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Can anyone tell me if I have made right thoughts in the attached task? To solve the system I just solve $x=A^{-1}b$. To find the 2-norm of matrix $A$ I use the "power method", because the 2-norm for a normal matrix like matrix $A$ is the largest eigenvalue for $A$?

I have found a formula for condition number for 2-norm of a normal matrix: $$ \mathrm{cond}(A)=\frac{\mathrm{eigenvalue}_\max(A)}{\mathrm{eigenvalue}_\min(A)} $$ To find $\mathrm{eigenvalue}_\min(A)$, I just find the eigenvalue for $A^{-1}$ with "inverse power method"? Have I understood it correctly?

Milten
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The inverse power method used on $A$ computes $\mathrm{eigenvalue}_\min(A)$. But remember that $A$ should be the input and not $A^{-1}$. $$ \frac{\mathrm{eigenvalue}_\max(A)}{\mathrm{eigenvalue}_\min(A)} = \mathrm{eigenvalue}_\max(A)\cdot \mathrm{eigenvalue}_\max(A^{-1}) = \Vert A\Vert_2 \cdot \Vert A^{-1}\Vert_2 =: \kappa(A) $$ so you formula is correct. The second equality is true when $A$ is normal.

Milten
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  • I don't understand. If it correct that eigenvalue_min(A)=eigenvalue_max(A^-1) then we must have: $$ \frac{\mathrm{eigenvalue}\max(A)}{\mathrm{eigenvalue}\min(A)} = \mathrm{eigenvalue}\max(A)\ / \mathrm{eigenvalue}\max(A^{-1}) = \Vert A\Vert_2 / \Vert A^{-1}\Vert_2 =: \kappa(A) $$? –  Oct 31 '19 at 15:15
  • $\mathrm{eigenvalue}\min(A) = \frac{1}{\mathrm{eigenvalue}\max(A^{-1})}$ – Milten Oct 31 '19 at 15:25
  • Thank you it makes sense. But how are the solutions for the equation system related to the condition-number? I have found that x=(2,0) for b=(2,2) and x=(1,1) for b=(2,2.001) and I have now found that cond(A)=4003.00075. I can't see anything relatable? –  Oct 31 '19 at 16:09
  • A large condition number means that even a small change in the b-vector can mean a large change in the solution vector. More precisely, the relative difference in x can be up to kappa times the relative difference in b. – Milten Oct 31 '19 at 16:14