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Show that if A is s pxp real matrix such that $||I_p-A|| <1$, then for any choice of b$\in\mathbb{R}^p$ and $u_0\in\mathbb{R}^p$ the vector $u_{n+1}=b+(I_p-A)u_n$ converge to a solution x of the equation Ax=b as n tends to infinity.

I have got a hint: Ax=b if and only if $b+(I_p-A)x=x$. Define $f(x)=b+(I_p-A)x$ and invoke fixed point theorem.

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Hint: Show that $f$ is a contraction. –  Tomás Jan 25 '13 at 13:30
    
@Tomás: could you please make it more detail? thank you. –  aneps Jan 25 '13 at 13:37

1 Answer 1

Note that $$|f(x)-f(y)|<|x-y|$$ i.e. $f$ is a contraction. Now you can use this.

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