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I have an iterative method \begin{eqnarray} X_{k+1}=(1+\beta)X_k-\beta X_k A X_k~~~~~~~~~~~~~~~~~ k = 0,1,\ldots \end{eqnarray} with initial approximation $X_0 = \beta A^*$ ($\beta$ is scalar lying between 0 to 1)converging to the moore-penrose inverse $A^+$. $\{X_{k}\}$ are sequence of approximations. I have to prove that $\{X_{k}\}$ defined by above equations satisfies $R(X_k)= R(A^*)$ and $N(X_k)=N(A^*)$ for $k\geq 0$. Where $R(X_k)$, $N(X_k)$ denotes the range and null space of $\{X_{k}\}$ .I am using mathematical induction to prove above result.This is what i did.

It trivially holds for $k = 0$ (since for $k = 0$,$X_0 = \beta A^*$). Suppose result is true for some k. Let $y\in N(X _k)$ be an arbitrary vector. From above equation we have \begin{eqnarray*} X_{k+1} y = (1+\beta)X_{k}y - \beta X_{k}AX_{k}y = 0 \end{eqnarray*} (since $X_{k}y = 0$). This gives $y\in ~N(X_{k+1})$ thus $N(X_k)\subseteq N(X_{k+1})$ holds true. Now my question is how the statement $R(X_k)\supseteq R(X_{k+1})$ can be proved analogously. I tried much but unable to figure out how to prove $R(X_k)\supseteq R(X_{k+1})$. Little hint will work for me. I would be very much thankful.

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Essentially, you are performing the (damped) Newton-Raphson method $$x_{k+1}=x_k-\beta\dfrac{f(x)}{f^\prime(x)}$$ to the equation $a=\dfrac1{x}$, but this time you are iterating with matrices instead of scalars. –  J. M. May 13 '12 at 4:42
    
@j.M. yes sir i am iterating using matrices. But i need little hint to prove that result. –  srijan May 13 '12 at 4:45
    
For searching purposes: the particular case of Newton-Raphson that OP is considering here is termed a (damped) Schulz iteration. See this for instance. –  J. M. May 13 '12 at 4:58
    
j.m. thank you sir that is helpful for me. –  srijan May 13 '12 at 5:04

1 Answer 1

up vote 2 down vote accepted

Here is a hint.

Let us first abstract things a little bit: one considers $Y=(1+\beta)X-\beta XAX$ and the aim is to prove that $R(Y)\subseteq R(X)$.

So, assume that $z$ is in $R(Y)$. This means that there exists $u$ such that $z=Yu$, that is, such that $z=(1+\beta)Xu-\beta XAXu$ and one wants to find $v$ such that $z=Xv$.

In a fit of inspiration, let us consider $v=\underline{\qquad}-\underline{\qquad}$.

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That mean further we can write $z=X(u+\beta u - \beta Axu)$ thus $v = (u+\beta u - \beta Axu)$ will serve our purpose. Am i right sir? And i feel i got my proof just needed your consent. –  srijan May 13 '12 at 8:06
1  
Bravo! $ $ $ $ –  Did May 13 '12 at 8:10
    
You made me very much happy today. I was trying this for the last few days. Thank you very much. –  srijan May 13 '12 at 8:12

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