Brandenburg Technical University Cottbus
Department 1, Institute of Mathematics
Chair for Numerical Mathematics an Scientific Computing
Prof. Dr. G. Bader, Dr. A. Pawell
Problem Session to the Course: Mathematics I
Environmental and Resource Management WS 2002/03
Solutions to Sheet No. 13 (Deadline: January, 27/28 2002)
Homework
H 13.1: Eigenvalues of A: 1 = -1, 2/3 = 1
Eigenvectors:
1 = -1:
ëÅ‚ öÅ‚
2 0 0
1
íÅ‚ Å‚Å‚
0 1 1 = 0, = (0, 1, -1)T , = (0, 1, -1)T
x x c1 "
2
0 1 1
2/3 = 1:
1
= (0, 1, 1)T , = (1, 0, 0)T
c2 " c3
2
ëÅ‚ öÅ‚
0 0 1
1 1
" "
íÅ‚ 0 Å‚Å‚
C = .
2 2
1 1
- " "
0
2 2
q( = T A = x2 + 2x2x3.
x) x x
1
B = C, CT AC = diag(1, -1, -1)
2 2 2
q(C = (C A(C = T CT AC = T diag(1, -1, -1) = y1 - y2 - y3.
y) y)T y) y y y y
ëÅ‚ öÅ‚
1 -  1 1
íÅ‚ Å‚Å‚
H 13.2: det(A - E) = det 0 1 -  5 =
0 -1 -1 - 
(1 - )[(1 - )(-1 - ) + 5] = (1 - )[4 + 2] = 0
Ò! 1 = 1, 2/3 = Ä…2i
Eigenvectors:
ëÅ‚ öÅ‚
1 -  1 1
íÅ‚ Å‚Å‚
0 1 -  5 = 0
x
0 -1 -1 - 
1 = 1:
ëÅ‚ öÅ‚
0 1 1
íÅ‚ Å‚Å‚
0 0 5 , = t(1, 0, 0)T
x
0 -1 -2
1 = 2i:
ëÅ‚ öÅ‚
1 - 2i 1 1
íÅ‚ Å‚Å‚
0 1 - 2i 5 = 0
x
0 -1 -1 - 2i
ëÅ‚ öÅ‚
T
1 - 2i 1 1
2i
íÅ‚ Å‚Å‚
0 -1 -1 - 2i = 0, = , -1 - 2i, 1
x x
1 - 2i
0 0 0
1 = -2i:
ëÅ‚ öÅ‚
1 + 2i 1 1
íÅ‚ Å‚Å‚
0 1 + 2i 5 = 0
x
0 -1 -1 + 2i
ëÅ‚ öÅ‚
T
1 + 2i 1 1
-2i
íÅ‚ Å‚Å‚
0 -1 -1 + 2i = 0, = , -1 + 2i, 1
x x
1 + 2i
0 0 0

1 -  1
H 13.3: det(A - E) = det = (1 - )(3 - ) - 1 = 2 - 4 + 2.
1 3 - 
Eigenvalues:
"
1/2 = 2 Ä… 2
"
Eigenvectors:  = 2 + 2
"
"
-1 - 2

"1 0, 1 (-1 + 2, 1)T
x = x =
"
1 1 - 2
4 - 2 2
"
 = 2 - 2
"
"
-1 + 2

"1 0, 1 (-1 - 2, 1)T
x = x =
"
1 1 + 2
4 + 2 2
2
ëÅ‚ " " öÅ‚
-1+ 2 -1- 2
" "
" "
4-2 2 4+2 2
íÅ‚ Å‚Å‚
B
1 1
" "
" "
4-2 2 4+2 2
"
2 + 2 0
"
B-1 = BT , B-1AB =
0 2 - 2
Additional Problems
ëÅ‚ öÅ‚
5 -  2 0
íÅ‚ Å‚Å‚
P 13.1: det(A - E) = det 2 5 -  0 =
0 0 3 - 

(3 - ) (5 - )2 - 4 = 2 - 10 + 21 = 0
Ò! 1 = 7, 2,3 = 3.
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
-2 2 0 1
1
íÅ‚ Å‚Å‚
"
2 -2 0 = 0, = 1
x x1 íÅ‚ Å‚Å‚
2
0 0 -4 0
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
2 2 0 1 0
1
íÅ‚ Å‚Å‚
"
2 2 0 = 0, = -1 , = 0
x x2 íÅ‚ Å‚Å‚ x3 íÅ‚ Å‚Å‚
2
0 0 0 0 1
ëÅ‚ öÅ‚
1 1 0
1
íÅ‚ Å‚Å‚
C = " 1 -1
"0
2
0 0 2
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
1 1 0 5 2 0 1 1 0
1
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
CT AC = 1 -1
"0 2 5 0 1 -1 "0 =
2
0 0 2 0 0 3 0 0 2
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
14 0 0 7 0 0 1 0 0
1
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
0 6 0 = 0 3 0 = 0 2 0
2
0 0 6 0 0 6 0 0 3
b)
q( = T A = 5x2 + 4x1x2 + 5x2 + x2
x) x x
1 2 3
ëÅ‚ öÅ‚
7 0 0
2 2 2
íÅ‚ Å‚Å‚
q( := q(C = q( = T 0 3 0 = 7y1 + 3y2 + 3y3
y) y) x) y y
0 0 6
The transformation = C is an orthogonal transformation of the coor-
x y
dinate system. det C = 1 Ò! the transformation is a rotation of the
coordinate system.
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
1 1 0 0 1 0 1 -1 0
1 1
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
C = " 1 -1
"0 = 1 0 0 "2 1 1 "0 =
2
0 0 2 0 0 1 0 0 2
3
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
Ä„ Ä„
0 1 0 cos - sin 0
4 4

Ä„ Ä„
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
1 0 0 sin cos 0
4 4
0 0 1 0 0 1
Ä„
Reflection about x2 = x1 and a rotation by around the origin.
4

0, i = j

P 13.2: C = ( . . . , T = , A = i
c1, cn), ci cj ci ci.
1, i = j
CT AC = CT (A . . . , A = CT (1 . . . , n =
c1, cn) c1, cn)
ëÅ‚ öÅ‚
1 T . . . n T
c1 c1 c1 cn
íÅ‚ Å‚Å‚
. . . = diag(1, . . . , n).
1 T . . . n T
cn c1 cn cn
P 13.3: The quadratic form q is given by

1 -1
q( = x2 - 2x1x2 + 2x2 = T A = T
x) x x x x
1 2
-1 2
Eigenvalues of A
(1 - )(2 - ) - 1 = 1 - 3 + 2 = 0
"
3 5
1/2 = Ä…
2 2
Eigenvectors:
T
" "
3 5 1 1 5
 = + Ò! = - , 1
c1
"
2 2 2 2
10 5
-
4 2
T
" "
3 5 1 1 5
 = - Ò! = + , 1
c1
"
2 2 2 2
10 5
+
4 2
C = ( c2)
c1,

" "
3 5 3 5
CT AC = diag + , -
2 2 2 2

" "
3 5 3 5
2 2
q(C = (C AC = T CT AC = + y1 + - y2
y) y)T y y y
2 2 2 2
Problems can be downloaded from the internet site:
http://www.math.tu-cottbus.de/<"pawell/education/erm/erm.html
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