Let $A \in GL(n,\mathbb C)$ be such that $0 \notin \{x^*Ax:x^*x=1\}$ ; then is it true that
$0\notin \{x^*A^{-1}x:x^*x=1\}$ ?
2026-04-24 02:27:01.1776997621
$A \in GL(n,\mathbb C)$ be such that $0 \notin \{x^*Ax:x^*x=1\}$ ; then is it true that $0\notin \{x^*A^{-1}x:x^*x=1\}$?
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Suppose $x^* A^{-1} x = 0$ with $x \neq 0$, then we can let $y = {1 \over \|A^{-1} x\|} A^{-1} x$, and then $y^* A y = {1 \over \|A^{-1} x \|^2 } (A^{-1} x)^* x = {1 \over \|A^{-1} x \|^2 } \overline{x^* (A^{-1}x)} = 0$, and since $y^*y = 1$, we have a contradiction.