Find the Jordan Canonical Form that is similar to the idempotent matrix $A$.
I know that since $A=A^2$ then $A(A-I)=0$ so the minimal polynomial is $m_A(\lambda)=\lambda(\lambda-1)$.
I also know that the largest Jordan block corresponding to $\lambda=0$ and $\lambda=1$ have the size $1 \mathrm x1$
But I don't really how to continue from here. I would really appreciate some help
Best Answer
Though cannot conclude about the minimal polynomial as you did, you have already done most of the work. From $A^2=A$ you can deduce, as you say, that the only possible eigenvalues are $0$ and $1$. Next, as you mention, you deduce that the Jordan blocks can only be $1\times 1$ (because otherwise they wouldn't be idempotents). So the Jordan form of $A$ has $1\times 1$ blocks (so it's diagonal) and the diagonal consists of $0$ and $1$.