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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.

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.

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).

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.

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

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

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.

Alright Bobbym thanks - I'm very grateful!

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.

**Anakin**- Replies: 23

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!

-----------------------------------------------------------------------------------------------

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.

Alright then, I'll go with that as well. Plus it fits the formula from the page you linked. 5 hours until class, I'm gonna catch some sleep.

Thanks Bobbym (and Bob) for the help, I'm very grateful.

That's exactly what I've got above as well. I'm not sure whether you're simply checking if my work was free of errors or whether that is the solution to the question. Could you clarify?

Also, I meant plugging the intervals into sqrt(1 + t^3) by canceling the derivative and integral out would yield a different answer from the one I got.

Hi Bobbym,

Okay, thanks. I appreciate it!

Hi,

By looking at the link that you last posted, I was able to get the following solution. Could you tell me if it seems right?

Also, it would seem that simply cancelling the integral and derivative would lead to a different solution - one that is lacking the 4x and sin(x) in front of the square roots.

Which is the right one then?

Bob: If one was to change the variable in the function, wouldn't the intervals on the integral also change?

Bobbym: I mean differentiating the definite integral, with respect to x. Or at least that's what I think the question appears to be asking for.

**Anakin**- Replies: 13

Hello! I was just working on a review assignment and in the 10% or so left, I've come across a question for which I have no idea on how to begin.

I've looked into the http://en.wikipedia.org/wiki/Leibniz_integral_rule but I'm not sure if that is the method I should be employing, as I've never come across it before. Or it is possible that I have but like I said, this is a review assignment and I haven't been in a math course for over 2 semesters so I may have forgotten it.

Any ideas on how to get started? I tried using substitution to express √(1+t^3) using only x but to no avail as the integral of √(1+t^3) itself looks far too long and complicated to be required for such a straightforward question (using a math engine).

I got the vector by inspection; though I suppose if the elements of B' were really complicated, we'd have to solve using a system of equations.

The matrix

|0 0|

|1 1|

has to be represented with the elements of basis B'. So I just used inspection.

The reason why I found T(1), T(x), T(x^2), T(x^3) is because those are the elements of the basis B and any polynomial whose linear transformation we want, that polynomial can be written in terms of these elements of basis B.

I'm gonna get a quick 8 hour nap before I power through all my notes one last time for my exam. I'll be back in a little while.

No problem.

It means that it is the matrix outputted by the linear transformation but needs to be converted with respect to ordered basis B', that would be = [0, -1, 1, 1]. The matrix itself that you showed has not been changed to B' coordinates until it is listed as [0, -1, 1, 1].

So essentially we want to be able to take any polynomial with degrees less than or equal to 3 (including the zero polynomial), these are in ordered basis B. So 3x^2 - 2x would be [0, -2, 3, 0] in basis B. We want to take their linear transformation, which outputs a matrix. Then we want to represent that output with respect to basis B'.

By creating the matrix [T] (subscript B,B') in part A of my question, we can simply take a polynomial's representation in basis B and multiply it with this matrix [T] (subscript B,B') to get the linear transformation in basis B'.

T(1) means that the values of a, b, c = 0 and d = 1. Thus the linear transformation results in a matrix

|0 0|

|1 1|

With respect to basis B', it means we need -1 of the 2nd matrix, 1 of the 3rd and 1 of the 4th to get it because

|0 -1| + |0 1| + |0 0|

|0 0| |1 0| |0 1|

=

|0 0|

|1 1|

Well the linear transformation is T(ax^3 + bx^2 + cx + d) = the matrix

|a-b b+c|

|c+d d-a|

If we try to find T(x^3 + x^2 - x + 1), it appears that the values of a, b, c, d are 1, 1, -1, 1 respectively.

The reason why we try to the find matrix representation relative to B,B' in the first part is so we can just do a matrix multiplication to find the answer without having to compute a-b, b+c, etc. and then convert it to basis B'.

With the matrix, we can just take the representation of the input (x^3 + x^2 - x + 1 in this case) with respect to basis B, which is [1, -1, 1, 1] in this case and do a matrix multiplication, and we get our result in basis B'.

Hi Bobbym!

Yup, second year of university and work make it nearly impossible to get online nowadays. Can't wait until the holidays start.

The computation is correct but my question was that is this how one would carry out the procedure to find the linear transformation of (x^3 + x^2 - x + 1) with basis B'?

Something leads me to believe that perhaps my final answer isn't wrong because say that we ignore the matrix representation of T relative to B, B'. Then if we do T(x^3 + x^2 - x + 1), we notice that a = 1, b = 2, c = -1, d = 1.

The linear transformation would be the matrix:

|1-1 1+(-1)|

|(-1)+1 1-1|

=

|0 0|

|0 0|

which basis B' would just be [0, 0, 0, 0] as I got as my answer to number 2.

Does that sound reasonable?

**Anakin**- Replies: 11

Hello everyone; it's been quite some time since I last posted. Anyway, I'm going for my Linear Algebra II exam on Saturday and all my studying is nearly done and prepared for.. except one question! I'll post the question and my work and see if someone can guide me:

I'm not sure I even did anything properly there as I somewhat missed that unit because I had a heavy week.

Thank you!

Hi gAr,

I appreciate you taking the time to find that link. Apparently my method was wrong (according to that source) and after some thinking, rightfully so. I've made the adjustments and I'm now confident in my answers.

Thanks once again.

Hi gAr,

I'm not actually certainly sure that my Q1 is right as it uses a one sided limit and it uses negative infinity, but I think it is by using the following reasoning:

Let N < 0 be arbitrary, and thus it is negative which makes sense as the limit is -∞.

We choose δ = 2^N.

If 0 < |x-3| < δ:

<=> -δ < x-3 < δ

<=> -δ < 3-x < δ (by multiplying each side by -1 and thus signs reverse)

=> 3-x < 2^N (δ = 2^N as chosen)

=> log(3-x) with base 2 < N (as 3-x > 0 since x is approaching 3 from the left side and the base >= 1 so the sign doesn't change).

Therefore, the function log(3-x) with base 2 has a limit of -∞ as it is always less than N when approaching 3 from the left side.

End of proof.