determined using the cofactors of the elements of the first row as
A
j j ¼ a
11C
11þ a
12C
12¼ a
11a
22a
12a
21If A is a 3 3 matrix defined as
A ¼
a
11a
12a
13a
21a
22a
23a
31a
32a
332 6 4
3 7 5,
the determinant of A in terms of the cofactors of the first row is given by
A j j ¼ X
3j¼1
a
1 jC
1 j¼ a
11C
11þ a
12C
12þ a
13C
13where
C
11¼ a
22a
23a
32a
33, C
12¼ a
21a
23a
31a
33, C
13¼ a
21a
22a
31a
32That is, the determinant of A is
A j j ¼ a
11a
22a
23a
32a
33a
12a
21a
23a
31a
33þ a
13a
21a
22a
31a
32¼ a
11ð a
22a
33a
23a
32Þ a
12ð a
21a
33a
23a
31Þ þ a
13ð a
21a
32a
22a
31Þ
ðA:13Þ
One can show that the determinant of a square matrix is equal to the determinant of
its transpose, that is, | A| ¼ |A
T|, and the determinant of a diagonal matrix is equal to
the product of the diagonal elements. Furthermore, the interchange of any two
columns or rows changes only the sign of the determinant. One can also show that
if a matrix has two identical rows or two identical columns, the determinant of this
matrix is equal to zero. This can be demonstrated by the example of Eq.
A.13, forexample, if the second and third rows are identical, a
21¼ a
31, a
22¼ a
32, and
a
23¼ a
33. Using these equalities in Eq.
A.13, one can show that the determinant ofthe matrix A in this special case is equal to zero. A matrix whose determinant is equal to zero is said to be a singular matrix. For an arbitrary square matrix, singular or nonsingular, it can be shown that the value of the determinant does not change if any row or column is added to or subtracted from another.
Inverse of a Matrix A square matrix A
1that satis fies the relationship
A
1A ¼ AA
1¼ I ðA:14Þ
where I is the identity matrix, is called the inverse of the matrix A. The inverse of the matrix A is defined as
A
1¼ C
tA
j j ðA:15Þ
where C
tis the adjoint of the matrix A. The adjoint matrix C
tis the transposed matrix of the cofactors C
ijof the matrix A.
Example A.3
Determine the inverse of the matrix
A ¼
1 1 1
0 1 1
0 0 1
2 6 4
3 7 5
Solution. The determinant of the matrix A is equal to one, that is, |A| ¼ 1. The cofactors of the elements of the matrix A are
C
11¼ 1, C
12¼ 0, C
13¼ 0, C
21¼ 1 C
22¼ 1, C
23¼ 0, C
31¼ 0, C
32¼ 1 C
33¼ 1
The adjoint matrix, which is the transpose of the matrix of the cofactor elements, is given by
C
t¼
C
11C
21C
31C
12C
22C
32C
13C
23C
332
4
3
5 ¼ 1 1 0
0 1 1
0 0 1
2 4
3 5
Therefore,
(continued)
A.2 Matrix Operations 385
A
1¼ C
tA j j ¼
1 1 0
0 1 1
0 0 1
2 4
3 5
It follows that
A
1A ¼
1 1 0
0 1 1
0 0 1
2 4
3
5 1 1 1
0 1 1
0 0 1
2 4
3
5 ¼ 1 0 0
0 1 0
0 0 1
2 4
3
5 ¼ AA
1Note that if A is the 2 2 matrix
A ¼ a
11a
12a
21a
22, the inverse of A can be simply written as
A
1¼ 1 A j j
a
22a
12a
21a
11where | A| ¼ (a
11a
12a
12a
21). If the determinant of A is equal to zero, the inverse of A does not exist. This is the case of a singular matrix.
If A and B are nonsingular square matrices, then AB
ð Þ
1¼ B
1A
1ðA:16Þ
It can also be veri fied that
A
1T
¼ A
T 1ðA:17Þ That is, the transpose of the inverse of a matrix is equal to the inverse of its transpose.
