Numerical Analysis using MATLAB. Find the condition number $\mu$

Question

Find the condition number $\mu = |A||A^{-1}|$ for the Hilbert Matrix $A$ using the uniform form.

$A = \left( \begin{array}{cccc} 1 &\frac{1}{2} & \frac{1}{3} &\frac{1}{4} \\ \frac{1}{2} & \frac{1}{3} &\frac{1}{4} &\frac{1}{5} \\ \frac{1}{3} &\frac{1}{4} &\frac{1}{5} &\frac{1}{6}\\ \frac{1}{4} &\frac{1}{5} &\frac{1}{6}&\frac{1}{7} \end{array} \right) $

So I have a couple of question:

1.What does uniform form mean?

2.What does $|A|$ stand for? Determinant?

3.Is all the problem about plugging the $A$ into the formula?

Does anyone know the algorithm of doing this in MATLAB?

Please help! Thank you.

Answer 1

Seems like this is a hand written question and either you or your instructor has made some minor mistakes. It's very likely that the original question should read:

Find the condition number $\mu=\|A\|\|A^{-1}\|$ for the Hilbert Matrix $A$ using the uniform norm.

Here $\|X\|$ denotes the uniform norm of a matrix $X=(x_{ij})_{i,j\in\{1,2,\ldots,n\}}$. The uniform norm, a.k.a. maximum row sum norm, is the matrix norm induced by the vector norm $\|x\|_\infty$. It is defined by $$ \|X\|=\max_{1\le i\le n}\,\sum_{j=1}^n|x_{ij}|. $$ In matlab, $\|X\|$ can be calculated as $\mathtt{max(sum(abs(X),2))}$.