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Let $K$ be an algebraic closed field and $M_n(K)$ the $K$-algebra of all matrices $n\times n$ over $K$. If $L$ and $M$ are two commutative isomorphic subalgebras of $M_n(K)$ it is true that there exista a regular matrix $S\in M_n(K)$ such that $SLS^{-1}=M$. That is the isomorphism is inner?

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Doesn't this answer the question (positively)? – darij grinberg Dec 30 '11 at 14:43
(Here $A=L$, $K=K$ and $B=M_n\left(K\right)$.) – darij grinberg Dec 30 '11 at 14:43
Is this homework? – Bruce Westbury Dec 30 '11 at 14:46
Take $n=4$. Let $L$ be the diagonal matrices whose upper left $2\times 2$-block and whose lower right $2\times 2$-block are scalar multiples of identity matrices. Let $M$ be the diagonal matrices whose upper left $3\times 3$-block is a scalar multiple of the identity matrix. – darij grinberg Dec 30 '11 at 15:49
If you aggree my answer, feel free to accept it. – Denis Serre Dec 30 '11 at 21:30

Two isomorphic subalgebra of $M_n(K)$ do not need to be conjugated. The following example is taken from Exercise 161 of my web site

Set $n=p+q$ with $q>p>0$. Then define $\mathcal A$ as the subset of $M_n(k)$ made of the matrices with block form $$\left(\begin{array}{cc} 0_p & 0_{p\times q} \\\\ A & 0_q \end{array}\right).$$ Likewise, ${\cal B}$ is made of the matrices $$\left(\begin{array}{cc} 0_q & 0_{q\times p} \\\\ B & 0_p \end{array}\right).$$ Both $\cal A$ and $\cal B$ are subalgebras of $M_n(k)$, with dimension $pq$ and the property that $MN=0_n$ for every two elements (of the same algebra). They are obviously isomorphic, because the algebra structure is trivial. But ${\cal A}$ and $\cal B$ are not conjugated in $M_n(k)$. However $\cal B$ is conjugated to ${\cal A}^T$ in $M_n(k)$.

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Dead link is dead. (Also, exorbis.pdf doesn't work either.) – darij grinberg Dec 30 '11 at 17:45
thank you Denis Serre. Your example responds to my question. Your answer made ​​me think of another question: two unital subalgebras (that is, containing the scalar matrices) are conjugated? – Miguel Dec 30 '11 at 18:05
Miguel, you could simply add the subspace $k\cdot I_n$ to each of the algebras - they would still be isomorphic but not conjugated. Also see my last comment for a different (diagonalizable) counterexample (which I hope to be correct). – darij grinberg Dec 30 '11 at 18:10
@darij. Sorry, I completed the link. – Denis Serre Dec 30 '11 at 21:19

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