EDIT:
A key error in my working is that it allows $M$ to be singular! I tried to solve $AM=MJ$, not realising that $M^{-1}AM=J$ requires $M^{-1}$ to exist and be completely linearly independent!
OP:
Reading Wikipedia, I saw examples of JNF but did not see a clear explanation for why it works, so I attempted to show it to myself – and I thought I had proven it, until I read this post on MSE, which claimed that the only way to express the matrix in JNF was with a very specific block count, and block size – the number of blocks with a certain eigenvalue $\lambda$ must apparently be the nullity of $A-\lambda I$, which clashes entirely with my "understanding" of why JNF works.
$A\in M_{n\times n}$ can be written in JNF by $M^{-1}AM=J\implies AM=MJ$, where $M$ is the "modal" matrix, and $J$ is the JNF of $A$. Wikipedia's example constructed the JNF by calculating the eigenvectors and generalised eigenvectors, creating Jordan chains of such vectors, and grouping the chains in correspondence with the Jordan blocks. My erroneous "proof" went as follows:
For convenience, let the matrix $B=A-\lambda I$. Let the Jordan chain of some such eigenvalue $\lambda$, for which the generalised eigenvector is of rank $m$, be written $\mathfrak{Ch}(\lambda)=\{x_m,x_{m-1},\dots,x_1\}$, and let $\mathfrak{Ch}_i(\lambda)=x_i$ in the chain, where all the $x_i$ are related by $B^mx_m=0;x_1=B^{m-1}x_m\neq0;x_{i}=Bx_{i+1}$. $J=\bigoplus_{i=1}^n J_k(\lambda_i)$, where $\lambda_i$ is the relevant eigenvalue depending on your chosen order of constructing $J$, and $k$ is the size of the relevant Jordan block, depending on $i$. $M$ is constructed (I thought) by using, as its column vectors, any series of vectors from $\mathfrak{Ch}(\lambda_i)$, such that the eigenvalue $\lambda_i$ in that section of $M$ corresponds with the Jordan block in $J$.
The product $MJ$ is expressible in the form of a sequence of column vectors $c_i$. Letting the Jordan block corresponding to that column $i$ have a relative index $k$ to represent the $k$th column of the Jordan block and also to represent the $k$th column in the series of chain-vectors in $M$ (as they must correspond – I think!):
Since the Jordan block at some column $k$ has the form $\begin{pmatrix}\lambda\\0\\0\\\vdots\end{pmatrix}$,$\begin{pmatrix}1\\\lambda\\0\\\vdots\end{pmatrix}$ or $\begin{pmatrix}\vdots\\0\\1\\\lambda\\\vdots\end{pmatrix}$ the product for $c_i$ evaluates to $\mathfrak{Ch}_{k-1}(\lambda)+\lambda\cdot\mathfrak{Ch}_k(\lambda)$, defining "$\mathfrak{Ch}_0(\lambda)$" to be zero to account for the case $k=1$, where there is no trailing one in the Jordan block to catch the previous column of $M$ in the product. By the chain identities above, $\mathfrak{Ch}_{k-1}(\lambda)+\lambda\cdot\mathfrak{Ch}_k(\lambda)=B\times\mathfrak{Ch}_k(\lambda)+\lambda I\times\mathfrak{Ch}_k(\lambda))=(B+\lambda I)\times\mathfrak{Ch}_k(\lambda)=A\times\mathfrak{Ch}_k(\lambda)$, which I believe is precisely the product in the column $i$ of $AM$, therefore asserting $MJ=AM$ to be true when (and I'd have thought only when) $M, J$ are constructed as I've described.
Evidently (since my method comprehensively failed to solve the question that I linked, and clashes with what the answerer said) my idea of why JNF works and how we construct it is wrong – but why?
Many thanks!
Best Answer
I suspect you have not worked through many matrices where the Jordan form is not diagonal. As with continued fractions, the development is alphabet soup; better to do examples.
These are from a book by Evar D. Nering, second edition. I will put both examples here, check for typing errors... the eigenvalues are integers,
Made up this one,
$$ C_{jagy} = \left( \begin{array}{ccc} 2&7&9 \\ 0&2&8 \\ 0&0&2 \\ \end{array} \right) $$
When everything is an integer, I like a sort of backwards order: the one denominator is in finding the inverse of a square matix. Here, $(C-2I)^3 = 0.$ To get the 1's above the diagonal, find column $p_3$ with $(C-2I)^2 p_3 \neq 0.$ Then column $p_2 = (C-2I)p_3.$ Finally $p_1 = (C-2I)p_2$ is a genuine eigenvector. Make the matrix $P$ from the columns $p_1, p_2, p_3.$ The determinant of your choice of $P$ is the denominator in finding $P^{-1}.$ Finally $J=P^{-1}C P$ is in Jordan form.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ A_{nering} = \left( \begin{array}{ccccc} 1&0&-1&1&0 \\ -4&1&-3&2&1 \\ -2&-1&0&1&1 \\ -3&-1&-3&4&1 \\ -8&-2&-7&5&4 \\ \end{array} \right) $$
$$ B_{nering} = \left( \begin{array}{ccccc} 5&-1&-3&2&-5 \\ 0&2&0&0&0 \\ 1&0&1&1&-2 \\ 0&-1&0&3&1 \\ 1&-1&-1&1&1 \\ \end{array} \right) $$