functional equation in Mn(C)

alekk

Prove that any endomorphism $f$ of $M_n(\mathbb C)$ such that
$f(X\cdot Y)=f(X)\cdot Y+X\cdot f(Y)$
for any matrices $X,Y$ is of the following form: $f(X)=AX-XA$ for a fixed matrix $A$ .

Edited by Myth

**Posted:** Sun Jan 09, 2005 10:47 am

_________________
They misunderestimated me!

Moubinool

Such $f$ is called derivation, there is way to solve this using
Kronecker product of matrices

$A\otimes B = \left[\begin{array}{ccc}a_{11}B&...&a_{1n}B\\...&...&...\\a_{n1}B&..&a_{nn}B\end{array}\...$

I don't know if there is another way to solve it.

alekk

what's this proof with Kronecker product ?

**Posted:** Sun Jan 09, 2005 3:02 pm

_________________
They misunderestimated me!

Moubinool

To each matrix $A$ we define a vector
$vec A=(a_{11},...,a_{n1},a_{12},...,a_{n2},...,a_{1n},...,a_{nn})^{t}$

The condition on $f$ can be written like

(1) $vec(f(X,Y))=vec(f(X)Y)+vec(Xf(Y))$

We use the

Lemma: $g:M_n(C)\rightarrow M_n(C)$ linear map, then there exista unique matrix
$K(T) \in M_{n^2}(C)$ such that $vec(f(X))=K(T)vec(X)$ for all $X\in M_q(C)$

Use also $vec(AB)=(I_n\otimes A)vec(B)$

(1) becomes

$K(T)(I_n\otimes X)vec(Y)=(I_n\otimes f(X))vec(Y)+(I_n\otimes X)K(T)vec(Y)$

(2) $K(T)(I\otimes X) - (I\otimes X)K(T)=(I\otimes f(X))$ for all $X\in M_n(C)$

use the block decomposition $K(T)=(K_{ij})$ wher $K_{ij}\in M_n(C)$

(2) implique

$f(X)=K_{ii}X-XK_{ii}$
$f(X)=K_{ij}X-XK_{ij}$ for $i\neq j$

$i=1$ give $f(X)=AX-XA$ , $A=K_{11}$

Matrix Analysis
Horn & Johnson
Cambridge University Press

Theorem4.3.4

**Posted:** Mon Jan 10, 2005 4:47 am