0
$\begingroup$

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 $\mathrm{Sym}(A)$?

$\endgroup$
3
  • $\begingroup$ 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

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.