Decomposition Via Generalized Eigenvectors

[@jonathansharir-smith6683]

At around 11:10, you mention that we "don't need to go any higher than the square power" because that's the highest algebraic multiplicity we're expecting. Now, by an earlier theorem in this video, I follow how we can justify that the highest algebraic multiplicity we will find will be 2 (sum of multiplicities equals dimension of V). But how do we know that we don't need to go any higher than the square power? Unless I'm forgetting or missing something, we've thus far proven that you must check up to the power of the dimension of the space you're in. The "square power" works in this case, but from the sounds of it you knew a priori that it would - I'm just wondering if that's a theorem we will see later perhaps? EDIT: On second thought, I guess this follows from earlier theorems on the equality of null spaces engendering equalities in null spaces for all higher powers. That is, if your statement in this video weren't true, and we had the square power adding no linearly independent vector to the null space from the first power, then the third power couldn't change anything because we already have an equality of null spaces of consecutive powers.

[@davidk7212]

Is it a typo on page 252 in the paragraph preceding 8.21 where it says "...each of which is a nilpotent operator plus a scalar multiple of the identity."? If this is referring to each restricted (T - lambda I), it seems to imply that T itself is nilpotent, when I believe it's the whole term (T - lambda I) that is nilpotent. Or am I confused? Thanks!

[@sheldonaxler5197]

No, the item you mention is not a typo. Each restricted T equals restricted (T - \lambda I) + \lambda I. The first term in this sum is nilpotent and the second is a scalar multiple of the identity.

[@davidk7212]

@@sheldonaxler5197 Thank you!