A First Course in Linear Algebra, 2017a

578 Selected Exercise Answers
Then a solution is
7.4.25
⎡
⎢
⎣
1
6√
6
1
3√
6
1
6√
6
0
⎡ √ √ √
6
1
2
6
1
⎤
√ √ 6 √ 6
√ 3
· ⎢ 0 2 2 3
5
18
2 3
√ ⎥
⎣
1
0 0 9√
3 37 ⎦
0 0 0
⎤
⎡
⎥
⎦ , ⎢
√
1
⎤ ⎡
6√
2 3
− 2 √
9√
2 3
√ ⎥
⎣ 5
18√
√ 2
√ 3 ⎦ , ⎢
⎣
2 3
1
9
⎡
[
x y
]
z ⎣
5
√
111√
3
√ 37
1
333√
3 37
− 17
22
333
√
333√
√ 3
√ 37
3 37
a 1 a 4 /2
⎤
a 5 /2
a 4 /2 a 2 a 6 /2 ⎦
⎡
⎣
x
y
z
⎤
⎦
⎤
⎥
⎦
7.4.26 The quadratic form may be written as
⃗x T A⃗x
where A = A T . By the theorem about diagonalizing a symmetric matrix, there exists an orthogonal matrix
U such that
U T AU = D, A = UDU T
Then the quadratic form is
⃗x T UDU T ⃗x = ( U T ⃗x ) T (
D U T ⃗x )
where D is a diagonal matrix having the real eigenvalues of A down the main diagonal. Now simply let
⃗x ′ = U T ⃗x
9.1.20 The axioms of a vector space all hold because they hold for a vector space. The only thing left to
verify is the assertions about the things which are supposed to exist. 0 would be the zero function which
sends everything to 0. This is an additive identity. Now if f is a function, − f (x) ≡ (− f (x)). Then
( f +(− f ))(x) ≡ f (x)+(− f )(x) ≡ f (x)+(− f (x)) = 0
Hence f + − f = 0. For each x ∈ [a,b], let f x (x) =1and f x (y) =0ify ≠ x. Then these vectors are
obviously linearly independent.
9.1.21 Let f (i) be the i th component of a vector⃗x ∈ R n . Thus a typical element in R n is ( f (1),···, f (n)).
9.1.22 This is just a subspace of the vector space of functions because it is closed with respect to vector
addition and scalar multiplication. Hence this is a vector space.
9.2.1 ∑ k i=1 0⃗x k =⃗0
9.3.29 Yes. If not, there would exist a vector not in the span. But then you could add in this vector and
obtain a linearly independent set of vectors with more vectors than a basis.