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Describe all differentiation on matrix algebra $M_n(F)$ over an associative commutative ring with identity $F$

Well as I understand that task I have Leibniz notation which says that every differentiation must satisfy $D(F,G) = F*D(G)+D(F)*G$. Also we have differentiation that looks like $D_A(B) = A*B-B*A$. So to describe I should to find how matrix A looks like. But how to do that?

Lobster
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1 Answers1

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Let $Z$ be the matrix of $M_n(F)$ such that $Z$ contains only the zero of $F$ everywhere.

Let $E_{i,j}$ be the matrix of $M_n(F)$ such that $E_{i,j}$ contains only the identity of $F$ in $(i,j)$ and the zero of $F$ everywhere else.

We are looking for $D : M_n(F) \rightarrow M_n(F)$ such that \begin{align} D(A+B)=&\;D(A)+D(B) & (1)\\ D(AB)=&\;AD(B)+D(A)B & (2) \end{align}

  1. Show that $D(Z)=Z$ (using $(2)$).
  2. Show that $\forall i\; D(E_{i,i})=Z$ (using $(2)$ and $E_{i,i}=E_{i,i}.E_{i,i}$ and $Z=E_{i,i}.E_{j,j}$ for $i\neq j$).
  3. Show that $\forall i,j\;D(E_{i,j})=c_{i,j}.E_{i,j}$, for some $c_{i,j}\in F$ (using $(2)$ and $E_{i,j}=E_{i,i}.E_{i,j}=E_{i,j}.E_{j,j}$).
  4. Show that $\forall i,j,k\; c_{i,j}+c_{j,k}=c_{i,k}$ (using $(2)$ and $E_{i,k}=E_{i,j}.E_{j,k}$).
  5. Show that $D$ is entirely defined by $c_{i,i+1}$ (and (1)), and such a $D$ is indeed a differentiation (using (1), (2) and previous results).
user26857
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Xoff
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  • I have only one question - in (2) . (point) means multiplication? – Lobster Feb 03 '15 at 17:27
  • @Lobster yes, it means equation (2). – Xoff Feb 03 '15 at 18:11
  • It's been eight years but if you still remember I have a question. How did you find that $D(E_{ii})=Z$? We know that $D(E_{ii})=D(E_{ii}\cdot E_{ii})=E_{ii}D(E_{ii})+D(E_{ii})E_{ii}$. However, this doesn't imply that $D(E_{ii})=Z$. It may be of the following form $$\begin{bmatrix} &&&1&&&\&&&\vdots&&& \ &&&1&&&\1&\cdots&1&0&1&\cdots&1\&&&1&&&\ &&&\vdots&&&\&&&1&&& \end{bmatrix},$$ ($i$th row and $i$th column). – confused Jul 10 '23 at 09:58
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    @confused. But for all $i\neq j$ $Z=D(Z)=D({,}.{,})={,}.D({,})+D({,}).{,}$ – Xoff Jul 17 '23 at 21:10
  • Ohh now I see. But what exactly is the matrix $A$ given in the question? It seems like $A\neq (c_{i,j}){i,j}$. Because otherwise, say $B=(b{i,j})$, $(D(B)){i,j}=b{i,j}c_{i,j}$ but $(AB-BA){i,j}= \sum c{i,t}e_{t,j}-\sum e_{i,t}c_{t,j}$. – confused Jul 19 '23 at 10:50