A square matrix A is said to be orthogonal if
A
TA ¼ AA
T¼ I ðA:18Þ
In this case, A
T¼ A
1. That is, the inverse of an orthogonal matrix is equal to its transpose. An example of orthogonal matrices is
A ¼ cos θ sin θ sin θ cos θ
ðA:19Þ
For this matrix, one has
A
TA ¼ cos θ sin θ
sin θ cos θ
cos θ sin θ sin θ cos θ
¼ cos
2θ þ sin
2θ 0 0 sin
2θ þ cos
2θ
Using the trigonometric identity cos
2θ + sin
2θ ¼ 1, one obtains A
TA ¼ I, and the matrix A defined by Eq.
A.19is indeed an orthogonal matrix.
A.3 Vectors
An n-dimensional vector a is an ordered set
a ¼ a ð
1; a
2; . . . ; a
nÞ ðA:20Þ of n scalars. The scalar a
i,i ¼ 1,2,. . .,n, is called the ith component of a.
An n-dimensional vector can be considered as an n 1 matrix that consists of only one column. Therefore, the vector a can be written in the following column form
a ¼ a
1a
2⋮ a
n2 6 6 4
3 7 7
5 ðA:21Þ
The transpose of this column vector de fines the n-dimensional row vector a
T¼ [a
1a
2. . . a
n]. The vector a of Eq.
A.21can also be written as
a ¼ a ½
1a
2. . . a
nTðA:22Þ By considering the vector as special case of a matrix with only one column or one row, the rules of matrix addition and multiplication apply also to vectors. For example, if a and b are two n-dimensional vectors, defined, respectively, as
a ¼ a ½
1a
2. . . a
nT, b ¼ b ½
1b
2. . . b
nT, then aþb is defined as
a þ b ¼ a ½
1þ b
1a
2þ b
2a
nþ b
nTTwo vectors a and b are equal if and only if a
i¼ b
ifor i ¼ 1,2,. . ., n.
The product of a vector a and scalar α is the vector
αa ¼ αa ½
1αa
2. . . αa
nTðA:23Þ
A.3 Vectors 387
The dot, inner, or scalar product of two vectors a ¼ a ½
1a
2. . . a
nTand b ¼ b ½
1b
2. . . b
nTis de fined by the following scalar quantity
ab ¼ a
Tb ¼ a ½
1a
2. . . a
nb
1b
2⋮ b
n2 6 6 4
3 7 7 5
¼ a
1b
1þ a
2b
2þ þ a
nb
nðA:24Þ
which can be written as
ab ¼ a
Tb ¼ X
ni¼1
a
ib
iðA:25Þ
It follows that a b ¼ b a.
The length of a vector a, denoted as jaj, is defined as the square root of the dot product of a with itself, that is,
a
j j ¼ ffiffiffiffiffiffiffiffi a
Ta
p ¼ a
21þ a
22þ þ a
2n1=2
ðA:26Þ
The terms modulus, magnitude, norm, and absolute value of a vector are also used to denote the length of a vector.
Example A.4
Let a and b be the two vectors
a ¼ 0 1 3 2 ½
T, b ¼ 1 0 2 3 ½
Tthen
a þ b ¼ 0 1 3 2 ½
Tþ 1 0 2 3 ½
T¼ 1 1 5 5 ½
TThe dot product of a and b is
a b ¼ a
Tb ¼ 0 1 3 2 ½
1 0 2 3 2 6 6 6 6 4
3 7 7 7 7 5
¼ 0 þ 0 þ 6 þ 6 ¼ 12 The length of the vectors a and b is defined as
(continued)
a
j j ¼ ffiffiffiffiffiffiffiffi a
Ta
p ¼ 0 h ð Þ
2þ 1 ð Þ
2þ 3 ð Þ
2þ 2 ð Þ
2i
1=2¼ ffiffiffiffiffi p 14
3:742 b
j j ¼ ffiffiffiffiffiffiffiffi b
Tb
p ¼ 1 h ð Þ
2þ 0 ð Þ
2þ 2 ð Þ
2þ 3 ð Þ
2i
1=2¼ ffiffiffiffiffi p 14
3:742
Differentiation In many applications in mechanics, scalar and vector functions that depend on one or more variables are encountered. An example of a scalar function that depends on the system velocities and possibly the system coordinates is the kinetic energy. Examples of vector functions are the coordinates, velocities, and accelerations that depend on time.
