A First Course in Linear Algebra, 2017a

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Now to write in terms of (a,b), note that a/ √ a 2 + b 2 = cosθ,b/ √ a 2 + b 2 = sinθ. Now plug this in to the
above. The result is [ a 2 −b 2
2 ab
a 2 +b 2 a 2 +b 2
= 1 [ a 2 − b 2 ]
2ab
a 2 + b 2 2ab b 2 − a 2
]
2 ab b 2 −a 2
a 2 +b 2 a 2 +b 2
Since this is a unit vector, a 2 + b 2 = 1 and so you get
[ a 2 − b 2 ]
2ab
2ab b 2 − a 2
5.5.5
⎡
⎣
1 0
0 1
0 0
⎤
⎦
5.5.6 This says that the columns of A have a subset of m vectors which are linearly independent. Therefore,
this set of vectors is a basis for R m . It follows that the span of the columns is all of R m . Thus A is onto.
5.5.7 The columns are independent. Therefore, A is one to one.
5.5.8 The rank is n is the same as saying the columns are independent which is the same as saying A is
one to one which is the same as saying the columns are a basis. Thus the span of the columns of A is all of
R n and so A is onto. If A is onto, then the columns must be linearly independent since otherwise the span
of these columns would have dimension less than n and so the dimension of R n would be less than n .
5.6.1 If ∑ r i a i⃗v r = 0, then using linearity properties of T we get
0 = T (0)=T (
r
∑
i
a i ⃗v r )=
r
∑
i
a i T (⃗v r ).
Sinceweassumethat{T⃗v 1 ,···,T⃗v r } is linearly independent, we must have all a i = 0, and therefore we
conclude that {⃗v 1 ,···,⃗v r } is also linearly independent.
5.6.3 Since the third vector is a linear combinations of the first two, then the image of the third vector will
also be a linear combinations of the image of the first two. However the image of the first two vectors are
linearly independent (check!), and hence form a basis of the image.
Thus a basis for im(T ) is:
⎧⎡
⎪⎨
V = span ⎢
⎣
⎪⎩
2
0
1
3
⎤
⎡
⎥
⎦ , ⎢
5.7.1 In this case dim(W)=1andabasisforW consisting of vectors in S can be obtained by taking any
(nonzero) vector from S.
⎣
4
2
4
5
⎤⎫
⎪⎬
⎥
⎦
⎪⎭