Thread: should I use real numbers ??

I have a maths question which reads:

Let M be the set of all real 2 * 2-matrices of the form:
(a 0)
(0 b)

Show that this set of matrices is closed under matrix multiplication, and that matrix multiplication is commutative on this set.



Should I use real numbers to solve this problem or use what the question gave??

I forgot to mention.. the attached image is the matrix given in the question.

Originally Posted by kgpretty

I have a maths question which reads:

Let M be the set of all real 2 * 2-matrices of the form:
(a 0)
(0 b)

Show that this set of matrices is closed under matrix multiplication, and that matrix multiplication is commutative on this set.



Should I use real numbers to solve this problem or use what the question gave??

Are you asking if you should replace a snd b with particular numbers?

If so the answer is no.

That M is closed under matrix multiplication is equivalent asking you
to prove that for any real numbers a, b, c, d, there exist two real numbers
e,f such that:

Code:

[a 0] [c 0] = [e 0]
[0,b] [0 d]   [0 f]

and that multiplication is comutative on M equivalent asking you
to prove that for any real numbers a, b, c, d that:

Code:

[a 0] [c 0] = [c 0] [a 0]
[0,b] [0 d]   [0 d] [0 b]

RonL

Originally Posted by CaptainBlack

Are you asking if you should replace a snd b with particular numbers?

If so the answer is no.

That M is closed under matrix multiplication is equivalent asking you
to prove that for any real numbers a, b, c, d, there exist two real numbers
e,f such that:

Code:

[a 0] [c 0] = [e 0]
[0,b] [0 d]   [0 f]

and that multiplication is comutative on M equivalent asking you
to prove that for any real numbers a, b, c, d that:

Code:

[a 0] [c 0] = [c 0] [a 0]
[0,b] [0 d]   [0 d] [0 b]

RonL

Thankyou for your help.

what exactly does a closed matrix mean? I'm confused.

I thought it meant.. in this case, that when I did the multiplication.. the resulting matrix would be the same size.

Originally Posted by kgpretty

what exactly does a closed matrix mean? I'm confused.

I thought it meant.. in this case, that when I did the multiplication.. the resulting matrix would be the same size.

Given a set S with an algebraic binary operation *
Take a subset R of S such that,
a*b in R for all a,b in R
Then we say R is closed under the binary operation *

--
Informally, closed mean whenever you take two numbers and you operate them you get the same number in the set.

For example,
If your matrix ended up being,
[0 a]
[b 0]
For a and b nonzero then your binary operation on this set would not have been closed because you have a diffrenet number (in this case a different matrix).

This is my 25th Post

Here's how I'm going to completely answer the question:

Code:

Let a,b,c,d,e,f be real numbers:

[a 0] [c 0] = [e 0]
[0 b] [0 d] = [0 f]

.: M is closed under matrix multiplication because the product of two M members is also in the M set.

Let A = [a 0]
        [0 a]

     B = [b 0]
         [0 b]

AB = [ab   0]
     [0   ab]

BA = [ab   0]
     [0   ab]

AB = BA

.: matrix multiplication is commutative on this set.

I think you meant
[a 0]
[0 b]
Not,
[a 0]
[0 b]

Anyways that is not the biggest problem, you need tos show that,
[ab 0]
[0 ba] is an element of these matricies.
Why?
Because a,b are real numbers and ab is closed under multiplication. That is the important step you omitted.
Because you need to show that,
[ab 0]
[0 ab]
is in the set which is true since ab is an element of the set.

Originally Posted by ThePerfectHacker

I think you meant
[a 0]
[0 b]
Not,
[a 0]
[0 b]

Anyways that is not the biggest problem, you need tos show that,
[ab 0]
[0 ba] is an element of these matricies.
Why?
Because a,b are real numbers and ab is closed under multiplication. That is the important step you omitted.
Because you need to show that,
[ab 0]
[0 ab]
is in the set which is true since ab is an element of the set.

Congrats on your 2500th post!
How do I show that ab is in M set? Do I just say let a and b be real numbers?

To prove commutativity:

Code:

 
Let A = [a 0]
        [0 b]
 
     B = [c 0]
         [0 d]
 
AB = [ac   0]
     [0   bd]
 
BA = [ca   0]
     [0   db]

But ordinary multiplication is commutative hence AB=BA,
and so matrix multiplication is commutative on this set.


RonL