0

Let $K$ be a field and $D:M_n(K)\to M_n(K)$ be a map defined by the properties

i) $D(X+Y)=D(X)+D(Y)$

ii) $D(XY)=D(X)Y+XD(Y)$

iii) $D(k)=0$

for $X,Y\in M_n(K)$ and $k\in K$. In other words, $D$ is the differentiation map over the $K$-algebra $M_n(K)$. Find a matrix $A\in M_n(K)$ such that $D(X)=AX-XA$ for all $X\in M_n(K)$.

I know that as a $K$-algebra, $M_n(K)$ is generated by the matrices $E_{ij}$ but I do not know how to follow. Any help/hint would be appreciated. Thanks in advance...

confused
  • 499
  • Show that $D$ is a derivation of the Lie algebra of $M_n(K)$, and hence that $D$ is an inner derivation, i.e., $D(X)=[A,X]=AX-XA$. For an elementary proof see also here. – Dietrich Burde Jul 09 '23 at 13:36
  • On the link you attached, I didn't understand why $D(E_{ii})=0$. I wrote a comment to that link too. Can you explain me where I am confused? – confused Jul 10 '23 at 16:19

0 Answers0