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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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There must be a vector that is unchanged by multiplication with the matrix, I don't really think there is much more to it... –  Jaime Dec 11 '12 at 0:56
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In relation to what Jaime said, think about how matrix multiplication works with a column vector X, that is, what has to be true about the system of equations A represents? This only gets you so far though, really there's only one thing this gives you from the linear algebra properties (the things you already mentioned). –  Vilid Dec 11 '12 at 1:01
    
$A-Id$ would then have eigenvalue $0$, hence would be nonivertible. –  Berci Dec 11 '12 at 1:06
    
@berci, that's not technically true, consider I, it has eigenvalue 1, and is most definitely non-singular. Ah missed what you were saying, never mind. –  Vilid Dec 11 '12 at 1:08
    
@Vilid: but $I-I = 0$ is most definitely singular. –  Robert Israel Dec 11 '12 at 1:10

1 Answer 1

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. $\text{Ran}(A^T - I) \ne {\mathbb F}^n$
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