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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hi there! I have a relatively straightforward question that I'm stuck at. I've tried a few ways of trying to go ahead with it but none have worked so far, but I'll post them anyway as a means of providing a head start.

**Question: **If A is a square matrix such that AA = A and A != I (the identity matrix), show that det(A) = 0.

**First solution attempt:**

AA = A

AA - A = 0 (the zero matrix)

det(AA-A) = det(0)

det(A) * det(A-I) = 0

This implies that det(A) = 0 or det(A-I) = 0. I'm stuck at this part because of the det(A-I) = 0 part.

**Second solution attempt:**

AA = A

det(AA) = det(A)

det(A) * det(A) - det(A) = 0

det(A) * [det(A) - 1] = 0

This implies that det(A) = 0 or det(A) = 1. Once again, stuck with proceeding from the second choice of det(A).

**Third solution attempt:**

Assume that A is invertible, as in det(A) != 0.

Then, AA = A

A^(-1) AA = A^(-1) A

A = I

But since we know that A != I, then our assumption that det(A) != 0 is wrong. Therefore, det(A) must be equal to 0.

However, this solution using contradiction seems like it's fundamentally wrong, lol.

Any hints? Thank you!

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Also: I'm inexperienced with functions of multiple variables. How does one restrict their domains to make them one-to-one or their codomains to make them onto?

For instance, let's say we have the function f(x,y) = 1 / √(x+y). The domain is {(x,y) in the reals | x + y > 0} and the range obviously are all the positive real numbers. But how would one go about restricting their domains and codomains to make them one-to-one and onto?

I have a test on this in a couple of weeks so I'd like to get it down.

*Last edited by Anakin (2012-10-04 19:42:48)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Hi Anakin;

Playing around with the general 2 x 2 the determinant can be either 0 or 1.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hi Bobbym,

I think what you're trying to tell me is that for a general 2 by 2 matrix satisfying the condition that AA = A, the determinant can be either 0 or 1. But how do I relate that intuition into a proof? And for any n by n matrix as well.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Hi;

Yes, I can prove it for a for any 2 x 2 but I do not see how to generalize it to an n x n.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

I've tried to do a few examples with 3 x 3 matrices and the property holds but it is just difficult to prove. How would you go about proving it just for a general 2 x 2 matrix?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Hi;

So solve the matrix equation:

This yields 4 simultaneous non linear equations:

The solutions are:

The first one is a parametric solution showing there are an infinite number of such matrices. From this

it is trivial to show that the det is 1 or 0.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Alright Bobbym thanks - I'm very grateful!

*Last edited by Anakin (2012-10-04 21:37:28)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Hi;

Do not thank me, all I have done is made it more confusing by showing a dead end

approach. See post #6.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Oops, I meant to thank you before you posted the solution (for your effort to do so) but I guess you edited your original post.

But you're right, that just makes it more confusing. I suppose I'll go with my 2nd attempt solution and explain in words that the second case only exists when A = I and when A != I, then the det(A) = 0. If I recall correctly, the TA does not do his own work very "formally" so he may let it slide.

But thank you for at least trying to come up with something. I've searched all over but couldn't find a formal proof for the question, I suppose the informal will have to do.

However, do you have any general advice about restricting domains and codomains in R^2? Specifically, in order to make the function 1-1 or onto.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Hi Anakin;

I was unable to do that. Maybe someone else can help more.

Then, AA = A

A^(-1) AA = A^(-1) A

A = I

But I can say this there is an error in your second line of your 3rd attempt.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

That's alright, Bobbym. I'll see if someone else has something to add.

But thank you very much - as always - for the aid!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Did you see the error in line 2?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Oops you edited your post again before I was able to read it.

I don't believe there is an error except that I'm assuming that A is invertible, as in A^(-1) A = I. Which I suppose is the wrong way of doing it. For instance if we want to prove that X => Y. We can't assume (not Y) and assuming a part of X (in this case that AA = A) and then say that since another part of X (the A = I part) is not satisfied , we have a contradiction. Thus Y must be true.

