$A_{n\times n}$ that implies : $A^2-2A+I=0$ Proof $1$ is an eigevalue of $A$ I have the following question :
Let $A_{n \times n}$ that implies : $A^2-2A+I=0$ 


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*Proof $1$ is an eigevalue of $A$


I don't really know how to approach this this what I manage to do (its not much though):


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*$A(A-2I)=-I$ 


We know that if $\lambda$ is an eigenvalue then $Ax=\lambda x$ $(x \neq 0)$
$$A(A-2I)=I$$
Can I say now that since $Ax=\lambda x$ and Let $x=(A-2I)$ but $x$ is vector, not a matrix.
I don't really understand what to do next.
Any help will be be dearly appreciated, Thanks.

Hint: For any nonzero vector $x$, $(A - I)^2 x = 0$.  If $(A-I) x = 0$ then ...
If not, then ...
