81904899
2.3 Matrix Calculus
The derivative or integral of a matrix is defined as differential or integration of each element of
the matrix. Other matrix functions, such as powers, polynomials and exponentials, can be
conveniently computed in terms of the eigenvalues of the matrix.
Consider a square matrix A(t) of order n with elements that are functions of time. The derivative
of A is defined as
 (t) = dA (t ) =  daij (t ) 
A
 dt 
dt


 t2
sin t 
 (t) =
Example: A(t) = 
 A
2 
cos
t
exp(
t
)


 2t
cos t 

2 
 sin t 2t exp( t )
The integral of A is
 A (t)dt
=
 a (t )dt 
ij
 t2
sin t 
Example: A(t) = 
 
cos
t
exp(
t
)


 t3

 cos t 

(t)dt
=
A

3
sin

 t exp( t ) 
A polynomial function of a matrix A is defined to be a function of the form
Pn(A) = anAn + an-1An-1 + ... + a0I
where the are scalars and A2 = AA, A3 = A2A, ..., An = An-1A
The exponential of a matrix A is defined in terms of the Maclaurin series for exp(x).

eA =
A A2
Ak
=
I
+
+
+ ....

1!
2!
k 0 k!
Cayley-Hamilton theorem: Every square matrix satisfies its own characteristic equation |A - I|
= 0.
Example 2.3-1 _____________________________________________________
Show that the matrix A where
 4  5
A= 

1  2
satisfies its own characteristic equation
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81904899
Solution
f() = |A - I| =
4
5
1
2
= 2 – 2 – 3 = 0
f(A) = A2 – 2A – 3I
 4  5
 1 0
 4  5
f(A) = 
– 2
– 3



0 1
1  2
1  2
2
11  10 8  10 3 0 0 0
f(A) = 
 – 
 – 
= 

 2  1  2  4  0 3 0 0
Example 2.3-2 _____________________________________________________
Show that the matrix A where
 0 1  i
A= 
1 
1  i
satisfies its own characteristic equation
Solution
f() = |A - I| =

1 i
1 i 1 
= – (1 – ) – (1 + i) (1 – i) = 2 –  – 2 = 0
f(A) = A2 – A – 2I
 0 1  i
 1 0
 0 1  i
f(A) = 
– 
– 2



1 
1  i
0 1
1  i 1 
2
 2 1  i   0 1  i   2 0   0 0
f(A) = 
–
–
=
3  1  i
1  0 2 0 0
1  i
The Halmilton-Cayley theorem may be used to determine the inverse of a nonsingular matrix A.
Let f() = n + an-1n-1 + ... + a1 + a0 = 0
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81904899
be the characteristic equation of A.
Since A is a nonsingular matrix, i  0; that is, every eigenvalue is nonzero, and (1)na0 = |A| 
0.
A n + an-1A n-1 + ... + a1A + a0I = 0
I =
1
(A n + an-1A n-1 + ... + a1A)
a0
A-1 = 
1
(A n-1 + an-1A n-2 + ... + a1I)
a0
Example 2.3-3 _____________________________________________________
Find the inverse of A where
 4  5
A= 

1  2
Solution
f() = |A - I| =
4
5
1
2
= 2 – 2 – 3 = 0
f(A) = A2 – 2A – 3I
1
1
I =  (A2 – 2A)  A-1 =  (A – 2I)
3
3
1 0  2 / 3  5 / 3
1 4  5
A-1 =  
 2


 = 
3  1  2
0 1  1 / 3  2 / 3
Some Properties of Eigenvalues
Consider the following eigenvalue problem with a square matrix A
Ax = x
Premultiplying both sides of the above equation by A gives
AAx = Ax = 2x
Similarly
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Amx = mx
Thus if  is an eigenvalue of A, m is an eigenvalue of Am where m is a positive integer.
Premultiplying both sides of (Ax = x) by A-1 yields
A-1Ax =  A-1x
1
x = -1x
or
A-1x =
Similarly
A-mx = -mx

Thus if  is an eigenvalue of A, -m is an eigenvalue of A-m where m is a positive integer. In
general
Aix = ix
where i can be positive or negative integer. Multiplying both sides of the above equation by i,
iAix = iix
Since the above equation is valid for any integer i, the summation of the equation over a range of
integer from  M to N is also valid.
N

iAix =
i M
iix
i M
N
The expression
N


ii is a polynomial Q() in terms of , and the matrix polynomial Q(A) is
i M
N
defined as Q(A) =

