All idempotent elements in $M_n(F)$

Question

I know that every idempotent matrix is diagonalizable. Thus, I can search diagonal matrices for all the idempotent elements. In this case, $(n,0), (n,1),..,(n,n)=2^n$ diagonal matrix can be obtained. It means there are $2^n$ idempotent elements in $M_n(F)$. But, I need to show that just $1$ can be on the diagonal. I can not understand why just $1$?

Could someone help me out please?

Answer 1

An idempotent matrix represent a projection onto a certain subspace, hence the only eigenvalues that can appear are 0 and 1. A diagonal matrix that represent a projection can have only ones as nonzero elements.

Answer 2

But, I need to show that just 1 can be on the diagonal. I can not understand why just 1?

This is all very elementary. Suppose $T$ is an idempotent transformation and $\lambda$ and eigenvalue with eigenvector $x$.

Then $\lambda^2 x=T^2(x)=T(x)=\lambda x$. From $(\lambda^2-\lambda)x=0$ you know $\lambda^2-\lambda=0$, and hence the eigenvalues are solutions to $\lambda(\lambda-1)=0$. Either $\lambda =0$ or $\lambda =1$.

It means there are $2^n$ idempotent elements in $M_n(F)$.

Yes there are at least that many elements... but not only that many. There could be infinitely many more.

The final step (which I'm not sure you're seeing) is to recognize that after you've written down all the diagonal matrices with all combinations of $0$ and $1$ in the set $E$, then you can conclude that the set of all idempotent matrices is

$\{u^{-1}eu\mid e\in E, u\in GL(n)\}$

where $GL(n)$ denotes the set of invertible matrices of $M_n(F)$.

That is, idempotent matrices aren't diagonal they are just diagonalizable, that is, similar to a diagonal matrix (of a certain type.) There are $2^n$ similarity classes, but each of those classes can (when $F$ is infinite) contain infinitely many members.

Apologies: a correction was pointed out to me by loup blanc: the similarity classes of the $2^n$ idempotents overlap! This is because the idempotents which share the same number of $1$'s are similar. Since there are $n+1$ possible counts for the number of $1$'s, there are only $n+1$ equivalence classes.

Answer 3

We consider the algebraic set $Z=\{A\in M_n(\mathbb{C});A^2=A\}$. The best indicator of the size of an algebraic set is its dimension, i.e. the number of complex parameters on which its elements depend.

Since there are only a finite number of similarity classes, it suffices to find a class of maximum dimension.

A class is in the form $\{P^{-1}diag(I_p,0_{n-p})P;P\in GL_n\}$. Its stabilizer (for the conjugation action) is the commutant of $diag(I_p,0_{n-p})$, that is, a vector space of dimension $p^2+(n-p)^2$.

The minimum for $p$ is obtained when $p\approx n/2$ and is $\approx n^2/2$. Then the required maximum dimension is $\approx dim(GL_n)-n^2/2=n^2/2$. More generally, one has

$\textbf{Proposition.}$ Let $Y=\{A\in M_n(\mathbb{C});p(A)=0\}$, where $p$ is a polynomial of degree $2$ with simple complex roots.

Then $Y$ is an algebraic set of dimension $\approx n^2/2$.

$\textbf{Remark.}$ i) The previous proposition applies to symmetries $S^2=I_n$.

ii) For comparison, a nilpotent matrix depends on many more parameters because the dimension of their set is $n^2-n$.