Let us first consider a scalar function f that depends on several variables q
1, q
2, . . ., q
nand the parameter t, that is,
f ¼ f q ð
1; q
2; . . . q
n; t Þ ðA:27Þ where q
1, q
2, . . ., q
nare functions of t, that is, q
i¼ q
i(t).
The derivative of f with respect to t is df
dt ¼ ∂f
∂q
1dq
1dt þ ∂f
∂q
2dq
2dt þ þ ∂f
∂q
ndq
ndt þ ∂f
∂t ðA:28Þ
which can be written using vector notation as
df dt ¼ ∂f
∂q
1∂f
∂q
2∂f
∂q
ndq
1dt dq
2dt
⋮ dq
ndt 2 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 5
þ ∂f
∂t ðA:29Þ
This equation can be written compactly as df dt ¼ ∂f
∂q d q
dt þ ∂f
∂t ðA:30Þ
in which ∂f/∂t is the partial derivative of f with respect to t and q ¼ q ½
1q
2q
nT∂f
∂q ¼ f
q¼ ∂f
∂q
1∂f
∂q
2∂f
∂q
n9 >
=
> ; ðA:31Þ
The second equation in this equation de fines the partial derivative of a scalar function with respect to a vector as a row vector. Note that if f is not an explicit function of time ∂f/∂t ¼ 0.
A.3 Vectors 389
Example A.5
Consider the function
f q ð
1; q
2; t Þ ¼ q
21þ 3q
32t
2where q
1and q
2are functions of the parameter t. The total derivative of f with respect to the parameter t is
df dt ¼ ∂f
∂q
1dq
1dt þ ∂f
∂q
2dq
2dt þ ∂f
∂t where
∂f
∂q
1¼ 2q
1, ∂f
∂q
2¼ 9q
22, ∂f
∂t ¼ 2t Hence,
df dt ¼ 2q
1dq
1dt þ 9q
22dq
2dt 2t
¼ 2q
19q
22dq
1dt dq
2dt 2 6 6 4
3 7 7 5 2t
where ∂f/∂q can be recognized as the row vector
∂f
∂q ¼ f
q¼ 2q
19q
22Consider the case of vector functions that depend on several variables. These vector functions can be written as
f
1¼ f
1q
1; q
2; . . . ; q
n,t f
2¼ f
2q
1; q
2; . . . ; q
n,t
⋮
f
m¼ f
mð q
1; q
2; . . . ; q
n; t Þ 9 >
> >
=
> >
> ;
ðA:32Þ
where q
i¼ q
i(t),i ¼ 1,2,. . ., n. Using the procedure previously outlined in this section,
the total derivative of an arbitrary function f
jcan be written as
df
jdt ¼ ∂f
j∂q d q
dt þ ∂f
j∂t , j ¼ 1,2, . . . , m ðA:33Þ in which ∂f
j/ ∂q is the row vector
∂f
j∂q ¼ ∂f
j∂q
1∂f
j∂q
2∂f
j∂q
nðA:34Þ
Consequently,
d f dt ¼
df
1dt df
2dt
⋮ df
mdt 2 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 5
¼
∂f
1∂q
1∂f
1∂q
2∂f
1∂q
n∂f
2∂q
1∂f
2∂q
2∂f
2∂q
n⋮ ⋮ ⋱ ⋮
∂f
m∂q
1∂f
m∂q
2∂f
m∂q
n2 6 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 7 5
dq
1dt dq
2dt
⋮ dq
ndt 2 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 5
þ
∂f
1∂t
∂f
2∂t
⋮
∂f
m∂t 2 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 5
ðA:35Þ
where
f ¼ f ½
1f
2f
mTðA:36Þ Equation
A.35can be written compactly as
d f dt ¼ ∂f
∂q d q
dt þ ∂f
∂t ðA:37Þ
where the m n matrix ∂f/∂q, the n-dimensional vector dq/dt, and the m-dimen- sional vector ∂f/∂t can be recognized, respectively, as
∂f
∂q ¼ f
q¼
∂f
1∂q
1∂f
1∂q
2∂f
1∂q
n∂f
2∂q
1∂f
2∂q
2∂f
2∂q
n⋮ ⋮ ⋱ ⋮
∂f
m∂q
1∂f
m∂q
2∂f
m∂q
n2 6 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 7 5
, d q dt ¼ q
t¼
dq
1dt dq
2dt
⋮ dq
ndt 2 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 5
, ∂f
∂t ¼ f
t¼
∂f
1∂t
∂f
2∂t ⋮
∂f
m∂t 2 6 6 6 6 6 6 6 6 6 4
3 7 7 7 7 7 7 7 7 7 5
ðA:38Þ
If f
jis not an explicit function of the parameter t, then ∂f
j/ ∂t is equal to zero.
