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I want to proof that $f(x)=Ax+b$ with \begin{align} A = \begin{bmatrix} 0 & -\frac{1}{8} & \frac{1}{4} \\ 0 & \frac{1}{3} & 0 \\ -\frac{1}{2} & -\frac{5}{22} & \frac{3}{4} \end{bmatrix} && b=[1,2,3]^{T} \end{align}

is a contraction mapping, that means $||f(x)-f(y)||\leq q ||x-y||$ with constant $q < 1$. But I don't know how I can get that constant.

Any ideas?

Thank you!

TheWaveLad
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    Look at $A^TA$, find its largest eigenvalue. Take the square root. – Daniel Fischer Nov 24 '13 at 15:21
  • @DanielFischer so $q=\sqrt{A^TA}$? Why? – TheWaveLad Nov 24 '13 at 15:34
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    What you want is $$\lVert A\rVert = \sup_{\lVert x\rVert = 1} \lVert Ax\rVert = \sqrt{\sup_{\lVert x\rVert = 1} \lVert Ax\rVert^2} = \sqrt{\sup_{\lVert x\rVert = 1} \langle Ax,Ax\rangle} = \sqrt{\sup_{\lVert x\rVert = 1} \langle x, A^TA x\rangle}.$$ – Daniel Fischer Nov 24 '13 at 15:38
  • What I have now is: $||f(x)-f(y)||=||Ax+b-Ay-b||=||A(x-y)||\leq||A|| \cdot ||x-y|| = \sqrt{\sup_{||x||=1}<x,A^TAx>}||x-y||$. The problem is that $A^TA$ is very hard to calculate (e.g. $a_{11}=\frac{17424}{69696}$..) So is there any other way to show that $||A|| < 1$? – TheWaveLad Nov 24 '13 at 16:17
  • What about $\rho(A)$? Isn't $\rho(A)\leq||A||$ for any matrix natural matrix norm? $\rho(A)=\frac{1}{8}(3+\sqrt{17})$ so: $||f(x)-f(y)||\leq \rho(A)||x-y||$? – TheWaveLad Nov 24 '13 at 16:20
  • @user2668777 If the spectral radius is strictly less than $1$, this will prove that it is a contraction mapping with respect to some norm on $\mathbb R^3$. If you are using a specific norm, the spectral radius might not work. It looks like the other people are answering for the Euclidean norm. Then $q = \sqrt{\rho{A^T A}}$, just like they said. – Stephen Montgomery-Smith Nov 24 '13 at 21:11

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