Compute the condition number of a $3 \times 3$ matrix.

Compute the condition number of the matrix $B$ :

$$B=\begin{bmatrix}3&7&1\\5&8&0\\6&3&2\end{bmatrix}$$ in terms of $\|\cdot\|_{1}$ and $\|\cdot\|_{\infty}$.

Important: Our definition of the condition number:

Consider a linear map $A : X \rightarrow Y$, then we define the condition number as follows: $$C(A) = \frac{\sup\limits_{x \in X, \|x\|_X=1} \|Ax\|_Y}{\inf\limits_{x \in X, \|x\|_X=1} \|Ax\|_Y}$$ with $\|\cdot\|_X$ and $\|\cdot\|_Y$.

So I've read that if $A \in \operatorname{isometry}(X,Y)$ then $C(A) = \|A\|_{L(X,Y)} \|A^{-1}\|_{L(Y,X)}$ with $L(X,Y)$. Do I have to use this? And if I have to use $C(A) = \|A\|_{L(X,Y)} \|A^{-1}\|_{L(Y,X)}$, can you explain what $L(X,Y)$ and $L(Y,X)$ are in this exercise?

• Please type $3\times 3$ for $3\times 3$ instead of 3 $\times$ 3 for 3 $\times$ 3 because, after all, $3$ is a mathematical symbol. – Shaun Jun 30 '18 at 11:33
• $L(X,Y)$ is the vector space of linear maps from $X$ to $Y$; in this case $X=Y = \mathbb{R}^3$ of course. If $X$ and $Y$ have norms (as is common) then $L(X,Y)$ carries the so-called operator norm, usually: $\|A\|_{L(X,Y)} = \sup \{\{\|Ax\|_Y : x \in X, \|x\|_X = 1 \}$ – Henno Brandsma Jun 30 '18 at 11:58
• – LutzL Jun 30 '18 at 12:20
• @JackJ. It is good that you try to improve posts by suggesting edits. However, it would be great if you would make your suggestions more thorough. Take a look at Xander Henderson's edit here for inspiration. – Daniel Fischer Jun 30 '18 at 19:16
• @ Daniel Fischer♦ Thank you. I will do it! – Jack J. Jun 30 '18 at 19:23

Let us compute for the $L^1$-norm. All "modules" applied on a vector $$x=\begin{bmatrix}a\\b\\c\end{bmatrix} \ne 0$$ reflect the $1$-norm for a while.

Then \begin{aligned} \sup_{|x|=1}|Bx| &=\sup_{x\ne 0}\frac{|Bx|}{|x|} \\ &=\sup_{(a,b,c)\ne 0}\frac{|3a+7b+c|+|5a+8b|+|6a+3b+2c|}{|a|+|b|+|c|} \\ &\le \sup_{(a,b,c)\ne 0} \frac {3|a|+7|b|+|c|+5|a|+8|b|+6|a|+3|b|+2|c|} {|a|+|b|+|c|} \\ &= \sup_{(a,b,c)\ne 0} \frac {14|a|+18|b|+3|c|} {|a|+|b|+|c|} \\ \\ &\le \sup_{(a,b,c)\ne 0} \frac {18|a|+18|b|+18|c|} {|a|+|b|+|c|} \\ &=18\ . \end{aligned} And the value $18$ is taken for $(0,1,0)$. Note that we are taking the maximal value for the sum of the modules of the column entries of the given matrix, this is the second column, the result is $\max(3+5+6, \ 7+8+3,\ 1+0+2)=18$. For the infimum... Let $C$ be the matrix $$C = B^{-1} =\begin{bmatrix}3&7&1\\5&8&0\\6&3&2\end{bmatrix}^{-1} =\frac 1{55} \begin{bmatrix} -16 & 11 & 8 \\ 10 & 0 & -5 \\ 33 & -33 & 11 \end{bmatrix} \ .$$ Then \begin{aligned} \inf_{|x|=1}|Bx| &=\inf_{x\ne 0}\frac{|Bx|}{|x|} \\ &\qquad\text{Substitution: Bx=y...} \\ &=\inf_{y\ne 0}\frac{|y|}{|Cy|} \\ &=\left(\sup_{y\ne 0}\frac{|Cy|}{|y|}\right)^{-1} \\ &=\left(\max\ \left(\ \frac 1{55}(16+10+33),\ \frac 1{55}(11+0+33),\ \frac 1{55}(8+5+11),\ \right)\ \right)^{-1} \\ &=\left(\frac{59}{55}\right)^{-1} =\frac{55}{59}\approx 0.932203389830508\dots \ . \end{aligned} We have only to take the quotient now, $$C(B)=\frac{18}{55/59}= \frac {1065}{55}\approx 19.3090909090\dots\ .$$

Same game for the $\infty$-norm. All "modules" on a non-zero vector are denoting in the following this norm. \begin{aligned} \sup_{|x|=1}|Bx| &=\sup_{\substack{x=(a,b,c)^T\\|a|,|b|,|c|\le 1}}|Bx| \\ &=\sup_{|a|,|b|,|c|\le 1}\max(\ |3a+7b+c|,\ |5a+8b|,\ |6a+3b+2c|\ ) \\ &\le \sup_{|a|,|b|,|c|\le 1}\max(\ 3+7+1,\ 5+8,\ 6+3+2\ ) \\ &=\max(\ 3+7+1,\ 5+8,\ 6+3+2\ )=13\ . \end{aligned} This time we are taking the maximal value of the sums of the absolute values on the rows. The second row won. The maximum is taken for $a=b=c=1$.

For the infimum value, we pass again to $C=B^{-1}$, the corresponding supremum is $\frac 1{55}(33+33+11)=\frac 15(3+3+1)=\frac 75$, so the infimum needed for $B$ is its inverse, $$\frac 57\approx 0.\,714285\,714285\dots\$$ So $$C(B) = \frac{13}{5/7}=\frac{91}5=18.2 \ .$$

Computer simulation. This experience should not be underestimated. It is often important to be able to check in a second such results. Here using sage:

Code:

import random
sup1, inf1, supoo, infoo = 0, oo, 0, oo    # init, we will make some of them bigger, the other smaller...

n1  = lambda  a,b,c:         abs(a) + abs(b) + abs(c)
noo = lambda  a,b,c:    max( abs(a) , abs(b) , abs(c) )

B = matrix(QQ, 3, 3, [3,7,1, 5,8,0, 6,3,2] )

for trial in xrange(10**6):
a, b, c = [ random.uniform(-1,1) for _ in (1,2,3) ]
s, t, u = B*vector( [a,b,c] )

norm1 = n1( s,t,u ) / n1( a,b,c )
if norm1 > sup1:    sup1 = norm1
if norm1 < inf1:    inf1 = norm1

normoo = noo( s,t,u ) / noo( a,b,c )
if normoo > supoo:    supoo = normoo
if normoo < infoo:    infoo = normoo

print "1-norm:", sup1, inf1, sup1/inf1
print "oo-norm:", supoo, infoo, supoo/infoo


Results this time:

1-norm: 17.9935919413 0.936889759239 19.2056661564
oo-norm: 12.9999942319 0.716228500506 18.1506240295

• Thanks, i am correcting it! – dan_fulea Jul 1 '18 at 12:16