Note also that the partial derivative of an m-dimensional vector function f with respect to an n-dimensional vector q is the m n matrix f
qde fined in the preceding equation.
A.3 Vectors 391
Example A.6
Consider the vector function f defined as
f ¼ f
1f
2f
32 6 4
3 7 5 ¼
q
21þ 3q
32t
28q
213t 2q
216q
1q
2þ q
222 6 4
3 7 5
The total derivative of the vector function f is
d f dt ¼
df
1dt df
2dt df
3dt 2 6 6 6 6 6 6 4
3 7 7 7 7 7 7 5
¼
2q
19q
2216q
10
4q
16q
2ð Þ ð 2q
26q
1Þ 2
6 4
3 7 5
dq
1dt dq
2dt 2 6 4
3 7 5 þ
2t
3 0 2 6 4
3 7 5
where the matrix f
qcan be recognized as
f
q¼
2q
19q
2216q
10
4q
16q
2ð Þ ð 2q
26q
1Þ 2
6 4
3 7 5
and the vector f
tis
∂f
∂t ¼ f
t¼ 2t 3 0 ½
TIn the analysis of mechanical systems, one may also encounter scalar functions in the form Q ¼ q
TAq. Following a similar procedure to the one previously outlined in this section, one can show that ∂Q/∂q ¼ q
T( AþA
T). If A is a symmetric matrix, that is, A ¼ A
T, one has ∂Q/∂q ¼ 2q
TA.
Linear Independence The concepts to be introduced here are of fundamental importance in the development presented in this book. Their use is crucial in formulating many of the dynamic relationships presented in several chapters of this text.
The vectors a
1, a
2, . . ., a
nare said to be linearly dependent if there exist scalars e
1, e
2, . . . , e
n, which are not all zeros, such that
e
1a
1þ e
2a
2þ þ e
na
n¼ 0 ðA:39Þ
Otherwise, the vectors a
1, a
2, . . ., a
nare said to be linearly independent. In the case
of linearly independent vectors, none of these vectors can be expressed in terms of
the others. On the other hand, if Eq.
A.39holds, and not all the scalars e
1, e
2, . . ., e
nare equal to zero, one or more of the vectors a
1, a
2, . . ., a
ncan be expressed in terms of the other vectors.
Equation
A.39can be written in matrix form as
a
1a
2a
n½
e
1e
2⋮ e
n2 6 6 6 4
3 7 7
7 5 ¼ 0 ðA:40Þ
which can be written compactly as Ae ¼ 0, where e ¼ e ½
1e
2e
nTand the columns of the coef ficient matrix A are the vectors a
1, a
2, . . ., a
n, that is, A ¼ a ½
1a
2a
n. If the vectors a
1, a
2, . . ., a
nare linearly dependent, the system of homogeneous algebraic equations Ae ¼ 0 has a nontrivial solution. On the other hand, if the vectors a
1, a
2, . . ., a
nare linearly independent vectors, then A must be a nonsingular matrix since the system of homogeneous algebraic equations Ae ¼ 0 has only the trivial solution e ¼ A
10 ¼ 0. Consequently, in the case where the vectors a
1, a
2, . . ., a
nare linearly dependent, the square matrix A must be singular. The number of linearly independent columns in a matrix is called the column rank of the matrix, while the number of independent rows is called the row rank of the matrix. It can be shown that for any matrix, the row rank is equal to the column rank and is equal to the rank of the matrix. Therefore, a square matrix which has a full rank is a matrix which has linearly independent rows and linearly independent columns. One concludes, therefore, that a matrix which has a full rank is a nonsingular matrix. Consequently, if a
1, a
2, . . ., a
nare n-dimensional linearly independent vectors, any other n-dimensional vector can be expressed as a linear combination of these vectors. For instance, let b be another n-dimensional vector.
