# Sets of matrices...

1. Dec 2, 2015

### cdummie

1. The problem statement, all variables and given/known data
If there are two sets of matrices $S = \begin{Bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} | a, b, c, d \in \mathbb{C} \end{Bmatrix}$ and
$M = \begin{Bmatrix} \begin{bmatrix} a & b \\ -\overline{b} & \overline{a} \end{bmatrix} | a, b \in \mathbb{C} \wedge |a| + |b| \neq 0 \end{Bmatrix}$

Then, for every $X, Y \in S$

$X \rho Y \Longleftrightarrow ( \exists A \in M) AXA^{-1}= Y$

It's AXA-1 up here, even though -1 doesn't seem like exponent.
Mod note: It's fixed now. In LaTeX, when an exponent consists of more than one character, put braces around the exponent. IOW, like this A^{-1}.
Prove that $\rho$ is equivalence relation.

2. Relevant equations

3. The attempt at a solution
Now, i know that in relation is equivalence if it's reflexive, symmetric and transitive, but i got the problem at the very beginning, i mean if it's reflexive then:

AXA-1=X

multiplying both sides by A on the right side i have

AX=XA but that proves nothing since multiplying of matrices isn't commutative. Any ideas?

Last edited by a moderator: Dec 2, 2015
2. Dec 2, 2015

### Samy_A

The condition for two matrices to be equivalent is $X \rho Y \Longleftrightarrow ( \exists A \in M) AXA^{-1}= Y$
See the $\exists A$?
$AXA^{-1}= Y$ doesn't have to be true for all matrices $A$ for $X$ and $Y$ to be equivalent.

Last edited: Dec 2, 2015
3. Dec 2, 2015

### cdummie

I know, finding at least one matrix from M that fits into this would be enough, but, how can i do that?

4. Dec 2, 2015

### Samy_A

Given a matrix $X$, you need to find one matrix $A$ satisfying $AXA^{-1}=X$. I think you are overthinking this one: actually, you can pick the "simplest" matrix there is, if we exclude the 0 matrix.

5. Dec 2, 2015

### cdummie

Oh i see it's unit matrix, but is there a way to find it algebraically?

Now, for the symmetry, i have:

$(\forall X, Y \in S) X \rho Y \Longleftrightarrow (\exists A \in M) AXA^{-1}=Y \Longleftrightarrow AX=YA \Longleftrightarrow X=A^{-1}YA$

And that's all i have, and this isn't proving symmetry.

6. Dec 2, 2015

### Samy_A

Yes, the unit matrix will do. I don't think you can find it algebraically in this context.
Well, you are very close. Having $A \in M$ satisfying $AXA^{-1}=Y$, you need to find a matrix $B \in M$ satisfying $BYB^{-1}=X$.
$B$ can be a different matrix than $A$...

7. Dec 2, 2015

### cdummie

This is what i thought of when you said that i am very close (correct me if i am wrong):

Since i have $AXA^{-1}=Y$ and i know that matrices in set M are all regular since $|a| + |b| \neq 0$ i could maybe think of matrix that is inverse matrix of A, knowing the following rule that works for all 2x2 matrices: $A^{-1} = \frac{1}{detA} \begin{bmatrix} a & -b \\ \overline{b} & \overline{a} \end{bmatrix}$
Now, since all of these are complex numbers it doesn't matter what determinant is as long as it isn't zero, and it is not, otherwise A wouldn't be in the set M in the first place so inverse matrix of A could be the matrix i am looking for.

Now, for the transitivity:

$(\forall X, Y, Z \in S) X \rho Y \land Y \rho Z \Longleftrightarrow (\exists A \in M) AXA^{-1} =Y \land (\exists B \in M) BYB^{-1} =Z \Longrightarrow BAXA^{-1}B^{-1}=Z$

Since by multiplying B and A and A-1 and B-1 resulting matrices remain in the M, i calculated for AB and it can be equal to some matrix C so C-1 is A-1B-1 which fits perfectly and proves that this relation is transitive and therefore it is equivalence relation. Now, thanks a lot for your help, and please correct me if i made any mistakes here.

8. Dec 2, 2015

### Staff: Mentor

What you implicitly used is that $M$ is a group, i.e. $1 ∈ M$, $(A ∈ M ⇒ A^{-1} ∈ M)$ and $(A,B ∈ M ⇒ AB ∈ M)$. Whether this can be assumed or has to be proven depends on where you start at. The first one is obvious and the second one is almost shown by you. (I think you made a little mistake with the inverse matrix: change $a$ and $\bar{a}$.) So maybe you have to show the third property, too.

Last edited: Dec 2, 2015
9. Dec 2, 2015

### Samy_A

Yes, that is correct (except the little mistake in the inverse as noted by @fresh_42).
That the product of two matrices in M is also an element of M can be shown by a straightforward computation.