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!

]]>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.

]]>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.

]]>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.

]]>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

what is "A!" ?]]>

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).

]]>I meant

A^(-1) AA

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

]]>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.

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

]]>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.

]]>