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Oct
22
awarded  Famous Question
Oct
17
comment Let $A''=\operatorname{End}_{A'}(V)=\operatorname{End}_{\operatorname{End}_A(V)}(V)$. Show that $A''$ is a $k$-algebra.
@quid You are right, just forgot it, updated it now.
Oct
17
revised Let $A''=\operatorname{End}_{A'}(V)=\operatorname{End}_{\operatorname{End}_A(V)}(V)$. Show that $A''$ is a $k$-algebra.
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Oct
17
asked Let $A''=\operatorname{End}_{A'}(V)=\operatorname{End}_{\operatorname{End}_A(V)}(V)$. Show that $A''$ is a $k$-algebra.
Oct
17
awarded  Taxonomist
Oct
16
accepted Let $L,K\in \operatorname{End}_k(V)$ such that $L\circ K =0$. Is there an easy way to see that $\operatorname{Im}(L) \cap \operatorname{Im}(K)=0$?
Oct
16
comment Let $L,K\in \operatorname{End}_k(V)$ such that $L\circ K =0$. Is there an easy way to see that $\operatorname{Im}(L) \cap \operatorname{Im}(K)=0$?
Ah okay, now I understand also my confusion, because my book said. Blabla. Therefore $LK=0$ and hence $\I(L)\cap \I(K)=0.$ So I thought there was some magical connection.
Oct
16
comment Let $L,K\in \operatorname{End}_k(V)$ such that $L\circ K =0$. Is there an easy way to see that $\operatorname{Im}(L) \cap \operatorname{Im}(K)=0$?
I understand the first part, thanks for that :) But the second part, I don't understand. You end with that $LK=0=KL$, but that is already our assumption right ?
Oct
16
asked Let $L,K\in \operatorname{End}_k(V)$ such that $L\circ K =0$. Is there an easy way to see that $\operatorname{Im}(L) \cap \operatorname{Im}(K)=0$?
Oct
16
comment Is a representation of a $k$-algebra a $k$-vector space?
@TobiasKildetoft
Oct
16
asked Can we say anything about the unit of a $k$-algebra $A$ in terms of the unit $1\in k$?
Oct
15
accepted Is it true that: $\sum _{i=0}^\infty\left( \prod _{j=0}^{i-1}p_{j}-\prod _{j=0}^{i}p_{j} \right)=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$?
Oct
15
revised Is it true that: $\sum _{i=0}^\infty\left( \prod _{j=0}^{i-1}p_{j}-\prod _{j=0}^{i}p_{j} \right)=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$?
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Oct
15
asked Is it true that: $\sum _{i=0}^\infty\left( \prod _{j=0}^{i-1}p_{j}-\prod _{j=0}^{i}p_{j} \right)=\left(1-\prod _{j=0}^{\infty }p_{j}\right)$?
Oct
13
comment Recurrence relation $x_0=1, x_n=p x_{n+1} + q x_{n-1}$
@Semiclassical The model here is the chance $x_n$ to get bankrupt if you begin with a stock of $n$ dollar. And have a change $p$ of winning a dollar, and chance $q$ of loosing a dollar.
Oct
13
revised Recurrence relation $x_0=1, x_n=p x_{n+1} + q x_{n-1}$
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Oct
13
comment Recurrence relation $x_0=1, x_n=p x_{n+1} + q x_{n-1}$
@Semiclassical I think there is no unique solution indeed, but is that a problem ? I forgot to mention that $x_n \in [0,1]$.
Oct
13
revised Recurrence relation $x_0=1, x_n=p x_{n+1} + q x_{n-1}$
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Oct
13
comment Recurrence relation $x_0=1, x_n=p x_{n+1} + q x_{n-1}$
@Semiclassical $x_0=1$ and $x_1=px_2+q$
Oct
13
revised Recurrence relation $x_0=1, x_n=p x_{n+1} + q x_{n-1}$
added 24 characters in body