In many diciplines that utlizes mathematics, we often see the equation $$y=y(x)$$ where $y$ might be other replaced by whichever letter that makes the most sense in ...

May 2, 2013 · 3 You could modify the proof of Sylvester's determinant theorem to show that $$\text {det} \left ( \lambda I_m - XY \right) = \text {det} \left ( \lambda I_n - YX \right)$$ for all $\lambda \neq 0$. …

Jul 8, 2018 · Claim : $H$ and $K$ are two normal subgroups of group $G$ such that $H\cap K = \ {e\}$ then $ xy = yx $ for $x \in H$ and $y \in K$. Proof : Let $x \in H$ and $y\in K$ then I need to prove …

Jan 22, 2015 · Verify my proof of $G$ is nilpotent iff $xy=yx \forall x,y\in G$ such that $ (o (x),o (y))=1$. Ask Question Asked 10 years, 10 months ago Modified 10 years, 10 months ago

Jul 3, 2024 · You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I …

Jun 28, 2017 · i guess you use morphisms, which i haven't learned yet. was trying to make a proof with some basic properties

Aug 3, 2022 · I am showing that if for some group $G$, $xy=yx$ for every $x, y \in G$ then $$ (xy)^n=x^ny^n.$$ I claim this holds by induction on $n$. So base case if $n=1$, we ...

Nov 20, 2019 · Is the sum of these two matrices of rank $1$, i.e. is the matrix $xy^ {\mathrm T}+yx^ {\mathrm T}$ of rank $1$? I think the answer is positive and I am trying to come up with an argument …

Nov 26, 2024 · My question is: $$\frac {\ln (y)\cdot y^x-yx^ {y-1}} {\ln (x)\cdot x^y-xy^ {x-1}}\stackrel {?} {\equiv}\frac {xy\ln (y)-y^2} {xy\ln (x)-x^2}\stackrel {?} {\equiv}\frac {\ln (y)-\frac {y} {x}} {\ln (x)-\frac {x} …

When beginning to study multivariate calculus, you almost immediately come across the Schwarz-Clairaut Theorem which gives sufficient conditions to guarantee that mixed partials are equal. For the ...