Science:Math Exam Resources/Courses/MATH307/December 2008/Question 07/Solution 1

The infinity norm of a matrix is the infinity norm of the vector whose entries are the 1-norm of each row, so it will be

${\Vert A\Vert }_{\mathrm{\infty }}=\max \{2+1+0,1+2+1,0+1+2\}=4.$

The 2-norm of a matrix is the largest of the absolute values of the eigenvalues. So we need to calculate the eigenvalues of A as the roots of the characteristic polynomial

$\begin{array}{rl}\det (A-\lambda I)& =\det \left|\begin{array}{ccc}-2-\lambda & 1& 0\\ 1& -2-\lambda & 1\\ 0& 1& -2-\lambda \end{array}\right|\\ & =(-2-\lambda )\det \left|\begin{array}{cc}-2-\lambda & 1\\ 1& -2-\lambda \end{array}\right|-(1)\det \left|\begin{array}{cc}1& 0\\ 1& -2-\lambda \end{array}\right|\\ & =(-2-\lambda )((-2-\lambda )(-2-\lambda )-1)-(-2-\lambda )\\ & =(-2-\lambda )({\lambda }^2+4\lambda +2)\end{array}$

Hence, the eigenvalues of A are ${\lambda }_1=-2$ and ${\lambda }_{2,3}=\frac{-4\pm \sqrt{16-8}}{2}=-2\pm \sqrt{2}$. The 2-norm of A is the largest of the absolute values of these and therefore ${\Vert A\Vert }_2=|-2-\sqrt{2}|=2+\sqrt{2}$.