Question about conjugate Jordan Form matrices. This was an example done in class last week. I'm struggling with the details of the argument, but have included my questions at the very end.
Let $(t-4)^7$ be the characteristic polynomial for a matrix in $\mathcal{M}_7(\mathbb{C}).$ Let $J_{n,\lambda}$ denote the $n\times n$ Jordan block with eigenvalue $\lambda$.
Let 
$$A=\begin{pmatrix}
4&1&&&&&\\
&4&1&&&&\\
&&4&&&&\\
&&&4&1&&\\
&&&&4&&\\
&&&&&4&1\\
&&&&&&4\\
\end{pmatrix}=J_{3,4}\oplus J_{2,4}\oplus J_{2,4}$$ and
$$
B=\begin{pmatrix}
4&1&&&&&\\
&4&1&&&&\\
&&4&&&&\\
&&&4&1&&\\
&&&&4&1&\\
&&&&&4&\\
&&&&&&4\\
\end{pmatrix}=J_{3,4}\oplus J_{3,4}\oplus J_{1,4}$$
Observe that the sum of the Jordan blocks for $A$ is $7=3+2+2$ and for $B$, $7=3+3+1$.
The claim made in class was that these two matrices are not conjugate. We were instructed that the way to determine if two matrices were conjugate in Jordan normal form was to compare the dimension of $\underbrace{\ker[(A-4I)^k]}_{A'}$ and $\underbrace{\ker[(B-4I)^k]}_{B'}$ for $k=0,1,\ldots 7$ by constructing a chart, like the one below.
\begin{array}{c|c}
k&\dim[\ker(A')^k]\\
\hline
7&\dim(\bar{0})=7\\
6&7\\
5&7\\
4&7\\
3&7=3+2+2\\
2&6=2+2+2\\
1&3=1+1+1\\
0& \dim(\ker(I))=0\\
\end{array}
I was told that we can think of the partitions of $7$ above can be thought of as each blocks contribution to the kernel of $A'$.
However, I confused on the following points


*

*How are we able to quickly read off the dimension of the kernel of $A'$? Are we to look at each individual Jordan block and rase it to some power, or look at the matrix as a whole and raise it to a power? 

*Is there anyway to circumvent this table method? For example, suppose we want to determine if two large matrices are conjugate, how would we approach this? Are there any global properties of the Jordan form that hint at the dimension of the kernel? 

*If it takes a list like this to show two matrices are not conjugate, then do we just produce two lists and show that the dimensions of their kernel agree? Or could we just produce a permutation matrix $P$ s.t. $AP=B$ or vice versa? 

 A: Each of your blocks will be in the form $$\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix} $$ for however many dimensions (drawn with three here). Observe that this matrix takes the vector $$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\to\begin{pmatrix}x_2\\x_3\\0\end{pmatrix}$$ so its kernel is anything where $x_2=x_3=0.$ Thus its kernel is one-dimensional. If you applied it twice, you'd get $$\begin{pmatrix}x_3\\0\\0\end{pmatrix} $$ so the kernel of $(A-\lambda I)^2$ is two-dimensional, and so on.
In order to get the dimension of the kernel of the whole matrix, you add up the kernel dimensions of the blocks (since the blocks act independently on orthogonal subspaces). Each block of size $a$ starts with a kernel dimension of $1$ and as the power $k$ increases, it increases by one until it hits $a$ and the block is zero.
This method you show is an easy way of showing that two matrices with different block structures of their Jordan normal form are not similar (at some $k$ the kernel dimensions of $(A-\lambda I)^k$ will become different). In fact, it's true that two matrices are similar if and only if they have the same Jordan normal form (up to permutation of the blocks). 
