Show/ prove: (C, +) is isomorphic to (H, +) where H is a subset of M2(R) (2x2 matrices with real numbers) of the form
[ a -b
b a].
My thought was to define a function f: C -> M2(R) by f(x) =
[ cos(ø) -sin(ø)
sin(ø) cos(ø)]
since x = cos(ø) + i sin(ø)
but I don't really know if that makes sense.
[ a -b
b a].
My thought was to define a function f: C -> M2(R) by f(x) =
[ cos(ø) -sin(ø)
sin(ø) cos(ø)]
since x = cos(ø) + i sin(ø)
but I don't really know if that makes sense.
-
Keep it even simpler.
Define f: C --> M2(R) by f(a + bi) = [a -b; b a].
(i) f is a homomorphism.
Let a+bi, c+di be in C (with a,b,c,d in R).
So, f(a + bi) + f(c + di) =
[a -b]....[c -d]
[b a].+.[d c].=
[a+c -(b+d)]
[b+d a+c] = f((a+c) + (b+d)i).
---------------
(ii) f is onto.
Given A =
[a -b]
[b a] in M2(R), note that a+bi is in C, and f(a+bi) = A, as required.
---------------
(iii) f is 1-1.
Note that ker f = {0}, because
ker f = {a+bi in C : f(a+bi) = 0 (matrix)}
.......= {[a -b; b a] = [0 0; 0 0]}
.......= {0 + 0i}, equating like entries (a = 0 and b = 0).
---------------
By (i)-(iii), we conclude that f yields an isomorphism.
I hope this helps!
Define f: C --> M2(R) by f(a + bi) = [a -b; b a].
(i) f is a homomorphism.
Let a+bi, c+di be in C (with a,b,c,d in R).
So, f(a + bi) + f(c + di) =
[a -b]....[c -d]
[b a].+.[d c].=
[a+c -(b+d)]
[b+d a+c] = f((a+c) + (b+d)i).
---------------
(ii) f is onto.
Given A =
[a -b]
[b a] in M2(R), note that a+bi is in C, and f(a+bi) = A, as required.
---------------
(iii) f is 1-1.
Note that ker f = {0}, because
ker f = {a+bi in C : f(a+bi) = 0 (matrix)}
.......= {[a -b; b a] = [0 0; 0 0]}
.......= {0 + 0i}, equating like entries (a = 0 and b = 0).
---------------
By (i)-(iii), we conclude that f yields an isomorphism.
I hope this helps!