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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

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

I have no idea how to actually prove b.

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You can use the relation $\det(XY) = \det(X) \det(Y)$ for a), that is possible to prove for $2\times 2$ matrices yourself. For b), all you have to do is give a formula for such a matrix with one unknown in it which, despite this, still has determinant 0 or 1, there are plenty of candidates. – jp26 Feb 18 '13 at 20:54
up vote 4 down vote accepted

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 $\pmatrix{1&r\\0&0}$.

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

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Why are the eigenvalues in $ \{0,1 \}$ ? – Bujanca Mihai Feb 18 '13 at 20:53
Because the minimal polynomial is $\rm X(X - 1)$. – Damien L Feb 18 '13 at 20:54
Even though your conclusion is correct you only know that the minimal polynomial divides $X(X-1)$. . – Sean Ballentine Feb 18 '13 at 21:00

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 $\det M=1$.

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