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Consider two matrices $M$ and $N$ of same dimension, such that $L=MN+NM$ is the symmetric sum in the sense that interchanging $M$ and $N$ does not change $L$. Is there any transformation $T[M]=M^\prime$ and $T[N]=N^\prime$ such that $M^\prime N^\prime+N^\prime M^\prime=MN+NM$? In other words, the symmetric sum is invariant under the transformation $T$.

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  • $\begingroup$ Yes, the identity. Surely you want something else? $\endgroup$
    – Saegusa
    Oct 7, 2022 at 20:00

2 Answers 2

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Not a full answer, but have you considered looking at what happens if the dimension is low? Suppose $M,N\in\mathbb{R}^\ast=\text{GL}_1(\mathbb{R})$ (we don't need to consider $0$, otherwise $L=0$). Then we have $MN+NM=2MN$ by commutativity.

Suppose $T$ is a linear transformation from $\mathbb{R}$ to itself. Then, your condition translates to $2MN=T(M)T(N)$. Since $T$ is linear, we have $T(M)T(N)=MT(1)NT(1)=MNT(1)^2$. Thus, $L=2MNT(1)^2$.

Since $M,N\in\mathbb{R}^\ast$, we can divide by $2MN$ on both sides to get $T(1)^2 = 1$, so $T(1) = \pm 1$, which means $T(x)=\pm x$ works. This already gives you two possible transformations, $\pm \text{Id}_n$. Using a basis (assuming your vector spaces are finite), I think those are the only two possible such transformations but you'd need to check by hand, which seems a bit tedious. Maybe someone with a better idea comes along, but at least this gives you two basic transformations that do what you want.

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Suppose the elements of the matrices are taken from a field whose characteristic is not $2$. Then the only possible choices of $T$ are $\pm\operatorname{Id}$.

Let $A=T(I)$. The given condition implies that $$ T(X)A+A\,T(X)=2X\tag{1} $$ for every matrix $X$. In particular, if we put $X=I$, we obtain $A^2=I$. Hence $A$ is diagonalisable and its only possible eigenvalues are $1$ and $-1$.

Suppose both $1$ and $-1$ are eigenvalues of $A$. Then $v^TA=-v$ and $Au=u$ for some nonzero vectors $u$ and $v$. But then we may pick two vectors $x$ and $y$ such that $v^Tx=y^Tu=1$ and with $X=xy^T$, $(1)$ implies that $$ 2=2v^T(xy^T)u=v^T\left(T(X)A+A\,T(X)\right)u =v^TT(X)(Au)+(v^TA)T(X)u=0, $$ which is a contradiction. Therefore $A=\pm I$ and $(1)$ implies that either $T(X)=X$ for all $X$ or $T(X)=-X$ for all $X$.

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