Math Is Fun Forum

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.

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.

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?

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,

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?

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'.

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 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.