We show that this vector has a unique representation in terms of the linearly independent vectors a
1, a
2, . . ., a
n. To this end, we write b as
b ¼ x
1a
1þ x
2a
2þ þ x
na
nðA:41Þ where x
1, x
2, . . ., x
nare scalars. In order to show that x
1, x
2, . . ., x
nare unique, Eq.
A.41can be written as b ¼ Ax, where A ¼ a ½
1a
2. . . a
nis a square matrix and x ¼ x ½
1x
2x
nT. Since the vectors a
1, a
2, . . ., a
nare assumed to be linearly independent, the coef ficient matrix A has a full row rank and, therefore, it is nonsingular. The system of algebraic equations Ax ¼ b has a unique solution x which can be written as x ¼ A
1b. That is, an arbitrary n-dimensional vector b has a unique representation in terms of the linearly independent vectors a
1, a
2, . . ., a
n. A familiar and important special case is the case of three-dimensional vectors.
One can show that the three vectors
a
1¼ 1 0 0 2 4
3 5, a
2¼
0 1 0 2 4
3
5, a
3¼ 0 0 1 2 4
3 5,
A.3 Vectors 393
are linearly independent. Any other three-dimensional vector b ¼ b ½
1b
2b
3Tcan be written in terms of the linearly independent vectors a
1, a
2, and a
3as b ¼ b
1a
1+ b
2a
2+ b
3a
3, where the coef ficients x
1, x
2, and x
3can be recognized in this special case as x
1¼ b
1, x
2¼ b
2, and x
3¼ b
3. The coef ficients x
1, x
2, and x
3are called the coordinates of the vector b in the basis defined by the vectors a
1, a
2, and a
3.
Example A.7
Show that the vectors
a
1¼ 1 0 0 2 6 4
3 7 5, a
2¼
1 1 0 2 6 4
3
7 5, a
3¼ 1 1 1 2 6 6 4
3 7 7 5
are linearly independent. Find also the representation of the vector b ¼ 1 3 0 ½
Tin terms of the vectors a
1, a
2, and a
3.
Solution. In order to show that the vectors a
1, a
2, and a
3are linearly indepen- dent, we must show that the relationship
e
1a
1þ e
2a
2þ e
3a
3¼ 0 holds only when e
1¼ e
2¼ e
3¼ 0. To this end, we write
e
11 0 0 2 6 4
3 7 5 þ e
21 1 0 2 6 4
3 7 5 þ e
31 1 1 2 6 4
3 7 5 ¼ 0
which leads to
e
1þ e
2þ e
3¼ 0 e
2þ e
3¼ 0 e
3¼ 0 Back substitution shows that
e
3¼ e
2¼ e
1¼ 0
That is, the vectors a
1, a
2, and a
3are linearly independent.
In order to find the unique representation of the vector b in terms of these linearly independent vectors, we write
(continued)
b ¼ x
1a
1þ x
2a
2þ x
3a
3which can be written in a matrix form as b ¼ Ax, where
A ¼
1 1 1
0 1 1
0 0 1
2 6 4
3 7 5, b ¼
1 3 0 2 6 4
3 7 5
Therefore, the vector of coordinates x can be obtained as
x ¼ x
1x
2x
32 6 4
3
7 5 ¼ A
1b ¼
1 1 1
0 1 1
0 0 1
2 6 4
3 7 5
1 3 0 2 6 4
3 7 5 ¼
4 3 0 2 6 4
3 7 5
A.4 Eigenvalue Problem
In the analysis of structural systems, we often encounter a system of homogeneous algebraic equations in the form
Ay ¼ λy ðA:42Þ
where A is a square matrix, y is an unknown vector, and λ is an unknown scalar.
