# Symmetric part of the inverse matrix

Let $$A\in\mathbb{R}^{d\times d}$$ be an invertible matrix with inverse $$A^{-1}$$ and symmetric part $$\mathrm{Sym}(A):=\tfrac{1}{2}(A + A^\intercal).$$

Do you know of any reasonable relations between $$\mathrm{Sym}(A^{-1})$$ and $$\mathrm{Sym}(A)$$?

• What have you tried ?
– EDX
Jun 27 at 18:36
• @EDX I'm asking if there are any relations that you would know, as I myself don't think there is much that can be said (hence the question). Jun 27 at 18:39
• When $d=2$ the cofactor matrix (determinant times inverse) of Sym$(A^{-1})$ is the same as the cofactor matrix of Sym$(A)^{-1}$. But that doesn't seem to generalize. Jun 27 at 18:53

In case $$\operatorname{Sym}(A)$$ is positive definite, we have $$\operatorname{Sym}(A^{-1})\preceq\operatorname{Sym}(A)^{-1}$$.
Let $$A=P(I-K)P$$ where $$P=\operatorname{Sym}(A)^{1/2}\succ0$$ and $$K$$ is skew-symmetric. Let $$K=QBQ^T$$ be an orthogonally block-diagonalisation of $$K$$ to its real Jordan form $$B$$. Then $$\operatorname{Sym}\left((I-B)^{-1}\right)\preceq I$$ because $$\operatorname{Sym}\left(\pmatrix{1&k\\ -k&1}^{-1}\right) =\operatorname{Sym}\left(\frac{1}{1+k^2}\pmatrix{1&-k\\ k&1}\right) =\frac{1}{1+k^2}I_2 \preceq I_2$$ for each $$2\times2$$ diagonal sub-block of $$\operatorname{Sym}\left((I-B)^{-1}\right)$$. It follows from $$A^{-1}=\left[PQ(I-B)Q^TP\right]^{-1} =P^{-1}Q(I-B)^{-1}Q^TP^{-1}$$ that $$\operatorname{Sym}(A^{-1})\preceq\operatorname{Sym}(A)^{-1}$$.