A First Course in Linear Algebra, 2017a

555
It follows that the expression reduces to 0. You can also do the following.
which implies the expression equals 0.
‖⃗u ×⃗v‖ 2 = ‖⃗u‖ 2 ‖⃗v‖ 2 sin 2 θ
= ‖⃗u‖ 2 ‖⃗v‖ 2 ( 1 − cos 2 θ )
= ‖⃗u‖ 2 ‖⃗v‖ 2 −‖⃗u‖ 2 ‖⃗v‖ 2 cos 2 θ
= ‖⃗u‖ 2 ‖⃗v‖ 2 − (⃗u •⃗v) 2
4.9.17 We will show it using the summation convention and permutation symbol.
(
(⃗u ×⃗v)
′ ) i
= ((⃗u ×⃗v) i ) ′ = ( ε ijk u j v k
) ′
and so (⃗u ×⃗v) ′ =⃗u ′ ×⃗v +⃗u ×⃗v ′ .
4.10.10 ∑ k i=1 0⃗x k =⃗0
4.10.40 No. Let ⃗u =
4.10.41 No.
⎡
⎢
⎣
1
0
0
0
⎡
⎢
⎣
π
20
0
0
⎤
⎤ ⎡
⎥
⎦ ∈ M but 10 ⎢
⎣
4.10.42 This is not a subspace.
= ε ijk u ′ j v k + ε ijk u k v ′ k = ( ⃗u ′ ×⃗v +⃗u ×⃗v ′) i
⎥
⎦ .Then2⃗u /∈ M although ⃗u ∈ M.
⎡
⎢
⎣
1
0
0
0
1
1
1
1
⎤
⎥
⎦ /∈ M.
⎤
⎡
⎥
⎦ is in it. However, (−1) ⎢
⎣
1
1
1
1
⎤
⎥
⎦ is not.
4.10.43 This is a subspace because it is closed with respect to vector addition and scalar multiplication.
4.10.44 Yes, this is a subspace because it is closed with respect to vector addition and scalar multiplication.
4.10.45 This is not a subspace.
⎡
⎢
⎣
0
0
1
0
⎤
⎡
⎥ ⎦ is in it. However (−1) ⎢
⎣
0
0
1
0
⎤ ⎡
⎥
⎦ = ⎢
⎣
0
0
−1
0
⎤
⎥
⎦ is not.
4.10.46 This is a subspace. It is closed with respect to vector addition and scalar multiplication.
4.10.55 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.