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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):

  • $A(A-2I)=-I$

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.

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    $\begingroup$ Don't get used to using the word "implies" like that, it is incorrect. As stated, a simple counter example is any matrix that isn't 2x2 with a determinant that isn't 1. You should say "let $A$ be an $n \times n$ matrix with the property $\dots$". $\endgroup$ – DanielV Aug 19 '15 at 20:04
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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 $A$.

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

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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.

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  • $\begingroup$ Can you explain why $(A^2-2a+I)x=(\lambda^2-2\lambda+1)x=0$ I don't understand this equality $\endgroup$ – JaVaPG Aug 19 '15 at 19:49
  • $\begingroup$ the equality implies that when you expand and simplify the expression, there's $Ax$ and $\lambda x$ on respective sides. $\endgroup$ – Eemil Wallin Aug 19 '15 at 19:55
  • $\begingroup$ @EemilWallin I don't quite understand how? $$Ax=\lambda X$$ If we multiply by $A$ then : $$A^2x=\lambda A x$$ I don't understand how you get it by expand the expression. $\endgroup$ – JaVaPG Aug 19 '15 at 19:56
  • $\begingroup$ I think you should first justify that $A$ has an eigenvector. That's why I like Robert's answer better. $\endgroup$ – Jyrki Lahtonen Aug 19 '15 at 20:04
  • $\begingroup$ @JyrkiLahtonen Of course that does need doing - I got interrupted ... and that is why Robert's answer is definitely better - I've done some editing accordingly. $\endgroup$ – Mark Bennet Aug 19 '15 at 20:58
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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 $1$.

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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.

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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.

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  • $\begingroup$ Can you explain why $(A-I)^2 \implies (x-1)^2$? $\endgroup$ – JaVaPG Aug 19 '15 at 19:51
  • $\begingroup$ I meant that when you substitute $A$ for $x$ in this polynomial, you obtain the null matrix. Thus $(x-1)^2 \in\{p(x)\in\mathbf R[x]\mid p(A)=0\}$, and these are made up of the multiples of the minimal polynomial of $A$. $\endgroup$ – Bernard Aug 19 '15 at 20:10

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