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?!


1 Answer 1


I do not understand your question about part b.

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 $$ K_2(A) = \|A\|_2\cdot \|A^{-1}\|_2 = \sqrt{\lambda_n} \cdot \sqrt{\frac 1{\lambda_1}} = \sqrt{\frac{\lambda_n}{\lambda_1}}. $$


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