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EEL3105 Homework 5 Solutions
A
^
B
^
(i)
AB
^
BA
^
AB BA
^
^
trace AB trace BA trace AB BA
(ii)
11 12 1 21 22 2
1 2
n n
n n nn
a a a a a a A
a a a
^
11 12 1 21 22 2
1 2
n n
n n nn
b b b b b b B
b b b
^
1 1 1
1 2 2
1
.. .. .. ..
. .. .. .. ..... .....
.. ....
n j j^ j n j j^ j
n j nj^ jn
a b
a b
AB
a b
^
11 1
1 2 2
1
.. .. .. ..
. .. .. .. ..... .....
.. ....
n j j^ j n j j^ j
n j nj^ jn
b a
b a
BA
b a
1 1 1 1 1 1
1 2 2 1 2 2
1 1
n n j j^ j^ j j^ j n n j j^ j^ j j^ j
n n j nj^ jn^ j nj^ jn
a b b a
a b b a
AB BA
a b b a
n n n n n n
trace AB BA j a bj j j b aj j j a j bj j b j a j j a bnj jn j b anj jn
(iii)
11 12 1 21 22 2
1 2
n n
m m mn
a a a a a a A
a a a
(A, mxn matrix)
11 12 1 21 22 2
1 1
m m
n n nm
b b b b b b B
b b b
(B, nxm matrix)
1 1 1
1 2 2
1
.. .. .. ..
. .. .. .. ..... .....
.. ....
n j j^ j n j j^ j
n j mj^ jm
a b
a b
AB
a b
11 1
1 2 2
1
.. .. .. ..
. .. .. .. ..... .....
.. ....
m j j^ j m j j^ j
m j nj^ jn
b a
b a
BA
b a
Note that AB is mxm and BA is nxn.
1 1 2 2 1 1 1 1 1
n n n m n j j j j nj jn ij ji j j j i j
trace AB a b a b a b a b
For matrix B,det(B) 0 so an inverse matrix for B exist.
Using the above general formula of the inverse of a matrix B is
1
B
^
^
Let’s verify this answer using MATLAB. Below is the output from MATLAB command window:
===========================MATLAB Command Window==========================
A=[1 3 ‐1; 2 6 ‐2; ‐ 1 ‐ 3 1];
B=[2 4 6; 1 3 5; 3 5 8];
det(A)
ans =
0
det(B)
ans =
2
det(A*B)
ans =
0
det(B*A)
ans =
0
inv(B)
ans =
‐0.5000 ‐1.0000 1. 3.5000 ‐1.0000 ‐2. ‐2.0000 1.0000 1.
===============================================================================
- Eigenvalues of A ,B, AB , BA through Matlab are as below (where subscripted lambda represent the eigenvalue) =====================MATLAB Command Window Output========================
eig(A)
ans =
‐0.
eig(B)
ans =
‐0.0440 + 0.3884i
det( ) 0 80 ( ) 1313.0881+-0.0440 + 0.3884i+-0.0440 - 0.3884i det( ) 213.0881(-0.0440 + 0.3884i)(-0.0440 - 0.3884i) ( ) 5 5 0 0 det( ) 0 50 ( ) 5 5 0 0 ( ) de
trace A A trace B B trace AB AB trace BA trace AB
1 1
t( ) 0 50 det( ) 0.5 1/ det( ) ( ) : 0.0764 1/13. -0.2882 + 2.5418i 1/ (-0.0440 - 0.3884i) -0.2882 - 2.5418i=1/(-0.0440 + 0.3884i)
BA
B B
eig B
Note that eigenvalues of the inverse of B are reciprocals of eigenvalues of B. This is a general property of eigenvalues of inverse of a matrix and the original matrix, i. e., if is an eigenvalue of an invertible matrix M, then 1/ is an eigenvalue of M‐^1. Can you try to prove this fact? Note that the determinant of inverse of B is the reciprocal of the determinant of B. Can you try to prove this for a general invertible matrix? Hint: use A(A‐^1 ) = I.
a c A c b
^ ^
i. For A to be invertible applying the condition for a matrix to be invertible: det( A ) (^0) , ab c^2 0
ii. For finding the eigenvalues we can use the characteristic polynomial equation det( I A ) 0 , and find its roots: 2 2 2 2 2 2 2 1 2
det( ) 0 ( )( ) 0, ( ) 0 ( ) (( ) 4 ) ( ) (( ) 4 ) , 2 2
I A a b c a b ab c a b a b c a b a b c
Note that eigenvalue are real regardless of values of a, b, c. This makes sense since A is a symmetric matrix. As I taught in class, a symmetric matrix is guaranteed to have real eigenvalues. And we see that this is true. In general, a quadratic equation may have complex roots. But here, by the very nature of the specific quadratic equation arising from the characteristic polynomial of a symmetric matrix, this cannot happen!
iii. Condition of invertibility det( A ) 0 ab c^2 0 Note that the determinant is the product of eigenvalues of A. Let us see if we get the same condition for invertibility.
2 2 2 2 1 2
(^1) ( ) ( ) 4 det( ) 0 4
^ a b a b c ab c A
