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15h
comment Find the matrix representing T and Find the Image of T (as a span of vectors)
The matrix you found is correct. The image of $T$ is the span of the columns of $A$.
Feb
5
comment Check the proof of $||x||^2$ is not a norm
@ElChapo Her question is "Can someone check my proof for correctness"
Feb
5
comment Check the proof of $||x||^2$ is not a norm
Yes, that's all correct. You didn't have to do the third case: after 2) you already know that it is not a norm.
Jan
30
comment A question about a proof of Noetherian modules and exact sequences
I don't think a proof-verification can be a duplicate of a newer question. Unless someone posts the exact same proof with the exact same mistakes.
Jan
22
comment Is the space $C[0,1]$ complete?
@karhas Yes, that's right: it's definitely needed in the proof that $f$ is the uniform limit of the $f_n$.
Jan
10
comment $a\mid b$ and $b\mid a$ but $a$ and $b$ are not associates
I couldn't find a definition of associates (other what's in @frogeyedpeas' link and there it's the same as $a\mid b$ and $b \mid a$). Can you include a definition in your question?
Jan
9
comment Continuous function on a compact metric space is uniformly continuous
@user46944 No, it's not a typo.
Jan
1
comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$
Oh, I wasn't trying to make a point -- I was merely asking a question (to which I don't know the answer).
Jan
1
comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$
+1 by the way, for posting your solution, regardless of whether it's correct or not.
Jan
1
comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$
Isn't $L^1$ the space of Lebesgue integrable functions? If it is, could you justify why you can use the Riemann integral in your first equation (the one with the $\infty$ on the left hand side)?
Dec
29
comment Is this contradiction faulty?
@ReinhildVanRosenú It seems to be a very confused person who wrote that website. For example, if $S$ is all the values where $f$ is $\le 0$ the person makes a case distinction for values where $f$ is $>0$. Perhaps you can find better proofs in a book. Or on Wikipedia.
Dec
22
comment Modulo of a negative number
But then your answer states that method 1 is correct while method 2 is not. Or is it not?
Dec
20
comment Properties of dual spaces of sequence spaces
I will have to read the whole thread and this answer as it's not even clear to me what pairing here means exactly. Sounds like either $f$ or $g$ is fixed because we probably want to show that these angle brackets (in one argument) define an isomorphism. But I won't have time to do this before tomorrow night.
Dec
20
comment Properties of dual spaces of sequence spaces
@AnthonyPeter That's an excellent question. I will have to think about this as I do not know the answer off the top of my head.
Dec
16
comment Contractible vs. Deformation retract to a point.
This I explain in the first paragraph of my answer. Pick a point in $Y$ and an $\varepsilon$ ball around it. Can you find a neighbourhood inside the ball that is contractible? I don't think so -- it's not even connected!
Dec
8
comment Let $S$ be a non-empty set with an associative, cancellative operation and for each $a\in S$, $\{a^n\}$ is finite, must S be a group?
Possible duplicate of Confused about this exercise question: if a set with a certain binary operation is a group
Dec
3
comment Examples of absolutely continuous functions that are not Lipschitz.
@FardadPouran Yes, as I point out in my comment above.
Nov
28
comment If an abelian group has more than 3 elements of order 2 then it must have at least 7 elements with order 2.
Alan's answer has been undeleted.
Nov
23
comment Orders of the elements in $\mathbb{Z}/8\mathbb{Z}$
You're confusing additive and multiplicative notation I suspect. The neutral element here is $0$ not $1$.
Nov
23
comment Cauchy Residue Theorem Application
Craig, I left a comment in response to your comment to my answer in another thread. Since you probably won't be notified (as you seem to have deleted your comment) I am notifying you here.