Equation
A.42can be written in the form A λI
ð Þy ¼ 0 ðA:43Þ
where I is the identity matrix. Equation
A.43represents an algebraic system of homogeneous equations which have a nontrivial solution if and only if the coef fi- cient matrix ( A λI) is singular. That is, the determinant of this matrix is equal to zero. This leads to
A λI
j j ¼ 0 ðA:44Þ
This is called the characteristic equation of the matrix A. If A is an n n matrix, Eq.
A.44is a polynomial of order n in λ. This equation can be written in the following form:
a
nλ
nþ a
n1λ
n1þ þ a
0¼ 0 ðA:45Þ where a
i, i ¼ 0, 1, 2, . . ., n, are the scalar coefficients of the polynomial. The solution of Eq.
A.45de fines n roots λ
1, λ
2, . . ., λ
n. These roots are called the characteristic values or the eigenvalues of the matrix A. Associated with each of these eigenvalues,
A.4 Eigenvalue Problem 395
there is an eigenvector y
iwhich can be determined to within an arbitrary constant by solving the system of equations
A λ
iI
ð Þy
i¼ 0 ðA:46Þ
If A is a real-symmetric matrix, one can show that the eigenvectors associated with distinctive eigenvalues are orthogonal, that is,
y
iTy
j¼ 0 if i 6¼ j y
iTy
j6¼ 0 if i ¼ j
)
ðA:47Þ
Example A.8
Find the eigenvalues and eigenvectors of the matrix
A ¼ 4 1 2
1 0 0
2 0 0
2 4
3 5
Solution. The characteristic equation of this matrix is
A λI
j j ¼
4 λ 1 2
1 λ 0
2 0 λ
¼ 4 λ ð Þλ
2þ λ þ 4λ ¼ 0 This equation can be rewritten as
λ λ 5 ð Þ λ þ 1 ð Þ ¼ 0 which has the roots
λ
1¼ 0, λ
2¼ 5, λ
3¼ 1
The ith eigenvector associated with the eigenvalue λ
ican be obtained using the equation
A λ
iI ð Þy
i¼ 0 The solution of this equation yields
y
1¼ 0 2
1 2 4
3 5, y
2¼
5 1 2 2 4
3
5, y
3¼ 1
1 2 2 4
3
5
Problems
A.1. Find the sum of the following two matrices
A ¼ 3:0 8 :0 20:5 5 :0 11:0 13 :0 7 :0 20:0 0 2
4
3
5, B ¼ 0 3 :2 0
17:5 5:7 0
12 :0 6:8 10:0 2
4
3 5
Evaluate also the determinant and the trace of A and B.
A.2. Find the product AB and BA, where A and B are given in Problem 1.
A.3. Find the inverse of the following matrices:
A ¼ 1 2 1
2 1 0
0 1 1
2 4
3
5, B ¼ 0 3 5
2 2 3
6 2 0
2 4
3 5
A.4. Show that an arbitrary square matrix A can be written as A ¼ A
1þ A
2where A
1is a symmetric matrix and A
2is a skew-symmetric matrix.
A.5. Show that the interchange of any two rows or columns of a square matrix changes only the sign of the determinant.
A.6. Show that if a matrix has two identical rows or two identical columns, the determinant of this matrix is equal to zero.
A.7. Let
A ¼ A
11A
12A
21A
22be a nonsingular matrix. If A
11is square and nonsingular, show by direct matrix multiplications that
A
1¼ A
111þ B
1H
1B
2B
1H
1H
1B
2H
1where
B
1¼ A
111A
12, B
2¼ A
21A
111H ¼ A
22B
2A
12¼ A
22A
21B
1¼ A
22A
21A
111A
12Problems 397
A.8. Let a and b be the two vectors
a ¼ 1 0 3 2 5 ½
T, b ¼ 0 1 2 3 8 ½
TFind aþb, a b, |a| and |b|.
A.9. Find the total derivative of the function
f q ð
1; q
2; q
3; t Þ ¼ q
1q
33q
22þ 5t
5with respect to the parameter t. De fine also the partial derivative of the function f with respect to the vector q(t), where q t ð Þ ¼ q ½
1ð Þ q t
2ð Þ q t
3ð Þ t
T. A.10. Find the total derivative of the vector function
f ¼ f
1f
2f
32 4
3
5 ¼ q
21þ 3q
225q
34þ t
3q
22q
23q
1q
4þ q
2q
3þ t 2
4
3 5
with respect to the parameter t. De fine also the partial derivative of the function f with respect to the vector q ¼ q ½
1q
2q
3q
4T.
