Let $V$ be a $n$ dimensional vector space over a field $\mathbb{k}$, and let $P:V \rightarrow V$ be a linear map such that $P^2=P$. i.e., A linear map is a projection.
I want to find all possible characteristic polynomials and their corresponding minimal polynomial.
I know the form of characteristic polynomial $\phi_P(t) = \operatorname{det}(P-tI)$, and minimal polynomial divides characteristic polynomial.
From the assumption $P^2-P=0$ so the minimal polynomial $m_P(x)$ is of the following form $x, (x-1), x(x-1)$. i.e., eigenvalues of $P$ are $0$ or $1$.
But I have a problem with analyzing characteristic polynomial. Here is my trial.
Since $\phi_P(t) = \operatorname{det}(P-tI) = (-1)^n t^n + \cdots$, my guess of the form of characteristic polynomial is $n$ degree polynomial. Since minimal polynomial divides characteristic polynomial $t(t-1)$ should be a factor of $\phi_P(t)$… but what about other $n-2$ degrees of polynomials? The only eigenvalues of projection matrix are $0$ and $1$ so my naive guess is $\phi_P(t) = (-1)^n t^{m} (t-1)^{n-m}$ with $ 1< m <n$. Is this correct? If so how one can prove this?
Best Answer
You have a small error: you mention the minimal polynomial could be $x$ or $x-1$, but only consider the case $x(x-1)$ in the last paragraph.
After correcting this mistake, you will have shown that the characteristic polynomial must be of the form $(-1)^n t^m (t-1)^{n-m}$ for $0 \le m \le n$. It remains to check that all of these polynomials are the characteristic polynomial of some projection; to show this it suffices to consider diagonal matrices with $0$s and $1$s on the diagonal.