How to use Jordan normal form to determine $A^k?$

How the simplified way of the matrix $A$ on the Jordan normal form allows to determine easily this power?

  • 1
    $\begingroup$ Well, when $A$ is in Jordan normal form, it is much easier to actually multiply $A$ by itself through the normal computation of matrix multiplication, which is a particular facet of the usefulness of the form, because the matrix contains significantly more zeros than any other possible form of the matrix, which always makes it easier to multiply. Additionally, when you take $A^k$, you can achieve the desired multiplication by raising each Jordan block to the $k$, which makes the process easier. Is this what you are looking for? That it greatly simplifies the calculations? $\endgroup$ – Christian Jul 7 '16 at 20:22

$A = P J P^{-1}\\ A^k = P J^k P^{-1}$

What happens to a Jordan-normal matrix as you multiply it by itself?

The Jordan-normal matrix breaks down into blocks and each block is pretty regular in its behavior.

| cite | improve this answer | |

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.