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What would be the necessary condition for a matrix of any $n \times n$ to have eigenvalue 1? I know that it must have a corresponding eigenvector - that is obvious - I want to know things like how values must be, or things like that.

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    @Vilid: but $I-I = 0$ is most definitely singular.2012-12-11

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The necessary and sufficient condition for $A$ to have eigenvalue $1$ is that $A-I$ is singular. This is equivalent to any of

  1. $\det(A-I) = 0$
  2. $\text{Ker}(A-I) \ne \{0\}$
  3. $\text{Ran}(A-I) \ne {\mathbb F}^n$ (where we're working over the field $\mathbb F$)
  4. $\text{Ker}(A^T - I) \ne \{0\}$
  5. $\text{Ran}(A^T - I) \ne {\mathbb F}^n$