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Just a quick linear algebra question:

We have $E:V\rightarrow V$ where V is finite dimensional.

I am studying projection matrices, those non-identity matrices such that $E^2=E$, and am trying to figure out the minimal polynomial for such a matrix. I can see that the eigenvalues must all be 0, as $Ev=\lambda v=E(\lambda v)={\lambda}^2v=E^2v=Ev$, so $\lambda^2=\lambda$, so $\lambda =0,$ or $1$ and it cannot be 1 as otherwise $Ev=v$ and $E$ becomes the identity, which we have assumed it is not. Thus I get that the characteristic polynomial is $x^n$, where n is the dimension of V. Then, by the Cayley-Hamilton Theorem, I know that the minimal polynomial is $x^m$ for some $m\leq n$. Is there anything more I can say about the minimal polynomial from this information, or is this all I can know?

Thanks.

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  • $\begingroup$ You’ve made critical error in your reasoning: the eigenvalues can’t all be $1$ for the reason you state, but that doesn’t exclude $1$ from being an eigenvalue. $E$ is, after all, the identity map when restricted to its range. $\endgroup$ – amd May 29 '19 at 18:06
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Let $U$ be the image of $E$ and $W$ be the kernel of $E$. Then we have :

$V= U \oplus W$, $Ev=v$ for all $v \in U$ and $Ev=0$ for all $v \in W.$

Case 1: $U=\{0\}$, then $E=0.$ Hence $E$ has only one eigenvalue : $0$

Then the characteristic polynomial is given by ...... ?

Case 2: $W=\{0\}$, then $E=I.$ Hence $E$ has only one eigenvalue : $1$

Then the characteristic polynomial is given by ...... ?

Case 3:$U \ne \{0\}$ and $W \ne\{0\}$. Hence $E$ heas exactly two eigenvalues : $0,1$

Since $E^2-E=0$, the minimal polynomial is $p(x)=x^2-x.$

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