A First Course in Linear Algebra, 2017a

504 Vector Spaces
This clearly only occurs when a = b = c = d = 0 which means that
[ ] [ ]
a b 0 0
A = = = 0
c d 0 0
Hence T is one-to-one.
T is onto: Let
and define matrix A ∈ M 22 as follows:
⃗x =
A =
⎡
⎢
⎣
⎤
x 1
x 2
x 3
x 4
⎥
⎦ ∈ R4 ,
[ ]
x1 x 2
.
x 3 x 4
Then T (A)=⃗x, and therefore T is onto.
Since T is a linear transformation which is one-to-one and onto, T is an isomorphism. Hence M 22 and
R 4 are isomorphic.
♠
An important property of isomorphisms is that the inverse of an isomorphism is itself an isomorphism
and the composition of isomorphisms is an isomorphism. We first recall the definition of composition.
Definition 9.72: Composition of Transformations
Let V,W ,Z be vector spaces and suppose T : V ↦→ W and S : W ↦→ Z are linear transformations.
Then the composite of S and T is
S ◦ T : V ↦→ Z
andisdefinedby
(S ◦ T)(⃗v)=S(T(⃗v)) for all⃗v ∈ V
Consider now the following proposition.
Proposition 9.73: Composite and Inverse Isomorphism
Let T : V → W be an isomorphism. Then T −1 : W → V is also an isomorphism. Also if T : V → W
is an isomorphism and if S : W → Z is an isomorphism for the vector spaces V ,W,Z, then S ◦ T
defined by (S ◦ T )(v)=S(T (v)) is also an isomorphism.
Proof. Consider the first claim. Since T is onto, a typical vector in W is of the form T (⃗v) where ⃗v ∈ V.
Consider then for a,b scalars,
T −1 (aT(⃗v 1 )+bT(⃗v 2 ))
where⃗v 1 ,⃗v 2 ∈ V . Consider if this is equal to
aT −1 (T (⃗v 1 )) + bT −1 (T (⃗v 2 )) = a⃗v 1 + b⃗v 2 ?