Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following matrix: $$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ What is the norm of $A$? I need to show the steps, should not use Matlab...
I know that the answer is $\sqrt{\sqrt{5}/2+3/2}$. I am using the simple version to calculate the norm but getting different answer: $\sum_{i=0}^3\sum_{j=0}^3(a_{ij})^2=\sqrt{1+1+1+1}=2$ Maybe this is some different kind of norm, not sure.

This might help - i need to get a condition number of $A$, which is $k(A)=\|A\|\|A^{-1}\|$...that is why i need to calculate the norm of $A$.

share|improve this question
1  
The answer, as always, is Totti. No, seriously, a norm is just something that satisfies 3 axioms (positivity, homogeneity and triangle inequality). You need to specify what norm you are using for the question to make sense. –  anegligibleperson Dec 2 '12 at 6:43
    
Actually I do not have specific name for the norm. I guess the right type should be used to calculate MATRIX CONDITION NUMBER as i mentioned above. I know for sure that the answer is correct, just do not know how to get there (so that i would be able to do this for other matrices) :) Forza Roma :) –  ASROMA Dec 2 '12 at 6:50
add comment

2 Answers

up vote 1 down vote accepted

You are looking at the induced 2-norm of a matrix. Induced 2-norm of a matrix is given by \begin{align} ||A||_2=\max_{x\neq 0}~\frac{||Ax||_2}{||x||_2} \end{align}
There is a bit of theory behind it which will help you derive that induced 2-norm is infact the highest singular value of that matrix. To find the highest singular value, find $AA^T$ and find the highest eigenvalue of that matrix and take its square root. The condition number is nothing but the product of induced 2-norm of $A$ and its inverse. You can find all this stuff in any standard textbook on matrix analysis.

share|improve this answer
add comment

Here is how you find the norm of a matrix. Apply the definition of the norm of a matrix

\begin{align} ||A||_2 = \max_{||u||= 1}~||Au||_2. \end{align}

to the matrix you have been given. First, let's find $ ||Au||_2 $. Pick up an arbitrary vector $u=(x,y,z)^{T}$ such that $||u||_2 = 1$ and apply the given matrix to it

$$ Au= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \begin{bmatrix} x \\ y+z \\ z \\ \end{bmatrix} $$

$$ \implies ||Au||_2 = \sqrt{x^2+(y+z)^2+z^2} = \sqrt{(x^2 + y^2 + z^2) + 2yz + z^2}$$

$$ \sqrt{1 + 2yz + z^2}.$$

Now, we have

$$ ||Au||_2 = \sqrt{1 + 2yz + z^2} \implies ||A||= \max_{||u||_2=1 }~\sqrt{1 + 2yz + z^2}= \frac{1}{2}+\frac{\sqrt{5}}{2} . $$

share|improve this answer
    
A is not the identity matrix. Au =$\left( \begin{array}{c} x \\ y+z \\ z \end{array} \right)$ and not u itself. You're searching the maximum for $\sqrt{\frac{x^2+y^2+2 y z+2 z^2}{x^2+y^2+z^2}}$ –  kram1032 Dec 2 '12 at 11:38
    
@kram1032: Thank you. I'll correct this. –  Mhenni Benghorbal Dec 5 '12 at 3:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.