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 $\mathrm{Sym}(A)$?

  • $\begingroup$ What have you tried ? $\endgroup$
    – EDX
    Jun 27 at 18:36
  • $\begingroup$ @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). $\endgroup$
    – rmcerafl
    Jun 27 at 18:39
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 27 at 18:53

1 Answer 1


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 $\operatorname{Sym}(A^{-1})\preceq\operatorname{Sym}(A)^{-1}$.


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