Matrices and Eigenvalues of Norm

Let $$A$$ be a nonsingular $$n \times n$$ matrix, $$\|\cdot\|$$ be any natural norm, and $$K_{p}(A)=$$ $$\|A\|_{p}\left\|A^{-1}\right\|_{p} .$$ Let $$\lambda_{1}$$ be the smallest and $$\lambda_{n}$$ be the largest eigenvalues of the matrix $$A^{t} A$$

(b) Show that $$K_{2}(A)=\sqrt{\frac{\lambda_{n}}{\lambda_{1}}}$$. (Hint: One should use the $$\|A\|=\sqrt{\rho\left(A^{t} A\right)}=\sqrt{\rho\left(A A^{t}\right)}$$ relation)

For part $$(b)$$, it seems minimal eigenvalue of matrix $$A$$ will be the largest eigenvalue of the inverse of matrix $$A$$, so we have to use this idea but could not understand formally why square roots are needed for $$K_{2}(A)$$ and how spectral radius is necessary for this part?!

By definition, we have $$K_2(A) = \|A\|_p \cdot \|A^{-1}\|_p$$. Note that $$A^TA$$ and $$AA^T$$ have the same eigenvalues. With that, we have \begin{align} \|A^{-1}\|_2 &= \sqrt{\rho(A^{-T}A^{-1})} = \sqrt{\rho([AA^T]^{-1})} \\ & = \sqrt{\lambda_{\max}([AA^T]^{-1})} = \sqrt{\frac{1}{\lambda_{\min}(AA^T)}} = \sqrt{\frac 1{\lambda_1}}. \end{align} It follows that $$K_2(A) = \|A\|_2\cdot \|A^{-1}\|_2 = \sqrt{\lambda_n} \cdot \sqrt{\frac 1{\lambda_1}} = \sqrt{\frac{\lambda_n}{\lambda_1}}.$$