Math 222 - Selected Homework Solutions from
Chapter 4
Instructor - Al Boggess
Fall 1998
Section 4.3
11. If A and B are similar, then there is a nonsingular matrix S with
B = S ,1 AS . Therefore
det B = det(S ,1 AS )
= (det S ,1 )(det A)(det S )
where we have used the fact that the determinant of a product of
matrices is the product of their determinants. We also know that
det S ,1 = 1= det S (problem on Exam I). Therefore det B = det A, as
desired.
12. Suppose A and B are similar; then there is a nonsingular matrix S
with B = S ,1 AS . Taking the transpose of this equation, we obtain
B T = (S ,1 AS )T = S T AT (S ,1 )T
Now (S ,1 )T = (S T ),1 by problem 17 from section 1.3 (an earlier HW
problem). Let T = (S T ),1 . Then the above equation reads
B T = T ,1AT T
which shows that B T and AT are similar.
For part b), we have
B k = (S ,1AS )k
= (S ,1 AS )(S ,1 AS ) : : : (S ,1 AS ) (k times)
Each interior factor of S cancels with S 1 giving
B k = S ,1 Ak S
so B k and Ak are similar.
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13. Suppose A and B are similar; then there is a nonsingular matrix S
with B = S ,1 AS . The inverse is computed as
B ,1 = (S ,1AS ),1
= S ,1 A,1 (S ,1 ),1
= S ,1 A,1 S
The inverses of both A and S exist since they are nonsingular. This
shows that B ,1 exists and that B ,1 and A,1 are similar.
Section 5.1
1d Using the law of cosines
h,2; 3; 1i h1; 2; 4i = p 8p
cos = jh,
2; 3; 1ij jh1; 2; 4ij
14 21
Therefore the angle is
= arccos( p 8p ) 62:2o
14 21
7b The plane is given by the equation
hx , 4; y , 2; z + 5i h,3; 6; 2i = 0
or ,3x + 6y + 2z = ,10.
11. To show the triangle inequality, jju + vjj jjujj + jjvjj, we square the
right side and use the denition of jjjj2 in terms of the inner product:
jju + vjj2 = (u + v) (u + v)
= jjujj2 + 2u v + jjvjj2
jjujj2 + 2jjujj jjvjj + jjvjj2 by Schwarz inequality
= (jjujj + jjvjj)2
Taking square roots gives jju + vjj jjujj + jjvjj. Equality occurs when
u v = jjujj jjvjj, which happens when the angle between u and v is
zero, i.e. u and v point in the same direction.
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