That would just be counter-productive - which is why I am still unsatisfied with that solution. The other way I can do this is using the contrapositive:

det(A) != 0 (which means that A is invertible) IMPLIES that AA != A OR A = I.

But I didn't really get very far with that either.

*Last edited by Anakin (2012-10-04 22:41:26)*

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

I just found out (Google!) that there is a term for such matrices, they are called idempotent matrices. I'm going to look a little more into them now and see if something useful appears.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Hi;

I meant

A^(-1) AA

you have that reducing down to A, sholdn't it be IA?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hi Bobbym,

IA = A as one of the properties of the identity matrix is that IA = AI = A (for an n x n matrix; a little different for m x n).

*Last edited by Anakin (2012-10-04 23:39:02)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

Yes, I am sorry. Very obvious.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

sorry anakin

what is "A!" ?

30+2=28 (Mom's identity)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,355

(!=)That is computerese for not equal.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

ok thanks...

seems to me your (Anakin) third demonstration is right

call

p: AA=A;

p': A!=I;

q: detA=0

q': detA=1

we have that p=>(q OR q')

and q'=>NOT p' (or, equivalent, p'=>NOT q')

so p AND p' => (q OR q') AND (NOT q') = q

Do you agree?

EDIT: to be correct i think i should say:

q' AND p => not p' which is equivalent to p' => (NOT q' or NOT p) => NOT q'

because we assume p is true

*Last edited by Fistfiz (2012-10-05 07:35:58)*

30+2=28 (Mom's identity)

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Hi everyone,

Fistfiz: Yes, I think your formal logic work is correct. Actually, yes it is correct. My third demonstration, though correct, is a little incomplete. However, your post completes it. Luckily, I ended up using the same approach as you by using the 3rd demonstration with my second demonstration to make a complete proof. I'll post my solution below:

AA = A

det(AA) = det(A)

det(A) * det(A) - det(A) = 0

det(A) * [det(A) - 1] = 0

This implies that det(A) = 0 or det(A) = 1. We can accept the first case. As for the second case, it implies that A is invertible.

AA = A

A^(-1) AA = A^(-1) A

A = I

But one of our preconditions was that A != I, therefore we can ignore the second case because det(A) = 1 and AA = A <==> A = I. This concludes that if A != I, then det(A) must equal 0.

This is equivalent to what you wrote I think, just differently worded and without the use of formal logic.

*Last edited by Anakin (2012-10-05 17:16:29)*

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

Yes, it's exactly the same argument but wrote in a less pedant manner

In my opinion, your demonstration in post#1 is not incomplete: i've read lots of (basic level) demonstration where "formal" things

are left to intuition and they're still considered good demonstration

For example consider the classical demonstration that sqrt2 is irrational:

we start with two proposition which are:

p: (a/b)^2=2

q: a and b are irreducible

and we prove that p AND q => NOT q;

but we have to assume q to be true, because in each case it's false we get back to a case in which is true.

so NOT q is false... so p AND q is false, but since q is true, p must be false.

But i've never (for God's sake) read a math's book explaining that in proposition in these terms.

30+2=28 (Mom's identity)

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**Anakin****Member**- Registered: 2009-10-04
- Posts: 145

Interesting. I suppose it is rather complete then, isn't it? At least considering the familiar example that you posted.

I don't recall reading it in a textbook either but I have seen that exact example being proven the way that you've proven it - albeit it can look a little confusing at first.

Nevertheless, I did finish the question with all of your help and now it's onto this week's assignment! I'll be back if I get stuck.

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

Anakin wrote:

Nevertheless, I did finish the question with all of your help and now it's onto this week's assignment! I'll be back if I get stuck.

I hope so (not that you get stuck... that you post more interesting questions ) since i'm studying the same subject right now. Good work!

30+2=28 (Mom's identity)

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