Describe $n$ x $n$ matrix $A$ , such that it follows $A+A^T = 2A^{-1}$?

Question

$A+A^T = 2A^{-1}$ => $(A+A^T)A= 2I$ $$\sum_{k=1}^n a_{kj}(a_{ik}+a_{ki}) = 2I_{ij}$$ So if $i = j$: $$\sum_{k=1}^na_{ki}(a_{ik}+a_{ki}) = 2$$ And if $i != j$: $$\sum_{k=1}^n a_{kj}(a_{ik}+a_{ki}) = 0$$ I am now stuck at this stage and can't figure out how to simplify the conditions.

Answer 1

Suppose we found $A \in \Bbb {GL_n}$ such that $A+A^T = 2A^{-1}$. We see that $$ A+A^T = 2{A^T}^{{-1}}=2A^{-1} $$ so $$ A=A^T. $$ Using the first equation, we get $$ A=A^{-1} $$ so $$ A^2=I_n. $$ For the other way, if $A^2=I_n$ and $A=A^T$, we have $A+A^T=2A=2A^{-1}$. Thus $$ A+A^T=2A^{-1} \iff A^2=I_n \text{ and } A=A^T. $$

Answer 2

Any symmetric and involutory matrix $B$ will fulfill the identity $B+B^{T}=2B^{-1}$. For example, the first Pauli matrix: $ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.

Answer 3

Claim: $A$ must be symmetric.

Proof: We have $$ A + A^T = 2A^{-1} \implies A^{-1} = \frac 12[A + A^T]. $$ We see that $A^{-1}$ is symmetric since $$ [A^{-1}]^T = \left(\frac 12[A + A^T]\right)^T = \frac 12[A^T + A^{TT}] = \frac 12[A^T + A] = A^{-1}. $$ It follows that $$ A^T = [(A^{-1})^{-1}]^T = [(A^{-1})^{T}]^{-1} = (A^{-1})^{-1} = A. \quad \square $$

From there, we see that $$ A + A^T = 2A^{-1} \implies 2A = 2A^{-1} \implies A^2 = I. $$ $A$ will satisfy your condition iff $A$ is symmetric with $A^2 = I$.