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I have the following situation:

If $A=I+B$ is a symmetric positive definite (spd) matrix and $rank(B)=r$, then the cg-method converges at most in $r+1$ steps.

So first of all, I hope I do understand correctly that I have to show that the cg-method for the matrix above will not give us the exact solution before reaching $r+1$ steps. Is that right?

Regarding the question itself: I have learned that for any regular spd matrix $C \in \mathbb {R^{n \times n}}$, we will get the exact solution after $n$ steps. If the rank of $B$ is $r$, then the rank of $A$ should also be $r$. Then why should the claim above be true? Or is there something I'm simply misunderstanding? Can someone give me a hint in the right direction?

Thanks in advance.

Yasuduck
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    No, that is not the right understanding. It says at most. It may converge sooner (e.g., what if your initial guess was the solution?). – parsiad May 29 '17 at 17:33

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