Reputation
Next privilege 15,000 Rep.
Protect questions
Badges
2 11 53
Impact
~176k people reached

19h
comment Product of regular spaces
May be you can write what you have tried to understand the proposition..
19h
comment Recovering past years
This is definitely off topic but i can suggest some thin... I suggest you to read prefaces of some of the recommended books in your area of interest.. This would definitely motivate you... It worked for me some times... I recommend you to do that...
19h
comment Differentiable function has measurable derivative?
@MarkMcClure : I was just wondering how do we define differentiability of $f$ at $b$ for a function $f:[a,b]\rightarrow \mathbb{R}$.. Is it sufficient to see existence of its left derivative?
1d
comment Let A be a square matrix such that $A^3 = 2I$
@SharmaineSwee : You can appreciate by upvoting this answer... There are two arrows to the left of this answer.. One which points above one which points below.. If you like this, you click the one which points above if you do not like this you click the one which points below..
1d
comment Let A be a square matrix such that $A^3 = 2I$
You are welcome..
1d
comment Let A be a square matrix such that $A^3 = 2I$
This is gem of an answer assuming that the person who asked this knows what eigenvalues are..
1d
comment Let A be a square matrix such that $A^3 = 2I$
Yes Yes.. You are correct..
1d
answered Let A be a square matrix such that $A^3 = 2I$
1d
comment Let A be a square matrix such that $A^3 = 2I$
Is there any information about order of matrix $A$
1d
comment Differentiable function has measurable derivative?
@MarkMcClure : I do not usually edit answers.. But this one was very old one and i thought i can do.
1d
revised Differentiable function has measurable derivative?
added 1 character in body
1d
comment Differentiable function has measurable derivative?
I think there is a typo... $f_n\rightarrow f'$
Feb
10
comment When $\operatorname{im}(A) = \ker(A)$
@gbox : Nipotent matrices are such that $A^k=0$ for some $k$ which may not imply $A^2=0$... So, it may be true for all nilpotent matrices..
Feb
10
comment When $\operatorname{im}(A) = \ker(A)$
I have said nothing about rank and dimension... I was commenting on the image and kernel.. @Svetoslav
Feb
10
comment $G$ be a finite group and $f \in Aut (G)$ such that $f^3$ is identity and $f$ has unique fixed point , then any $p$-Sylow subgroup is normal?
It may be useful to say the background of the problem so that the one who wants to answer may have better idea what properties they can use
Feb
10
comment When $\operatorname{im}(A) = \ker(A)$
If rank is $1$ then you are saying something about the image and kernel.. I suggest you to check than one more time.. It may not be correct..
Feb
10
answered When $\operatorname{im}(A) = \ker(A)$
Feb
10
comment Anticommuting matrices and their eigenvalues
It is reasonable to think about some variation only when you know the proof of present theorem... It is just an advice...
Feb
10
comment Prove that every simple function is Riemann integrable.
Write down your argument...
Feb
10
comment Prove that every simple function is Riemann integrable.
yes.. Precisely that is what you are supposed to show..