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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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    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 :)2012-12-02

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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} . $