Math 2270 Spring 2004
Homework 14 Solutions
Section 5.3) 2, 3, 12, 13, 14, 15, 16, 17, 20, 22, 36, 38
(2,3) Let A be the matrix for the orthogonal transformation L from ℜn to ℜn . Then, by
Fact 5.3.3 A is orthogonal, and by Fact 5.3.7 AT A = In . Therefore,
L(~v ) · L(w)
~ = (A~v) · (Aw)
~ = (A~v )T (Aw)
~ = ~v T AT Aw
~ = ~v T (AT A)w
~ = ~v T w
~ = ~v · w
~
Therefore, orthogonal transformations preserve the dot product. The angle between two
nonzero vectors ~v and w
~ is defined by the formula
cos θ =
~v · w
~
||~v || ||w||
~
Since L is orthogonal, ||~v || = ||L(~v )|| and ||w||
~ = ||L(w)||,
~
so the angle between L(~v ) and
L(w)
~ is
cos θ =
~v · w
~
L(~v ) · L(w)
~
=
||L(~v )|| ||L(w)||
~
||~v || ||w||
~
Therefore, orthogonal transformations preserve angle.
√ 


1/ √2
2/3




(12) We must find a unit vector that is perpendicular to both  2/3  and  −1/ 2 . One
1/3
0
way to find a vector perpendicular to the given two is to take their cross product. Other
methods do exist. For example, one could form a matrix using the given vectors as the rows
and then find the kernel of the matrix. Here, I will use the cross product.
√
√ 
√  
 



0 +√
1/3 2
1/3√2
1/ √2
2/3


 




1/3
 2/3  x  −1/ 2  = 
√ 2 − 0 √  =  1/3 √2 
1/3
0
−2/3 2 − 2/3 2
−4/3 2

Now that we have a vector that is orthogonal to the other two we must normalize it.
√ 

1/3√2 s
1
16
1


+
+
=1
 1/3 2  =
√ 18 18 18
−4/3 2 The magnitude of the vector is already one. The desired matrix is:
√
√ 

2/3 1/ √2 1/3√2


1/3 √2 
 2/3 −1/ 2
1/3
0
−4/3 2
1
Math 2270 Spring 2004
Homework 14 Solutions
(13) From problems 2 and 3, we know
that orthogonal
transformations preserve angles
be





2
−3
2






tween vectors. Here we notice that  3  and  2  are perpendicular, whereas T  3  =
0
0
0






2
−3
3






 0  and T  2  =  −3  are not (look at the corresponding dot products). There0
0
2
fore, there is no orthogonal transformation with the given criteria.
(14) By definition, a matrix is symmetric if it is equal to its transpose.
(AT A)T = AT (AT )T = AT A
(AAT )T = (AT )T AT = AAT
Both AAT and AT A are symmetric.
(15) We are given that AT = A and B T = B. Therefore,
(AB)T = B T AT = BA
This is only equal to AB if the matrices A and B commute with one another. In general,
the product of symmetric matrices is not symmetric.
(16) We are given that AT = A. Therefore,
(A2 )T = (AA)T = AT AT = AA = A2
The square of a symmetric matrix is always symmetric.
(17) We are given that AT = A. Therefore,
(A−1 )T = (AT )−1 = (A)−1 = A−1
The inverse of a symmetric matrix is always symmetric.
2
Math 2270 Spring 2004
Homework 14 Solutions
(20) First, we must find an orthonormal basis for the given subspace. Then, we will use Fact
5.3.10.


1

1
 1 
w
~1 =  
2 1 
1
~v2 · w
~1 =




~v2 − 4w
~1 = 
1 9 5 3
+ − + =4
2 2 2 2


 
−1
2
1
 7


9 

  2 
=
−
−5   2   −7
1
2
3
w
~2 =




AAT = 




=

1 


10 
−1
7
−7
1










1/2 −1/10 "
#
1/2 1/2
1/2 1/2
1/2 7/10 


1/2 −7/10  −1/10 7/10 −7/10 1/10
1/2 1/10

26/100 18/100 32/100 24/100
18/100 74/100 −24/100 32/100
32/100 −24/100 74/100 18/100
24/100 32/100 18/100 26/100





=

1 


50 
13
9
16 12
9
37 −12 16
16 −12 37 9
12 16
9 13





(22) We can think of A2 as AA. Then A(A~x) first projects the vector ~x orthogonally onto
the subspace V defined by the columns of A. Then, we perform the orthogonal projection
again on the new vector projV ~x. But, projV ~x is already in the subspace V so the orthogonal
projection will return the same vector back. Therefore, A2 = A if A is the matrix of an
orthogonal projection.
Now, consider an orthonormal basis {~v1 , . . . ~vm } for the subspace V and define the matrix
B = [~v1 · · · ~vm ]. Then, by Fact 5.3.10 the matrix of the orthogonal projection onto V
is A = BB T . Therefore, A2 = (BB T )(BB T ) = B(B T B)B T since matrix multiplication
is associative. Since the matrix B is orthogonal, we know that B T B = Im . Therefore,
A2 = B(B T B)B T = BB T = A.
3
Math 2270 Spring 2004
Homework 14 Solutions
(36) We are given L(A) = AT . Let A and B be 2 x 3 matrices, and let k be a scalar. Then,
L(A + B) = (A + B)T = AT + B T = L(A) + L(B)
L(kA) = (kA)T = kAT = kL(A)
so L is a linear transformation. We see that L(L(A)) = L(AT ) = (AT )T = A, so the
transformation is its own inverse. Since L−1 exists, L is an isomorphism.
(38) We will use the general matrix A.




T (A) = A + AT = 




=
a1,1
a2,1
···
an,1
a1,2
a2,2
···
an,2
···
···
···
···
a1,n
a2,n
···
an,n
2a1,1
a1,2 + a2,1
a1,2 + a2,1
2a2,2
···
···
a1,n + an,1 a2,n + an,2


 
 
+
 
a1,1
a1,2
···
a1,n
a2,1
a2,2
···
a2,n
· · · a1,n + an,1
· · · a2,n + an,2
···
···
···
2an,n
···
···
···
···
an,1
an,2
···
an,n










Looking at the calculations above, we see that all matrices in the image are symmetric.
Therefore, the image of T is the set of all n x n symmetric matrices. The kernel is the set of
all matrices satisfying





2a1,1
a1,2 + a2,1
a1,2 + a2,1
2a2,2
···
···
a1,n + an,1 a2,n + an,2
· · · a1,n + an,1
· · · a2,n + an,2
···
···
···
2an,n









=
0
0
···
0
0
0
···
0
···
···
···
···
0
0
···
0





We see that the diagonal entries of A must equal zero. The off-diagonal entries must satisfy
ai,j = −aj,i for all i 6= j. Matrices that satisfy this criteria are called skew-symmetric, so the
kernel of T is the set of all n x n skew-symmetric matrices.
4