Suppose $A$ and $B$ are nonsingular matrices in $End(V)$ (the space of endomoprhisms for an n-dimensional vector space $V$, hence a space of nonsingular matrices). Let $[A, B] = AB - BA$. Is it possible to find matrices $A$ and $B$ such that $[A, B] = AB - BA = E_{ij}$ where $i \neq j$? I other words, is it possible to find such matrices so that $[A, B]$ is a matrix with zeros everywhere except in the $ij$-th position for $i \neq j$?
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Try two nonsingular symmetric matrices which do not commute. – jnyan Aug 05 '17 at 06:10
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@jynan: the result will be skew-symmetric. – Robert Lewis Aug 05 '17 at 06:27
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Doesn't it satisy the OP's requirement? It has zero in diagonals. Or am I wrong? – jnyan Aug 05 '17 at 06:56
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@jnyan If $A$ and $B$ are nonsingular noncommuting symmetric matrices then $[A,B]$ will be skew-symmetric. The OP asked for $[A,B] = E_{ij}$ which is not skew-symmetric. As explained in the question $E_{ij}$ is the matrix with zeros everywhere except in the $ij$'th position. That is not skew-symmetric. – Zach Teitler Aug 05 '17 at 07:10
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See Does the set of matrix commutators form a subspace? and Traceless matrices and commutators.
Here is an answer using the idea from the first of those links. Let $A$ be a diagonal matrix with distinct, nonzero diagonal entries $a_1,\dotsc,a_n$. Let $B = A + E_{ij}/(a_i-a_j)$. Then $A$ is invertible because it is diagonal with nonzero diagonal entries, and $B$ is invertible because it is upper or lower triangular with nonzero diagonal entries; and we have that $$ \begin{split} AB-BA &= A(A+E_{ij}/(a_i-a_j))-(A+E_{ij}/(a_i-a_j))A \\ &= (AE_{ij}-E_{ij}A)/(a_i-a_j) \\ &= (a_i E_{ij} - E_{ij} a_j)/(a_i-a_j) \\ &= E_{ij} , \end{split} $$ as desired.
Zach Teitler
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