Math 2270 Spring 2004
Exam 2 Extra Problems
(1) (a)




1 1 1
1 1 1




det(A) = det  1 2 2  = det  0 1 1  = −5
0 0 −5
2 2 −3
(2)
(b) det(AAT A)=det(A) · det(AT ) · det(A) = det(A)3 = (−5)3 = −125
(c) det(A − 3A)=det(−2A)=(−2)3 det(A)=(−8)(−5) = 40
1
= − 51
(d) det(A−1 ) = det(A)


0


x
x
x ′
x
x
T (e ) = (e ) − (e ) = e − e = 0 ⇒  0 
0




0 0 0

B= 0 1 0 

0 −1 1
0

′
T (x) = (x) − (x) = x − 1 ⇒  1 

−1


0


′
T (1) = (1) − (1) = 1 − 0 = 1 ⇒  0 
1
A basis for the kernel of T is {ex }. A basis for the image of T is {x, 1}. T is not an
isomorphism. There are several ways to see this. First, the kernel of T contains more than
the zero element. Second, the image of T is not the entire space V . Third, the B-matrix of
T is not invertible.
(2e)
< 2x − 1, 1 >=
Z
0
1
h
i1
((2x − 1) · 1) dx = x2 − x
0
= 1−1 =0
The inner product of (2x − 1) and 1 is zero, so they are orthogonal.
(2f)
||(x + 1) − 1||2 = ||x||2 =< x, x >=
Z
1
0
√
Therefore, the distance between x + 1 and 1 is 1/ 3.
1
x2 dx =
1 3
x
3
1
0
=
1
3
Math 2270 Spring 2004
Exam 2 Extra Problems
(3) Assume that W is an inner product space. Let V be a subspace of W with basis {f, g, h}.
(Notice that the information only says basis, not a special kind of basis.) Use the following
information to find the element of V that is closest to p, where p is an element of W :
< f, f >= 1
< f, g >= 0
< f, p >= 2
< g, g >= 4
< f, h >= 0
< p, g >= −8
< h, h >= 9
< g, h >= 0
< h, p >= 9
First, we must find an orthonormal basis for V . The second line of inner products tells
us that f , g, and h are orthogonal to one another. Therefore, if we normalize each one
(divide by their magnitudes)
we will have an orthonormal
basis. The length of f is already
√
√
1. The length of g is 4 = 2. The length of h is 9 = 3. An orthonormal basis for V is
{f, g/2, h/3}. Now, we find the orthogonal projection of p onto V .
g
h
h
g
projV p =< f, p > f + < , p > + < , p >
2
2
3
3
1
1
=< f, p > f + < g, p > g + < h, p > h
4
9
1
1
= 2f + (−8)g + (9)h = 2f − 2g + h
4
9
(4) We use the Corollary to Cramer’s rule to get:


1 −3 1
1
1

=
adj(A) =  1 5 −3 
det(A)
4
1 −7 5
A−1
(5a) FALSE: T (A) = AAT is not a linear transformation. We look at T (A + B) = (A +
B)(A + B)T = (A + B)(AT + B T ) = AAT + AB T + BAT + BB T . This is not equal to
T (A) + T (B) unless AB T = −BAT .
(5b) FALSE: The matrix AT A + A is not symmetric. We look at the transpose:
T
T
A A+A
=
T
T
A A
T
+A
= AT A + AT
This is not equal to the original matrix unless A is symmetric.
2
Math 2270 Spring 2004
Exam 2 Extra Problems
(5c) TRUE: If A and B are symmetric, and A and B commute with one another, then
their product AB must be symmetric. We look at the transpose of the product: (AB)T =
B T AT = BA = AB.
(5d) TRUE: The following is an inner product on ℜ2 : < ~x, ~y >= x1 y1 + 2x2 y2 . The function
satisfies all of the properties of inner products on ℜ2 . For all ~x, ~y and ~z in ℜ2 and all c in ℜ:
(i) < ~x, ~y >=< ~y , ~x >
(ii) < ~x + ~z , ~y >=< ~x, ~y > + < ~z , ~y >
(iii) < c~x, ~y >= c < ~x, ~y >
(iv) For all nonzero ~x in ℜ2 , < ~x, ~x > > 0
(5e) TRUE: The following is an orthonormal basis for ℜ4 :

√  
 
 

0
1/ √2
1/2
−1/2




 
 1/2  




0√
  −1/ 2   −1/2  

,

,
,






1/2
−1/
1/2
0 




√2

 1/2
1/2
1/ 2
0






 .





We notice that all of the vectors are unit vectors and they are all orthogonal to one another.
(5f) FALSE: The matrix is not orthogonal since the columns do not have unit length.
(5g) TRUE: Let {~v , w,
~ ~x, ~y , ~z} be an orthonormal collection of vectors. Then
||~v + 2w
~ + 3~x + ~y + ~z || = 4. We use the Pythagorean theorem:
||~v + 2w
~ + 3~x + ~y + ~z||2 = ||~v||2 + ||2w||
~ 2 + ||3~x||2 + ||~y ||2 + ||~z||2
=< ~v , ~v > + < 2w,
~ 2w
~ > + < 3~x, 3~x > + < ~y , ~y > + < ~x, ~z >
=< ~v , ~v > +4 < w,
~ w
~ > +9 < ~x, ~x > + < ~y , ~y > + < ~x, ~z >
= 1 + 4 + 9 + 1 + 1 = 16
√
||~v + 2w
~ + 3~x + ~y + ~z|| = 16 = 4
(5h) TRUE: Let T be a linear transformation from V to W where dim(V ) = 6, dim(W ) = 9
and the dimension of the orthogonal complement of ker(T ) is 2. The sum of the dimension
of a subspace and the dimension of its orthogonal complement is equal to the dimension of
the entire space. The kernel of T is a subspace of V . Therefore, the dimension of ker(T )
must equal 6 − 2 = 4. Then, by the rank-nullity theorem,
dim(im(T ))+dim(ker(T ))=dim(V ) ⇒ dim(im(T ))+4=6 ⇒ dim(im(T ))=2
3
Math 2270 Spring 2004
Exam 2 Extra Problems
(6)
||v + w||2 + ||v − w||2 =< v + w, v + w > + < v − w, v − w >
=< v, v > + < v, w > + < w, v > + < w, w > + < v, v > + < v, −w > + < −w, v > + < −w, −w >
=< v, v > +2 < v, w > + < w, w > + < v, v > −2 < v, w > + < w, w >
= 2 < v, v > +2 < w, w >= 2||v||2 + 2||w||2
4