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.
Best Answer
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 $p(x)=x^2-x.$