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

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Question 07
Consider the symmetric matrix $A={\begin{bmatrix}-2&1&0\\1&-2&1\\0&1&-2\end{bmatrix}}$ Compute the matrix norms $\left\|A\right\|_{\infty }$ and $\left\|A\right\|_{2}$

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Hint
The infinity norm of a matrix is simply the maximum absolute row sum of the matrix, that is, the largest 1-norm of its rows. The 2-norm for a symmetric matrix is the largest absolute value of the eigenvalues.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\displaystyle \left\|A\right\|_{\infty }=\max\{2+1+0,1+2+1,0+1+2\}=4.$ 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 ${\begin{aligned}\det(A-\lambda I)&=\det {\begin{vmatrix}-2-\lambda &1&0\\1&-2-\lambda &1\\0&1&-2-\lambda \end{vmatrix}}\\&=(-2-\lambda )\det {\begin{vmatrix}-2-\lambda &1\\1&-2-\lambda \end{vmatrix}}-(1)\det {\begin{vmatrix}1&0\\1&-2-\lambda \end{vmatrix}}\\&=(-2-\lambda )((-2-\lambda )(-2-\lambda )-1)-(-2-\lambda )\\&=(-2-\lambda )(\lambda ^{2}+4\lambda +2)\end{aligned}}$ Hence, the eigenvalues of A are $\displaystyle \lambda _{1}=-2$ 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 $\left\|A\right\|_{2}=|{-2}-{\sqrt {2}}|=2+{\sqrt {2}}$.

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