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Is $\dim(\ker(A-I))$ equal to $\text{rank}(A)-\text{rank}(A-I)$?

I mean does $$\dim(\ker(A-I))=\text{rank}(A)-\text{rank}(A-I)$$

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    Could you explain what you mean by $A-I$? Maybe you should tell more about your specific problem. – 2017-02-03

1 Answers 1

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I suppose that $A$ is a square matrix and $I$ is the identity martix.

Let $A=O$.

Then $\dim(ker(A-I))=0$, but $rank(A)-rank(A-I)=-rank(-I) \ne 0$