Endomorphism of $mathcal{M}_n(mathbb{R})$ such that $ f({}^t M)={}^t f(M)$

Question

I and a friend are trying to find all endomorphisms $f$ of $mathcal{M}_n(mathbb{R})$ such that $f({}^t M)={}^t f(M)$ for all $M$. We believe they are of the form $Mmapstolambda M+mu {}^t M$ for a fixed $(lambda,mu)inmathbb{R}^2$. Any help is appreciated, thank you.

user1551 · Accepted Answer

Let $f$ be a linear endomorphism on $M_n(mathbb R)$. Then
$$
f(M^T)=f(M)^Tquadforall Mtag{1}
$$
if and only if
$$
f(M)=frac12left(g(M)+g(M^T)^Tright)quadforall Mtag{2}
$$
for some endomorphism $g$.
Given $g$, it is straightforward to verify $(1)$ when $f$ is defined by $(2)$. Conversely, given any $f$ that satisfies $(2)$, condition $(1)$ is satisfied by $g=f$.
Alternatively, note that $(1)$ is satisfied if and only if $f$ preserves both symmetric and skew-symmetric matrices. Hence such an $f$ takes the form of $f(M)=hleft(frac{M+M^T}{2}right)+kleft(frac{M-M^T}{2}right)$ where $h$ is an endomorphism defined on the space $mathcal H_n$ of all symmetric matrices and $k$ is an endomorphism defined on the space $mathcal K_n$ of all skew-symmetric matrices. Therefore the dimension of the space of all such $f$s is
$$
dimoperatorname{End}(mathcal H_n)+dimoperatorname{End}(mathcal K_n)
=left[frac{n(n+1)}{2}right]^2+left[frac{n(n-1)}{2}right]^2
=frac{n^2(n^2+1)}{2}.
$$

Nick · Answer

First, consider a basis of the space $mathcal{M}_n$. For example, the $n^2$ elementary matrices $e_{ij}$ which are all zero's except a $1$ in the $(i,j)$ spot. An element $f in mathrm{End}(mathcal{M}_n)$ is determined by where it sends each $e_{ij}$. Call these $M_{ij}$, so that $f(e_{ij}) = M_{ij}$. Then for an arbitrary matrix $A$ with entries $a_{ij}$, you have
$$ f(A) = sum_{i,j} a_{ij} M_{ij} $$
Now, you want to consider when $f$ commutes with the transpose operation. Consider this in the special case of the basis elements $e_{ij}$. Notice that ${}^te_{ij} = e_{ji}$, so the condition becomes :
$$
begin {align*}
f({}^te_{ij}) &= {}^tf(e_{ij}) \
f(e_{ji}) &= {}^tM_{ij} \
M_{ji} &= {}^tM_{ij}
end {align*}
$$

Endomorphism of $mathcal{M}_n(mathbb{R})$ such that $ f({}^t M)={}^t f(M)$

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