Matrix set determination

Question

Let $ H=\{ A\in M_2(\mathbb{R}) | A^2=A \},x \in \mathbb{R} $

a) Prove that if $M \in H$ and $\det(M) \neq 0$ then $\det(M)=1$.

I tried this using the Hamilton-Cayley relationship, but didn't really help. $ M^2- \operatorname{Tr}(M)M- \det(M)I_2=O_2 \Leftrightarrow M-\operatorname{Tr}(M)\cdot M-\det(M)I_2=O_2$

Also, supposing $\det(M)=1$ the equation is even harder to prove in my opinion, because it is
$M(1-\operatorname{Tr}(M))-I_2=O_2$.

b) Prove that the set $H$ is infinite.

I have no idea how to actually prove b.

Answer 1

a) Use $\det(AB)=\det A\cdot \det B$ for any two square matrices of the same size $A$ and $B$.

b) Try examples like $\pmatrix{1&r\\0&0}$.

Answer 2

The eigenvalues are in $\{0,1\}$.

Answer 3

I tried this using the Hamilton-Cayley relationship, but didn't really help.

For the record, althouth this exerice can be done without using Cayley-Hamilton theorem, the theorem does help. First, we have \begin{equation} M^2- \operatorname{tr}(M)M + \det(M)I_2 = 0.\tag{1} \end{equation} As $M^2=M$, if $\det M\not=0$, then $(1)$ gives $\frac{\operatorname{tr}(M)-1}{\det M}\,M = I_2$. Since the RHS is nonzero, the LHS must be nonzero, too. Therefore $M=kI_2$ for some $k\not=0$. But then $M^2=M$ implies that $k^2=k$. Therefore $k=1$, i.e. $M=I_2$. So $\det M=1$.