$A_{n\times n}$ 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 : $A^2-2A+I=0$ 


*

*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.
 A: Hint: For any nonzero vector $x$, $(A - I)^2 x = 0$.  If $(A-I) x = 0$ then ...
If not, then ...
A: Note that
$(A - I)^2 = A^2 - 2A + I = 0; \tag{1}$
thus for all $n$-vectors $x$,
$(A - I)^2 x = 0; \tag{2}$
if
$(A - I)x = 0 \tag{3}$
for all $x$, then
$Ax = Ix = x \tag{4}$
for all $x$ as well; in this case, $A = I$ is the identity matrix, and every nonzero vector is an eigenvector of $A$ with eigenvalue $1$. If, on the other hand, there is some $x$ with
$(A - I)x \ne 0, \tag{5}$
setting
$y = (A - I)x, \tag{6}$
from (2) it follows that 
$(A - I)y = (A - I)(A - I)x = (A - I)^2x = 0, \tag{7}$
that is,
$Ay = Iy = y; \tag{8}$
which shows that $y \ne 0$ is an eigenvector of $A$ with eigenvalue $1$; thus we see that $1$ is always an eigenvalue of $A$.
A: Let $x$ be an eigenvector of $A$ so that $Ax=\lambda x$, then $$(A^2-2A+I)x=A(Ax)-2Ax+Ix=\lambda Ax-2\lambda x +x=(\lambda^2-2\lambda+1)x=0$$
And with $x\neq 0$ we have $\lambda=1$

As per comments below we need to know we have an eigenvector in the first place.
This pretty much gets straight back to Robert Israel's answer. 
We have $(A-I)^2=0$ so that for any vector $y$ we have $$(A-I)^2y=(A-I)[(A-I)y]=0y=0$$so that if $(A-I)y=z\neq 0$ $z$ is an eigenvector of eigenvalue zero of $A-I$ i.e. we have $$(A-I)z=0$$ whence $Az=z$ and otherwise ...

Comment
This was a little bit of a rushed attempt to work from the definition. Here it is much easier to work with the eigenvectors which are in plain sight - they don't actually need to be found. So this isn't a great answer, but I'm leaving it up as an example of how things can go wrong, just in case it helps others.
A: This means that the minimal polynomial of $A$ divides $x^{2}-2x+1=(x-1)^{2}$ and hence is either $(x-1)$ or $(x-1)^{2}$. So $1$ is a root of the minimal polynomial of $A$ hence an eigenvalue of $1$.
A: You should notice the factorization as a square as if $A$ were a real number:
$$
A^2-2A+I=(A-I)^2=\bf{0}
$$
where $\bf{0}$ denotes the zero matrix of dimension $n\times n$. If $A-I$ were injective, the zero map would have to be injective, but clearly it is not.
A: You have $(A-I)^2=0$, hence $(x-1)^2$ is a divisible by the minimal polynomial of $A$, which is therefore $x-1$ or $(x-1)^2$. In both cases, as the eigenvalues are the roots of the minimal polynomial, the only eigenvalue is $1$. In the first case, $A =I$; in the second case, the matrix is not diagonalisable.