iAi. Thus if  is an eigenvalue of A, Q() is an eigenvalue of Q(A),
i M
Q(A)x = Q()x
Sylvester’s Theorem
The matrix polynomial might be evaluated using Sylvester’s theorem. Consider a polynomial of
degree (n  1):
Q(x) = a1x n-1 + a2x n-2 +... + an-1x + an
where the ai are constants. The polynomial can be written in Lagrangian form:
Q(x) = c1(x  2) (x  3)... (x  n) + c2(x  1) (x  3)... (x  n) + ...
ci(x  1) (x  2) ... (x  i-1) (x  i+1) ... (x  n) + ...
cn(x  2) (x  3)... (x  n-1)
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where i, i = 1, 2, ..., n are arbitrary scalars, while the constants ci are related to the constants ai.
Example 2.3-4 _____________________________________________________
Write the polynomial Q(x) = x 2  4x + 3 in the Lagrangian form
Solution
The Lagrangian form for Q(x) = x 2  4x + 3 is
Q(x) = c1(x  2) (x  3) + c2(x  1) (x  3) + c3(x  1) (x  2)
where i, i = 1, 2, 3 are arbitrary scalars. Let 1 = 1, 2 = 2, 3 = 3, then
Q(x) = c1(x  2) (x  3) + c2(x  1) (x  3) + c3(x  1) (x  2)
The constants ci can be evaluated from the relations
c1 + c2 + c3 = 1
5c1 + 4c2 + 3c3 = 4
6c1 + 3c2 + 2c3 = 3
to obtain: c1 = 0, c2 = 1, and c3 = 0. The Lagrangian form for the polynomial is
Q(x) = (x  1)(x  3)
Let 1 =  2, 2 =  1, 3 = 2, then
Q(x) = c1(x + 1) (x  2) + c2(x + 2) (x  2) + c3(x + 2) (x + 1)
The constants ci can be evaluated from the relations
c1 + c2 + c3 = 1
 c1 + 0c2 + 3c3 =  4
 2c1  4c2 + 2c3 = 3
to obtain: c1 = 3.7500, c2 = -2.6667, and c3 = -0.0833. The Lagrangian form for the polynomial is
Q(x) = 3.7500 (x + 1) (x  2)  2.6667 (x + 2) (x  2)  0.0833 (x + 2) (x + 1)
A short form notation for Q(x) is
n
Q(x) =
 ci
i 1
n
 (x  
k 1,#i
k
)
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n
where
 (x  
k 1,#i
k
) denotes product of all terms (x  k), for k varying from 1 to n except i. Let x
= i then
Q(i) = ci(i  1) (i  2) ... (i  i-1) (i  i+1) ... (i  n)
The constant ci can be expressed as
Q ( i )
ci =
n
 (
k 1,#i
i
 k )
Now, consider the matrix polynomial of degree (n – 1), where A is a square matrix of order n
Q(A) = c1(A  2I) (A  3)... (A  nI) + c2(A  1I) (A  3I)... (A  nI) + ...
ci(A  1I) (A  2I) ... (A  i-1I) (A  i+1I) ... (A  nI) + ...
cn(A  2I) (A  3I)... (A  n-1I)
n
Q(A) =
 ci
i 1
n
 ( A   I)
k
k 1,#i
Since i are arbitrary scalars, we can let i = i the eigenvalues of A and assume A have simple
eigenvalues (no repeated values). Then
n
Q(A) =
 ci
i 1
n
 ( A   I)
k
k 1,#i
Postmultiplying both sides by xm, the eigenvector corresponding to m yields
n
n
Q(A)xm =
 c  (A   I) xm
i 1
i
k
k 1,#i
Since m is an integer between 1 and n and Axm = mxm, every term in the sum
n
 ci
i 1
n
 (
k 1,#i
m
 k ) xm
contains the term (m - m) except for the term that arises from i = m. Therefore
n
Q(A)xm = cm
 (
k 1,#m
m
 k ) xm
Q(A)xm = cm (m  1) (m  2) ... (m  m-1) (m  m+1) ... (m  n)xm
From the relation Q(A)xm = Q(m)xm
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81904899
Q(m)xm = cm
n
 (
k 1,#m
m
 k ) xm
The coefficient cm is then
Q ( m )
cm =
n
 (
 k )
m
k 1,# m
, m = 1, 2, ..., n
or
ci =
Q (i )
n
 (
i
k 1,#i
,
i = 1, 2, ..., n
 k )
Therefore
n
n
Q(A) =
 c  ( A   I)
i 1
i
k 1,#i
k
Q (i )
n
=

i 1
n
 (i  k )
n
 ( A   I)
k 1,#i
k
k 1,#i
Example 2.3-5 _____________________________________________________
Find A100 where
 4  5
A= 

1  2
Solution
Q(A) = A100  Q(m) = (m)100
A has eigenvalues 1 = 3, 2 = –1
A100 = (1)100
( A  1I)
( A  2I)
+ (2)100
= 1.0e+047
( 1  2 )
( 2  1 )
6.4422  6.4422
1.2884  1.2884 


For n = 3
Q(A) = Q(1)
( A  1 I)( A  2 I)
( A  2 I )( A  3 I)
( A  1I)( A  3I)
+ Q(2)
+ Q(3)
( 1  2 )( 1  3 )
( 2  1 )( 2  3 )
(3  1 )( 3  2 )
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