# Minimal Polynomial of Projection Matrices

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.

• 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. – amd May 29 '19 at 18:06

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