Linear Algebra, 2020a

188 Chapter Three. Maps Between Spaces
2.9 Example The space M 2×2 of 2×2 matrices is isomorphic to R 4 . With this
basis for the domain
( ) ( ) ( ) ( )
1 0 0 1 0 0 0 0
B = 〈 , , , 〉
0 0 0 0 1 0 0 1
the isomorphism given in the lemma, the representation map f 1 = Rep B , carries
the entries over.
⎛ ⎞
( ) a
a b
f
↦−→
1
b
⎜ ⎟
c d ⎝c⎠
d
One way to think of the map f 1 is: fix the basis B for the domain, use the
standard basis E 4 for the codomain, and associate ⃗β 1 with ⃗e 1 , ⃗β 2 with ⃗e 2 , etc.
Then extend this association to all of the members of two spaces.
⎛ ⎞
( )
a
a b
f
= a⃗β 1 + b⃗β 2 + c⃗β 3 + d⃗β
1
b
4 ↦−→ a⃗e1 + b⃗e 2 + c⃗e 3 + d⃗e 4 = ⎜ ⎟
c d
⎝c⎠
d
We can do the same thing with different bases, for instance, taking this basis
for the domain.
( ) ( ) ( ) ( )
2 0 0 2 0 0 0 0
A = 〈 , , , 〉
0 0 0 0 2 0 0 2
Associating corresponding members of A and E 4 gives this.
( )
a b
=(a/2)⃗α 1 +(b/2)⃗α 2 +(c/2)⃗α 3 +(d/2)⃗α 4
c d
f 2
↦−→ (a/2)⃗e1 +(b/2)⃗e 2 +(c/2)⃗e 3 +(d/2)⃗e 4 =
⎛ ⎞
a/2
b/2
⎜ ⎟
⎝c/2⎠
d/2
gives rise to an isomorphism that is different than f 1 .
The prior map arose by changing the basis for the domain. We can also
change the basis for the codomain. Go back to the basis B above and use this
basis for the codomain.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 0 0 0
0
D = 〈 ⎜ ⎟
⎝0⎠ , 1
⎜ ⎟
⎝0⎠ , 0
⎜ ⎟
⎝0⎠ , 0
⎜ ⎟
⎝1⎠ 〉
0 0 1 0