Compute the condition number of matrix A=$\left(\begin{array}{cc} 1 & -1\\0&-1\end{array}\right)$ in norm $||.||_2$.
Since $A^{-1}$ is same as $A$. How to dela with that?
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The condition number of $A$ is the ratio of the maximum and minimum singular values of $A$, and the singular values of $A$ are the square roots of the eigenvalues of $A^T A = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$.
The characteristic polynomial of $A^T A$ is $\lambda^2 - 3 \lambda + 1$, so the eigenvalues of $A^T A$ are $\frac{3 \pm \sqrt{5}}{2}$. Thus the condition number of $A$ is \begin{equation} \kappa = \sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}} = \frac{3 + \sqrt{5}}{2}. \end{equation}
Alternatively, we could note that the norm of $A$ is the maximum singular value of $A$: \begin{equation} \|A\| = \sqrt{\frac{3 + \sqrt{5}}{2}}. \end{equation} Using your observation that $A^{-1} = A$, we have $\kappa = \|A^{-1}\| \|A\| = \|A\|^2 = \frac{3 + \sqrt{5}}{2}$.
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