For $1-w\bar w$ is positive definite , there exists an $n × n$ matrix $a$ with complex entries such that $(1- w\bar w)\{a\}= 1$

Question

For $1-w\bar w$ is positive definite , there exists an $n × n$ matrix $a$ with complex entries such that $(1- w\bar w)\{a\}= 1$ where $1$ in the identity matrix and $w$ is $ n$ rowed complex symmetric matrix. Where $ a\{b\}=\bar b^t a b$ . How to prove this as I know that all the eigen values are real and positive. Does this imply anything else? please help.

Answer 1

This has nothing to do with the property that $w$ is complex symmetric. For any Hermitian matrix $P$, if there exists a square matrix $A$ such that $A^\ast PA=I$, then $P=(AA^\ast)^{-1}$, which is positive definite. Conversely, if $P$ is positive definite, by taking $A=P^{-1/2}$, we have $A^\ast PA=I$.