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For $A,B,L$ linear operators, when is there a linear operator $X\{A,B\}$ such that

$$ALA^{-1}+BLB^{-1}=2 XLX^{-1}$$

can be solved independently for all $L$ only depending on $A$ and $B$?

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Looks awfully like a generalized Sylvester equation... –  J. M. Aug 9 '11 at 8:08
    
If the equation $(a_{ij}b_{kl}-a_{kl}b_{ij})^2=0$ holds for all $i,j,k,l$, and the matrix entries are real or complex numbers, does it not follow that $B$ and $A$ linearly dependent? –  Jyrki Lahtonen Aug 9 '11 at 8:20
    
@JyrkiLahtonen: please disregard my 2x2 case, that was calculated when I omitted the $\frac12$ and assumed $L\neq1$. But maybe the answer still boils down to linear dependence –  Tobias Kienzler Aug 9 '11 at 8:36
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1 Answer

For any fixed $k \neq 2$, there does not exist any $X \in GL(n,\mathbb{C})$ such that $$ A L A^{-1} + B L B^{-1} = k X L X^{-1}$$ for all $L \in M_n(\mathbb{C})$, for if there were, then taking $L=1_n$ yields $$ 0 = A 1_n A^{-1} + B 1_n B^{-1} - k X 1_n X^{-1} = (2-k) 1_n,$$ which is impossible. In particular (unless I've misunderstood something), no such $X$ exists in your case, which is $k = \tfrac{1}{2} \neq 2$.


EDIT: In the case that $k = 2$, since every algebra automorphism of $M_n(\mathbb{C})$ is inner, and since $ALA^{-1} + BLB^{-1} = A(L + CLC^{-1})A^{-1}$ for $C = A^{-1}B$, your problem is equivalent to finding all $C \in GL(n,F)$ such that $L \mapsto \tfrac{1}{2}(L + CLC^{-1})$ is an algebra automorphism of $M_n(\mathbb{C})$. In the case that $C^2=1_n$, then you can check that this is true if and only if $C$ is a non-zero scalar multiple of the identity, in which case $\tfrac{1}{2}(L+CLC^{-1}) = L$, so that $B$ must be a non-zero scalar multiple of $A$, and you can just take $X=A$. Anything more general looks like a bit of a mess, though perhaps you can compute something explicit for $n=2$?

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The $k^{-1}$ in the displayed equation should ber $k$, no? –  coffeemath Jan 1 '13 at 17:11
    
Indeed. Thanks for the correction! –  Branimir Ćaćić Jan 1 '13 at 17:20
    
+1 Urgh, I must have slept when posting this question... Thanks for your input, please apologize that in that case I change the question into the one with correct factor... –  Tobias Kienzler Jan 1 '13 at 17:21
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I was actually wondering why you wanted a $\tfrac{1}{2}$ there, but with a $2$ everything makes sense. I've been trying to apply the necessary condition that $L \mapsto \tfrac{1}{2}(ALA^{-1}+BLB^{-1})$ be multiplicative, but all that seems to imply is something like $ALMA^{-1} + BLMB^{-1} = ALA^{-1}BMB^{-1} + BLB^{-1}AMA^{-1}$; plugging $L = 1$ or $M = 1$ yields nothing new, whilst plugging in, say, $L = A^{-1}$ and $M = A$ doesn't seem to yield anything helpful. –  Branimir Ćaćić Jan 1 '13 at 17:30
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