Numerical Analysis/Matrix norm exercises

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

Find ${\Vert A\Vert }_1$

Solution:

given: ${\Vert A\Vert }_1=\max {\mathrm{\backslash limits}}_{1\le j\le n}\sum _{i=1}^n|a_{ij}|$

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

${\Vert A\Vert }_1$ = the maximum absolute column sum of the matrix

 

Find ${\Vert B\Vert }_1$

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

${\Vert B\Vert }_1=\max [12,15,18]=18$

 

Find the infinite norm, ${\Vert A\Vert }_{\mathrm{\infty }}$

Solution:

given: ${\Vert A\Vert }_{\mathrm{\infty }}=\max {\mathrm{\backslash limits}}_{1\le i\le n}\sum _{j=1}^n|a_{ij}|$

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

${\Vert A\Vert }_{\mathrm{\infty }}$ = the maximum absolute row sum of the matrix

 

Find the infinite norm, ${\Vert B\Vert }_{\mathrm{\infty }}$

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

${\Vert B\Vert }_{\mathrm{\infty }}=\max [6,15,24]=24$

 

Find the Frobenius norm, ${\Vert A\Vert }_F$

Solution:

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

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

 

Find the Frobenius norm, ${\Vert B\Vert }_F$

Solution:

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