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: λ1 = −1, λ2/3 = 1
Eigenvectors:
λ1 = −1:

2
 0
0

0
1
1


0
1  x = 0,
1

x = (0, 1, −1)T ,

1 c1 = √ (0, 1, −1)T
2

λ2/3 = 1:
1
c2 = √ (0, 1, 1)T ,
2

C=

0
1
√
2
1
− √2

c3 = (1, 0, 0)T


0 1
0 .
0

1
√
2
1
√
2

q(x) = xT Ax = x2 + 2x2 x3 .
1
B = C,

C T AC = diag(1, −1, −1)

2
2
2 q(Cy) = (Cy)T A(Cy) = y T C T ACy = y T diag(1, −1, −1)y = y1 − y2 − y3 .



H 13.2:

1−λ det(A − λE) = det  0
0

1
1−λ
−1


1
=
5
−1 − λ

(1 − λ)[(1 − λ)(−1 − λ) + 5] = (1 − λ)[4 + λ2 ] = 0

⇒

λ2/3 = ±2i

λ1 = 1,

Eigenvectors:


1−λ
 0
0

1
1−λ
−1


1
x = 0
5
−1 − λ

λ1 = 1:



0
1
1
 0
0
5 ,
0 −1 −2

x = t(1, 0, 0)T

λ1 = 2i:



1 − 2i
1
1

x = 0
0
1 − 2i
5
0
−1
−1 − 2i




1 − 2i 1
1

0
−1 −1 − 2i  x = 0, x =
0
0
0

2i
, −1 − 2i, 1
1 − 2i

T

−2i
, −1 + 2i, 1
1 + 2i

T

λ1 = −2i:


1 + 2i
1
1

x = 0
0
1 + 2i
5
0
−1
−1 + 2i



1 + 2i 1
1

0
−1 −1 + 2i  x = 0, x =
0
0
0
det(A − λE) = det

H 13.3:

1−λ
1

1
3−λ

= (1 − λ)(3 − λ) − 1 = λ2 − 4λ + 2.

Eigenvalues: λ1/2 = 2 ±

√

2

Eigenvectors: λ = 2 +
−1 −

λ=2−

√

√

√

2

2
1

√1
1− 2

x = 0,

x=

2
1

√1
1+ 2

x = 0,

x=

1

√

1

√

√ (−1 +
4−2 2

2, 1)T

2

−1 +

√

2

√ (−1 −
4+2 2

2, 1)T


B

−1+
√

√

2
√
4−2 2
√ 1 √
4−2 2

B −1 = B T ,

√

−1− 2
√ √



√



4+2 2
1
√
4+2 2

√
2+ 2
0

B −1 AB =

0√
2− 2

Additional Problems


5−λ
2
det(A − λE) = det 
0

P 13.1:


2
0
5−λ
0 =
0 3−λ

(3 − λ) (5 − λ)2 − 4 = λ2 − 10λ + 21 = 0
⇒

λ1 = 7,

λ2,3 = 3.
 
1
1  
1
x1 = √
2
0




−2
2
0
 2 −2
0  x = 0,
0
0 −4


2
 2
0

2
2
0


0
0  x = 0,
0



1
1 
−1  , x2 = √
2
0




0
x3 =  0 
1



1
1
0
1 
1 −1 √0 
C=√
2
0
0
2


1
1
0
5
1
1 −1 √0   2
C T AC =
2
0
0
0
2

 
14 0 0
7
1
0 6 0 = 0
2
0 0 6
0

0
3
0

2
5
0



1
1
0
0
0   1 −1 √0  =
3
0
0
2

 

0 λ1 0
0
0  =  0 λ2
0 
6
0
0 λ3

b) q(x) = xT Ax = 5x2 + 4x1 x2 + 5x2 + x2
1
2
3


7 0 0
2
2
2
q(y) := q(Cy) = q(x) = y T  0 3 0  y = 7y1 + 3y2 + 3y3
0 0 6
The transformation x = Cy is an orthogonal transformation of the coordinate system. det C = 1 ⇒ the transformation is a rotation of the coordinate system.

 



1 −1
0
1
1
0
0 1 0
1 
1 
1
1 √0  =
1 −1 √0  =  1 0 0  √
C=√
2
2
0 0 1
2
2
0
0
0
0
3



0
 1
0


1 0 cos 0 0   sin
0 1

π
4
π
4

π
4
π
4

− sin cos 0

0

Reflection about x2 = x1 and a rotation by
P 13.2:

C = (c1 , . . . , cn ),


0
0 
1
π
4

around the origin.

0, i = j
,
1, i = j

cT cj = i Aci = λi ci .

C T AC = C T (Ac1 , . . . , Acn ) = C T (λ1 c1 , . . . , λn cn ) =



λ1 cT c1
1
λ1 cT c1 n ...
...
...

λn cT cn
1
λn cT cn n 
 = diag(λ1 , . . . , λn ).

P 13.3: The quadratic form q is given by q(x) = x2 − 2x1 x2 + 2x2 = xT Ax = xT
1
2

1 −1
−1 2

x

Eigenvalues of A
(1 − λ)(2 − λ) − 1 = 1 − 3λ + λ2 = 0 λ1/2 √
5
3
= ±
2
2

Eigenvectors:
√
3
5
λ= +
2
2

1

⇒ c1 =
10
4

√
3
5 λ= −
2
2

√

−

5
2

1

⇒ c1 =
10
4

√

+

5
2

√
5
1
,1
−
2
2

T

√
1
5
+
,1
2
2

T

C = (c1 , c2 )
T

C AC = diag

T

√
√
5
5 3
3
, −
+
2
2 2
2

T

T

q(Cy) = (Cy) ACy = y C ACy =

√
5
3
+
2
2

2 y1 +

√
5
3
−
2
2

Problems can be downloaded from the internet site: http://www.math.tu-cottbus.de/∼pawell/education/erm/erm.html 4

2 y2