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Suppose $A$ is a $n\times n$ real matrix and $det(A)<0$ .

Does there exixt a positive number $\delta$ such that $\|Ax\|\geq\delta\|x\|$, for all $x\in\mathbb{R}^n$ ?

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    every linear operator is bounded and continuous in finite dimensional space - linear operator can be represent as a matrix – 2017-01-24

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The question becomes easier if you square both sides (using nonnegativity) and divide by $x^Tx$ (which is again nonnegative, assuming $x \neq 0$): $$\frac{x^T A^TA x}{x^Tx} \geq \delta^2. $$ By the Rayleigh quotient: $$\frac{x^T A^TA x}{x^Tx} \geq \lambda_{\text{min}}(A^TA). $$ The matrix $A^T A$ ispositive definite as $x^T(A^TA)x = ||Ax||^2 > 0$ if $x \neq 0$. The smallest eigenvalue is therefore strictly positive. So, $\delta = \sqrt{\lambda_{\text{min}}(A^TA)}$ does the trick.