Numerical Analysis/Matrix norm exercises

Consider the general $3\times 3$ matrix $A={\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\\end{bmatrix}}$ and the specific example $B={\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\\\end{bmatrix}}$.

Find $\left\|A\right\|_{1}$

Solution:

given: $\left\|A\right\|_{1}=\max \limits _{1\leq j\leq n}\sum _{i=1}^{n}|a_{ij}|$

Solve for column 1: $\sum _{i=1}^{3}|a_{i1}|=|a_{11}|+|a_{21}|+|a_{31}|$

Solve for column 2: $\sum _{i=1}^{3}|a_{i2}|=|a_{12}|+|a_{22}|+|a_{32}|$

Solve for column 3: $\sum _{i=1}^{3}|a_{i3}|=|a_{13}|+|a_{23}|+|a_{33}|$

$\left\|A\right\|_{1}$ = the maximum absolute column sum of the matrix

 

Find $\left\|B\right\|_{1}$

Solution:

Solve for column 1: $\sum _{i=1}^{3}|b_{i1}|=|1|+|4|+|7|=12$

Solve for column 2: $\sum _{i=1}^{3}|b_{i2}|=|2|+|5|+|8|=15$

Solve for column 3: $\sum _{i=1}^{3}|b_{i3}|=|3|+|6|+|9|=18$

$\left\|B\right\|_{1}=\max[12,15,18]=18$

 

Find the infinite norm, $\left\|A\right\|_{\infty }$

Solution:

given: $\left\|A\right\|_{\infty }=\max \limits _{1\leq i\leq n}\sum _{j=1}^{n}|a_{ij}|$

Solve for row 1: $\sum _{j=1}^{3}|a_{1j}|=|a_{11}|+|a_{12}|+|a_{13}|$

Solve for row 2: $\sum _{j=1}^{3}|a_{2j}|=|a_{21}|+|a_{22}|+|a_{23}|$

Solve for row 3: $\sum _{j=1}^{3}|a_{3j}|=|a_{31}|+|a_{32}|+|a_{33}|$

$\left\|A\right\|_{\infty }$ = the maximum absolute row sum of the matrix

 

Find the infinite norm, $\left\|B\right\|_{\infty }$

Solution:

Solve for row 1: $\sum _{j=1}^{3}|b_{1j}|=|1|+|2|+|3|=6$

Solve for row 2: $\sum _{j=1}^{3}|b_{2j}|=|4|+|5|+|6|=15$

Solve for row 3: $\sum _{j=1}^{3}|b_{3j}|=|7|+|8|+|9|=24$

$\left\|B\right\|_{\infty }=\max[6,15,24]=24$

 

Find the Frobenius norm, $\left\|A\right\|_{F}$

Solution:

given: $\|A\|_{F}={\sqrt {\sum _{i=1}^{3}\sum _{j=1}^{3}|a_{ij}|^{2}}}$

$\|A\|_{F}={\sqrt {|a_{11}|^{2}+|a_{12}|^{2}+|a_{13}|^{2}+|a_{21}|^{2}+|a_{22}|^{2}+|a_{23}|^{2}+|a_{31}|^{2}+|a_{32}|^{2}+|a_{33}|^{2}}}$

 

Find the Frobenius norm, $\left\|B\right\|_{F}$

Solution:

$\|B\|_{F}={\sqrt {|1|^{2}+|2|^{2}+|3|^{2}+|4|^{2}+|5|^{2}+|6|^{2}+|7|^{2}+|8|^{2}+|9|^{2}}}={\sqrt {285}}\approx 16.9$