Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $\pmatrix{1&r\\0&0}$.

share|cite|improve this answer

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

share|cite|improve this answer
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 $\det M=1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.