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Let $A$ and $B$ be $n\times n$ real matrices.

When $n=2$, we have the equality $$A\Big(\mbox{Trace}(B)A-\mbox{Trace}(A)B\Big) B=B\Big(\mbox{Trace}(B)A-\mbox{Trace}(A)B\Big) A.$$

  1. Can we give an interpretation to this equality?

  2. Are there similar equalities when $ n = 3,4,...$?

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    $\begingroup$ Is this an actual identity? If you take A to have zero trace and B to have nonzero trace then the above implies AAB=BAA $\endgroup$ Jun 15, 2013 at 22:08
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    $\begingroup$ Ralph, when $n=2$ and $A$ has trace zero, its square is a scalar matrix, whence $A^2B=BA^2$. $\endgroup$ Jun 15, 2013 at 22:40
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    $\begingroup$ This sounds like a polynomial identity. There are many over $M_n(k)$, the simplest being that $S_{2n}(A^1,\ldots,A^{2n})=0_n$, with $S_m$ the standard non-commutative polynomial in $m$ variables. See the MO question mathoverflow.net/questions/38698 . $\endgroup$ Jun 16, 2013 at 10:41

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Let us rewrite it using the commutators $[P,Q]=PQ-QP$, as follows: $$ tr(B)[A^2,B]=tr(A)[A,B^2]. $$ Now, for matrices $X$ of size~$2$, we have $X^2=tr(X)X-det(X)I$ (a particular case of Cayley--Hamilton), so $$ tr(B)[tr(A)A-det(A)I,B]=tr(A)tr(B)[A,B]=tr(A)[A,tr(B)B-det(B)I], $$ since $I$ commutes with everything.

This might be the most economic proof of your identity; moreover, it is known (Procesi, Razmyslov) that every identity with traces for $n\times n$-matrices does follow from the Cayley--Hamilton identity, so generalisations of your identity should be obtained in a similar way.

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Let $A'$ and $B'$ be the classical adjoints of $A$ and $B$. Write

$$S(A,B)=(B+B')A^2B-AB^2(A+A')$$ $$T(A,B)=BA^2(B+B')-(A+A')B^2A$$ $$U(A,B)=B'A^2B-AB^2A'$$ $$V(A,B)=BA^2B'-A'B^2A$$

Then:

1) For any $n$ we have (trivially) $$S(A,B)-T(A,B)=U(A,B)-V(A,B)$$

2) For $n=2$ (but not otherwise), $S(A,B)$ is equal to your left-hand side and $T(A,B)$ is equal to your right-hand side.

3) For $n=2$ (but not otherwise) it's easy to check that $U(A,B)=V(A,B)$.

The proof of (2) above relies on the fact that, for $n=2$ (but not otherwise) we have $B+B'=trace(B)$, and therefore $B+B'$ commutes with everything. There's no good generalization of this to $n>2$.

With a little more patience and/or cleverness, one might hope to juggle the expressions for $S,T,U$ and $V$ a little so that 1) and 2) remain true as stated, while 3) becomes true for every $n$. This would essentially imply that your identity is equivalent to $B+B'=tr(B)$ and explain why we shouldn't expect it to generalize past $n=2$.

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  • $\begingroup$ What does it mean to say $B+B'=tr(B)$? The left-hand side is a matrix while the right-hand side is a number. $\endgroup$
    – user21162
    Jun 16, 2013 at 6:04
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    $\begingroup$ robninson1: The right hand side is the matrix $tr(B)I$ where $I$ is the identity matrix. $\endgroup$ Jun 16, 2013 at 6:22

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