0
$\begingroup$

I need a hint. My task is to proof that

  • ((a11, a12), (b11, b12))

  • ((a21, a22), (b21, b22))

    ∈ R ⇔ a11 + a12 + a21 + a22 = b11 + b12 + b21 + b22

R is equivalnce relation. My problem is that I have no idea how to deal with matrix.

I tried to work like in this post and just make something like A={{a11,a12},{a21,a22}} instead of A={0,1}, but I got stuck.

I would be grateful for any help.

$\endgroup$
6
  • 2
    $\begingroup$ just check reflex, symmetric and transitivity. It is clear $\endgroup$ Commented Nov 3, 2014 at 11:13
  • $\begingroup$ Why don't you write what is meant by an equivalence relation? And try your self. $\endgroup$ Commented Nov 3, 2014 at 11:38
  • 1
    $\begingroup$ I'm pretty sure I saw this exact same question asked here several hours ago? Did you delete it an ask again with no changes? $\endgroup$ Commented Nov 3, 2014 at 11:43
  • $\begingroup$ @Henning: math.stackexchange.com/questions/1003400/… also since this is a separate user account, you can't close as a duplicate (even though it is a word-for-word duplicate) unless there is an upvoted answer. $\endgroup$ Commented Nov 3, 2014 at 12:16
  • $\begingroup$ @Asaf: Of course I tried to google for a phrase that included the one word there's in difference between the two questions. Thanks -- we'll just close it the other way around then. $\endgroup$ Commented Nov 3, 2014 at 12:29

1 Answer 1

Reset to default
1
$\begingroup$

Hint:

If $f:X\rightarrow Y$ is a function then the relation $R$ on $X$ defined by $xRy\iff f(x)=f(y)$ is an equivalence relation. So find a proper function here on $2\times 2$-matrices.

(Why? Simply because $f(x)=f(x)$ i.e. reflexitivity, $f(x)=f(y)\Rightarrow f(y)=f(x)$ i.e. symmetry, and $f(x)=f(y)\wedge f(y)=f(z)\Rightarrow f(x)=f(z)$ i.e. transitivity)

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.