A.11. Let Q ¼ q
TAq, where A is an n n square matrix and q is an n-dimensional vector. Show that
∂Q
∂q ¼ q
TA þ A
TA.12. Show that the vectors
a
1¼ 0 0 1 2 4
3 5, a
2¼
0 1 1 2 4
3
5, a
3¼ 1 1 1 2 4
3 5
are linearly independent. Determine also the coordinates of the vector b ¼ 1 5 3 ½
Tin the basis a
1, a
2, and a
3.
A.13. Find the rank of the following matrices
A ¼
2 5 1 6 9 3 4 0 2 2
4
3
5, B ¼ 3 5 1 0
2 0 1 3
7 1 2 9
2 4
3
5
A.14. Find the eigenvalues and eigenvectors of the following two matrices
A ¼
2 1 0
1 2 1
0 1 1
2 4
3
5, B ¼ 6 2 0
2 2 3
4 4 3
2 4
3 5
A.15. Show that if A is a real-symmetric matrix, then the eigenvectors associated with distinctive eigenvalues are orthogonal.
Problems 399
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Index
A
Absolute nodal coordinate formulation (ANCF), 98, 281, 329–333 Analysis of higher modes, 169, 232 Angular acceleration, 6, 9, 12, 101 Angular oscillations, 23, 51, 58, 112, 135,
187, 200
Approximation methods, 201, 255–264, 334 Arbitrary forcing function, 1, 39–42, 265 Assumed displacementfield, 269, 279, 281,
286, 287, 289, 296, 313, 314, 323, 326 Assumed-modes method, 269–271
B
Beam element, 280, 285–287, 289, 291–293, 298, 307, 308, 316, 324, 330, 332, 337–342
Boolean matrix, 291, 296
Boundary conditions, 184, 188, 201, 205, 208–210, 212, 213, 216, 218–220, 226, 227, 229, 235, 238, 240, 241, 243, 245, 247, 248, 253, 256–260, 265, 266, 268, 269, 272–274, 313, 335
Brick element, see Solid element
C
Cayley–Hamilton theorem, 347, 348, 376 Characteristic equation, 17, 18, 20, 21, 119,
121, 137, 138, 159, 165, 184, 186, 189, 206, 264, 313, 314, 320, 355, 395 Characteristic matrix, 155, 159, 343, 346, 350,
351, 353, 354, 357, 360, 361, 376 Characteristic polynomial, 345–349, 356 Characteristic values, 119, 357, 395 Characteristic vector, 119
Coefficient of sliding friction, 26 Completeness, 287
Concentrated loads, 242 Condition for similarity, 350 Connectivity conditions, 290, 291 Connectivity of thefinite elements, 279,
290–296
Conservation of energy, 55, 90, 107, 140–143, 253, 255, 256
Conservation theorems, 55, 88–93 Conservative systems, 55, 88, 89, 93 Consistent-mass formulation, 296 Constant strain triangular element, 283 Constitutive equations, 301, 303, 333 Constrained motion, 7
Continuous systems, 1, 11, 16, 42, 192, 201–271, 279–281, 300, 314 Convergence of thefinite-element solution,
281, 313–317 Convolution integral, 40
Coordinate reduction, 270, 334, 335 Coordinate transformation, 125, 290, 329,
330, 335
Coulomb damping, 1, 26, 28
Critical damping coefficient, 18–24, 30, 35, 39, 50
Critically damped systems, 20, 21, 48
D
Damped natural frequency, 22, 38, 39, 48, 147
Damping
general viscous, 108, 152
proportional damping, 107, 146, 147, 161 viscous damping, 17, 24–26, 28, 108, 151 Damping factor, 18–24, 31, 33, 35, 38,
39, 48, 147, 148
Damping matrix, 146, 147, 151, 154, 310 Deformation modes, 134, 139, 176, 229, 239 Degree of freedom, 7
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