Let $T:\mathbb{R}^6\to \mathbb{R}^6$ be a linear operator with characteristic polynomial $p_T(t)=(t-a)^3(t-b)^3$ minimal polynomial $m_T=(t-a)^2(t-b)$

Question

Let $T:\mathbb{R}^6\to \mathbb{R}^6$ be a linear operator with characteristic polynomial $p_T(t)=(t-a)^3(t-b)^3$, minimal polynomial $m_T=(t-a)^2(t-b)$ and $a\neq b$. Find the Jordan form of $T$.

I'm trying to understand this and one more general case where we only have a characteristic polynomial and also the minimal polynominal and then we need to produce the possibles Jordam forms given only this, is it possible?

From the above information we can conclude that the diagonal of the jordam form consist of 3's "a" and 3's "b" but I don't know how many blocks it has.

How can I use the minimal polynomial to discover the amount of blocks? Can I determinate $\ker(T-aI)$ and $\ker(T-BI)$ only by seeing the characteristic and the minimal polynomial? Thanks!

Answer 1

The three $b$'s are three 1-dimensional blocks, since $b$ is a simple root of $m_T.$

Among the three $a$'s, two of them must belong to a 2-dimensional block (the remaining one being then a 1-dimensional block), since $a$ is a double root of $m_T.$