A First Course in Linear Algebra, 2017a

9.7. Isomorphisms 507
for ∑ n i=1 c i⃗v i an arbitrary vector of V,
T
( n∑
i=1c i ⃗v i
)
=
n
∑
i=1
c i T (⃗v i )=
n
∑
i=1
It is necessary to verify that this is well defined. Suppose then that
Then
n
∑
i=1
n
∑
i=1
c i ⃗v i =
n
∑
i=1
ĉ i ⃗v i
(c i − ĉ i )⃗v i = 0
and since {⃗v 1 ,···,⃗v n } is a basis, c i = ĉ i for each i. Hence
n
∑
i=1
c i ⃗w i =
n
∑
i=1
and so the mapping is well defined. Also if a,b are scalars,
(
)
n
n
T a c i ⃗v i + b ĉ i ⃗v i = T
∑
i=1
∑
i=1
= a
= aT
ĉ i ⃗w i
c i ⃗w i .
( n∑
i=1(ac i + bĉ i )⃗v i
)
n
n
∑ c i ⃗w i + b ∑
i=1 i=1
( n∑
)
i ⃗v i
i=1c
ĉ i ⃗w i
+ bT
=
n
∑
i=1
( n∑
i=1ĉ i ⃗v i
)
(ac i + bĉ i )⃗w i
Thus T is a linear map.
Now if
T
( n∑
i=1c i ⃗v i
)
=
n
∑
i=1
c i ⃗w i =⃗0,
then since the {⃗w 1 ,···,⃗w n } are independent, each c i = 0andso∑ n i=1 c i⃗v i =⃗0 also. Hence T is one to one.
If ∑ n i=1 c i⃗w i is a vector in W , then it equals
n
∑
i=1
c i T⃗v i = T
( n∑
i=1c i ⃗v i
)
showing that T is also onto. Hence T is an isomorphism and so V and W are isomorphic.
Next suppose these two vector spaces are isomorphic. Let T be the name of the isomorphism. Then
for {⃗v 1 ,···,⃗v n } abasisforV , it follows that a basis for W is {T⃗v 1 ,···,T⃗v n } showing that the two vector
spaces have the same dimension.
Now suppose the two vector spaces have the same dimension.
First consider the claim that 1. ⇒ 2. If T is one to one, then if {⃗v 1 ,···,⃗v n } is a basis for V, then
{T (⃗v 1 ),···,T (⃗v n )} is linearly independent. If it is not a basis, then it must fail to span W. But then