Jordan chain when matrix has only one eigenvalue.

Question

A $12\times 12$ matrix has sole eigenvalue $3$. It is given that the kernels of $A-3I$, $(A-3I)^{2}$, $(A-3I)^{3}$ and $(A-3I)^{4}$ have dimensions $4$, $7$, $9$ and $10$ respectively. What are the possible jordan forms of $A$?

I would have thought to approach this problem by constructing a chain diagram like this

using up $4$ vertical crosses to make $4$ in first column then the difference to $7$ of in the next column then

the difference of two to make up $9$ in the next then difference of one to give $10$, which would have resulted

in a diagram like this

xxxx

xxx

xx

x

And a corresponding jordan form of $j_{4}(3)\bigoplus j_{3}(3) \bigoplus j_{2}(3)\bigoplus j_{1}(3)$ reading down the chain diagram, however the answer is

apparently $j_{1}(3)\bigoplus j _{2} (3) \bigoplus j _{3}(3)\bigoplus j_{6}(3)$. What am I doing wrong?

Answer 1

Hint: What does "Kernel of $A-3I$ has dimension 4" tell you?

There are exactly 4 jordan blocks with.

Hint: Furthermore, What does "Kernel of $(A-3I)^2$ has dimension 7" tell you?

There are exactly $7-4=3$ jordan blocks with dimension at least 2.

Continue on, and you will reach the conclusion.

Answer 2

You are missing the last two to reach $12$, kernels of $(A-3I)^5$ and $(A-3I)^6$ should have dimensions $11$ and $12$.