3
$\begingroup$

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.

$\endgroup$
0
2
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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