Idempotent $17\times 17$ Matrix Jordan Normal Form

Question

I have a $17\times 17$ idempotent matrix $M$ with rank $10$

$M$ is idempotent so $M^2=M \rightarrow M(M-1)=0$. So the minimum polynomial is $x(x-1)$. The eigenvalues of $M$ are $x=0$ and $x=1$.

The rank of $M$ is $10$, so $\dim \ker M=7$.

But I'm stuck determining the number of blocks and the sizes of the blocks of the Jordan normal form.

Answer 1

It's apparent that all the blocks have to be idempotent, too.

Using this, you can see the blocks are all $1\times1$.

Alternatively, you could also have noted that $\ker(M)\oplus Im(M)$ is a decomposition of the space, and that $M$ is the identity on the image. You join a basis of the image with a basis of the kernel to get a basis of the whole space in which the matrix for the transformation of M is diagonal with ten 1's and seven 0's (hey look, a Jordan decomposition.)

Answer 2

Being general: if $m_i(t)$ is the minimum polynomial of your matrix M, you have $m_i(t- \alpha)^k$, where k is going to be the order of the largest block, hence the previous correct answer about the size of the blocks and the rest is routine.