# Need little hint to prove a theorem .

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 $X =A^+$. $\{X_{k}\}$ are sequence of approximations. Assume that after $sth$ iteration initiates $rank(X_{s})>rank(A)$. Then the method will diverge.

I have confusion in the following line of the proof.

Since $N(X_{s})\subseteq N(X_{s}AX_{s})$ and $rank(X_{s}AX_{s})\leq rank(A)<X_{s}$(no doubt till here)

Thus we conclude that the inclusion is strict(here i have confusion) i.e. $N(X_{s})\subset N(X_{s}AX_{s})$ and thus there exist a non zero vector $y\in N(X_{s}AX_{s})\setminus N(X_{s})$.

A little hint will be sufficient for me. Rest of the things i have done on my own. I am stucked here. I would be highly thankful.

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Hint: rank + dim Ker = ... – Did May 13 '12 at 13:43
@Didier Little more hint required sir. – srijan May 13 '12 at 13:51
Rank–nullity theorem. – Did May 13 '12 at 13:53
@Didier I am applying this theorem on $XAX$ and $X$ seperately. Then what i have to show then get the proof? I am confused here sir. – srijan May 13 '12 at 13:57
@Didier Dear sir i got this conclusion $N(XAX)>N(X)$ – srijan May 13 '12 at 14:02