Isomorphic Matrices: Proving Homomorphisms

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In summary: I'll try to fix it.In summary, we need to show that the set of matrices of the form \left[a\right]-b\left[b\right]a for a, b elements of the reals is isomorphic to the sets <complex, +> and <complex, .>. This can be done by showing one-to-one and onto mappings between the sets. To prove injectivity, we can use the fact that two matrices are equal if their corresponding entries are equal, and two complex numbers are equal if their real and imaginary parts are equal. To prove surjectivity, we can show that for an arbitrary matrix in H, there is a corresponding complex number that maps to it. Overall, this shows the isom
  • #1
kathrynag
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Homework Statement



Let H be the subset of [tex]M_{2}(\Re)[/tex] consisting of all matrices of the form
[tex]\left[a[/tex][tex]\right]-b[/tex]
[tex]\left[b[/tex][tex]\right]a[/tex] 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 [tex]\varphi[/tex](x+y)=[tex]\varphi[/tex](x)+[tex]\varphi[/tex](y)
Ok I'm having problems in what to use for [tex]\varphi[/tex] and I'm not entirely sure I remember matrix addition. Also, how do I define the complex - a+ib or just ib?
 
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  • #2
You're mapping the complex number a + bi to the matrix [tex]\left[ \begin{array}{cc}a & -b\\b & a \end{array} \right][/tex]
 
  • #3
Citan Uzuki said:
You're mapping the complex number a + bi to the matrix [tex]\left[ \begin{array}{cc}a & -b\\b & a \end{array} \right][/tex]

ok, so I wnat to show one to one and onto.
Then my problem is deifining [tex]\varphi[/tex](x) is it that matrix?
 
  • #4
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.
 
  • #5
Ok, so [tex]\varphi[/tex] is an isomorphism from <a+bi,+> to H.
So, first I need to show that [tex]\varphi[/tex](x)=[tex]\varphi[/tex](y) to show one to one.
This is where I start to get stuck because of the matrix. What hint would be helpful?
 
  • #6
No, [itex]\varphi[/itex] is an isomorphism from [itex]\mathbb{C}[/itex] to H (or more precisely, an isomorphism from [itex](\mathbb{C},\ +,\ \cdot)[/itex] to [itex](\mathrm{H},\ +,\ \cdot)[/itex]). It satisfies the equation:

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

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.
 
  • #7
Suppose [itex]\varphi(z_1) = \varphi(z_2)[/itex]
[itex]where z_1 = a_1 + b_1 i and z_2 = a_2 + b_2 i[/itex]
Isn't [itex]\varphi(z_1)[/itex] 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.
 
  • #8
Ok, so c+di I just switch the points for a=c and b=d
 
  • #9
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)
 
  • #10
Ok, I think I did something wrong for part 2.
I showed one to one by [tex]\varphi[/tex](a+ib)=
[tex]\left[a[/tex][tex]\right]-b[/tex]
[tex]\left[b[/tex][tex]\right]a[/tex]
[tex]\varphi[/tex](c+id)=
[tex]\left[c[/tex][tex]\right]-d[/tex]
[tex]\left[d[/tex][tex]\right]c[/tex]
Then a=b and b=d and this shows one - to one.

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

Then it's onto H.

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


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


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

My only problem is when doing [tex]\varphi[/tex](x)*[tex]\varphi[/tex](y) it shows that there is not an isomorphism, but I think there should be. Am I going wrong somewhere?
 
  • #11
kathrynag said:
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).
 
  • #12
kathrynag said:
Ok, I think I did something wrong for part 2.
I showed one to one by [tex]\varphi[/tex](a+ib)=
[tex]\left[a[/tex][tex]\right]-b[/tex]
[tex]\left[b[/tex][tex]\right]a[/tex]
[tex]\varphi[/tex](c+id)=
[tex]\left[c[/tex][tex]\right]-d[/tex]
[tex]\left[d[/tex][tex]\right]c[/tex]
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.
 
  • #13
Mark44 said:
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

Mark44 said:
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.
 

Related to Isomorphic Matrices: Proving Homomorphisms

1. What is an isomorphic matrix?

An isomorphic matrix is a square matrix that has the same number of rows and columns and follows a specific set of rules. These rules include being invertible, having the same number of non-zero elements in each row, and having a unique solution for every set of equations.

2. What is the significance of proving homomorphisms for isomorphic matrices?

Proving homomorphisms for isomorphic matrices is significant because it helps us understand the relationship between two different matrices. It allows us to determine if two matrices are isomorphic, which can help us solve problems involving these matrices more efficiently.

3. How do you prove homomorphisms for isomorphic matrices?

The process of proving homomorphisms for isomorphic matrices involves showing that the mapping between the two matrices preserves the operations of addition and multiplication. This is done by demonstrating that the output of the mapping for the sum of two matrices is equal to the sum of the individual mappings, and that the output of the mapping for the product of two matrices is equal to the product of the individual mappings.

4. What are the key properties of homomorphisms for isomorphic matrices?

The key properties of homomorphisms for isomorphic matrices include preserving the identity element, associativity, and distributivity. This means that the mapping between the two matrices should produce the same result as the original matrices when the identity element is used, when the operation is performed in a different order, and when the operation is distributed over another operation.

5. How can one use homomorphisms for isomorphic matrices in real-world applications?

Homomorphisms for isomorphic matrices can be used in real-world applications to simplify and solve complex problems involving matrices. For example, they can be used in cryptography to convert data into a different form while preserving certain properties, or in computer graphics to transform images or objects using matrix operations.

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