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On the matrix equation $X+A\sp TX\sp{-1}A=I$. (English) Zbl 0863.15005
It is shown that the matrix equation $(*)$ $X+A^TX^{-1}A=I$ has a positive definite solution $(X>0)$ if and only if $A$ has a factorization $A=W^TZ$, where the matrix $W$ is nonsingular and the columns of $[W^T,Z^T]^T$ are orthonormal. In this case $X=W^TW$. It is also proved that equation $(*)$ has a solution $X>0$ if and only if there exist orthogonal matrices $P$ and $Q$ and diagonal matrices $\Gamma>0$, $\Sigma\ge 0$ with $\Gamma^2+\Sigma^2=I$ such that $A=P^T\Gamma Q\Sigma P$. Finally, it is shown that if $(*)$ has a solution $X>0$ then the following relations are valid $X-AA^T>0$, $I-AA^T-A^TA>0$, $r(A)\le 1/2$, $r(A+A^T)\le 1$, $r(A-A^T)\le 1$, where $r(A)$ is the spectral radius of $A$.

##### MSC:
 15A24 Matrix equations and identities
##### Keywords:
matrix equation; positive definite solution; factorization
Full Text:
##### References:
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