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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 $A$.

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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
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.

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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

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

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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

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