TITLE: Inverse of the anti-commutator
QUESTION [3 upvotes]: Let $B \in M_{n \times n}(\mathbb{R})$ be a fixed $n \times n$ symmetric positive matrix. Consider the anti-commutator map
$$f:M_{n \times n}(\mathbb{R})\to M_{n \times n}(\mathbb{R})$$
via
$$f(C)=BC+CB$$
We can prove that $f$ is an injective linear operator and thus invertible. (Find a basis so that B is diagonal and everything is clear.) My question:
Is there a good expression for $f^{-1}$? 
I am looking for an expression for $f^{-1}(D)$ as some polynomial or power series in terms of $B, B^{-1}, D$.

REPLY [0 votes]: If $B$ has the eigenvalues $t_1,\dots,t_n$, then the eigenvalues of $f=f_B$ are $(t_i+t_j)_{1\le i\le j}$. So the characteristic polynomial of $f_B$ is $P(X)=\prod_{i,j}(X-t_i-t_j)$. This polynomial is known: $P(X)$ is (maybe up to sign) equal to the resultant of $\chi(Y)$ and $\chi(X-Y)$, where $\chi(X)=\prod_i(X-t_i)$ is the characteristic polynomial of $B$ and the resultant is computed viewing terms as polynomials on $Y$. Write $P(X)=\sum_{i=0}^{n^2}m_iX^i$ ($m_{n^2}=1$; the other coefficients can be explicitly computed using a formula on the resultant, or using roots: e.g. $m_0=(-1)^n\prod_{i,j}(t_i+t_j)$). Then $P(f_B)=0$. So
$$f_B^{-1}(C)=-m_0^{-1}\sum_{i=0}^{n^2-1}m_{i+1}f_B^{i}(C)$$
for all $C$.
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