Isomorphic Matrices: Proving Homomorphisms

[kathrynag]

Homework Statement

Let H be the subset of $$M_{2}(\Re)$$ consisting of all matrices of the form
$$\left[a$$$$\right]-b$$
$$\left[b$$$$\right]a$$ for a, b elements of the reals.


1. Show that <complex, +> is isomorphic to <H,+>
2. Show that <complex, .> is isomorphic to <H,.>

Homework Equations

The Attempt at a Solution

Ok, so I need to show one to one and onto.
Then I need to show that $$\varphi$$(x+y)=$$\varphi$$(x)+$$\varphi$$(y)
Ok I'm having problems in what to use for $$\varphi$$ and I'm not entirely sure I remember matrix addition. Also, how do I define the complex - a+ib or just ib?

 

[Citan Uzuki]

You're mapping the complex number a + bi to the matrix $$\left[ \begin{array}{cc}a & -b\\b & a \end{array} \right]$$

 

[kathrynag]

You're mapping the complex number a + bi to the matrix $$\left[ \begin{array}{cc}a & -b\\b & a \end{array} \right]$$

ok, so I wnat to show one to one and onto.
Then my problem is deifining $$\varphi$$(x) is it that matrix?

 

[Citan Uzuki]

What is this "x" you are referring to? Is it a+bi? If so, then yes. Otherwise, no.

Really, you seem to be just pushing symbols around without understanding what they mean.

 

[kathrynag]

Ok, so $$\varphi$$ is an isomorphism from <a+bi,+> to H.
So, first I need to show that $$\varphi$$(x)=$$\varphi$$(y) to show one to one.
This is where I start to get stuck because of the matrix. What hint would be helpful?

 

[Citan Uzuki]

No, $\varphi$ is an isomorphism from $\mathbb{C}$ to H (or more precisely, an isomorphism from $(\mathbb{C},\ +,\ \cdot)$ to $(\mathrm{H},\ +,\ \cdot)$). It satisfies the equation:

$$\varphi(a+bi) = \left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right]$$

To prove injectivity, note that two matrices are equal iff their corresponding entries are equal, and two complex numbers are equal iff their real and imaginary parts are equal. So represent the complex numbers in terms of their real and imaginary parts (for instance, let one of them be a+bi and the other be c+di). Then show that φ(a+bi) = φ(c+di) implies that a=c and b=d, which in turn implies that a+bi = c+di.

 

[Mark44]

Suppose $\varphi(z_1) = \varphi(z_2)$
$where z_1 = a_1 + b_1 i and z_2 = a_2 + b_2 i$
Isn't $\varphi(z_1)$ just the matrix with a1 and b1 in the appropriate places?
If so, it's pretty easy to show that z1 = z2 to establish one-to-one-ness.

For onto-ness, just show that for an arbitrary matrix in H, there is a complex number z that maps to it.

 

[kathrynag]

Ok, so c+di I just switch the points for a=c and b=d

 

[kathrynag]

ok, so ontoness, is confusing me. I'm just having trouble with finding the arbitrary matrix. So, I could just use the the complex number(a1+ib1)

 

[kathrynag]

Ok, I think I did something wrong for part 2.
I showed one to one by $$\varphi$$(a+ib)=
$$\left[a$$$$\right]-b$$
$$\left[b$$$$\right]a$$
$$\varphi$$(c+id)=
$$\left[c$$$$\right]-d$$
$$\left[d$$$$\right]c$$
Then a=b and b=d and this shows one - to one.

Then for onto, I let $$\varphi$$(a+ib)=
$$\left[a$$$$\right]-b$$
$$\left[b$$$$\right]a$$

Then it's onto H.

So, $$\varphi$$(xy)=
$$\left[xy$$$$\right]0$$
$$\left[0$$$$\right]xy$$


So, $$\varphi$$(x)=
$$\left[x$$$$\right]0$$
$$\left[0$$$$\right]x$$


So, $$\varphi$$(y)=
$$\left[y$$$$\right]0$$
$$\left[0$$$$\right]y$$

My only problem is when doing $$\varphi$$(x)*$$\varphi$$(y) it shows that there is not an isomorphism, but I think there should be. Am I going wrong somewhere?

 

[Mark44]

ok, so ontoness, is confusing me. I'm just having trouble with finding the arbitrary matrix. So, I could just use the the complex number(a1+ib1)

Not an arbitrary matrix, but an arbitrary matrix in H, namely
[x -y]
[y x]

Show that there is a complex number that maps to this matrix (easy).

 

[Mark44]

Ok, I think I did something wrong for part 2.
I showed one to one by $$\varphi$$(a+ib)=
$$\left[a$$$$\right]-b$$
$$\left[b$$$$\right]a$$
$$\varphi$$(c+id)=
$$\left[c$$$$\right]-d$$
$$\left[d$$$$\right]c$$
Then a=b and b=d and this shows one - to one.

No, for 1-1-ness, you want to show that phi(z1) = phi(z2) ==> z1 = z2. I.e., that a = c and b = d, using your choices for the two complex numbers.

BTW, your latex stuff is coming out pretty screwy.

 

[kathrynag]

Not an arbitrary matrix, but an arbitrary matrix in H, namely
[x -y]
[y x]

Show that there is a complex number that maps to this matrix (easy).

Ok so, x+iy

No, for 1-1-ness, you want to show that phi(z1) = phi(z2) ==> z1 = z2. I.e., that a = c and b = d, using your choices for the two complex numbers.

BTW, your latex stuff is coming out pretty screwy.

Ok, that makes sense.
Yeah, I'm not sure of the right Latex code for a matrix.