Your "proof" is non-sense. To find the minimal polynomial, you must first find $A$-invariant spaces. For example, $span(e_4)$ is invariant but $span(e_1,e_2,e_3)$ is not.
Here, it suffices to show that the minimal polynomial has degree $4$. Do as follows
i) Randomly choose a vector $x$.
ii) Considering the invariant space generated by $x$, we obtain the matrix $M=[x,Ax,A^2x,A^3x]$. If $\det(M)\not= 0$, then $x$ generates the whole space and the minimal polynomial of $A$ is the characteristic polynomial of $A$.
That works with $x=[1,1,1,1]^T$ (which is not very random; try another vector...).
EDIT. Answer to the OP.
For the minimal polynomial, you can't separate some characteristic space into two parts. For example, here, you may consider whole characteristic space $\ker((A-2I)^2)$ which is $span(e_3,e_4)$ -an invariant subspace-. The matrix of the restriction is $\begin{pmatrix}2&0\\1&2\end{pmatrix}$ and its minimal polynomial is $(x-2)^2$.
Now you consider $\mathbb{R^4}/span(e_3,e_4)$; the associated matrix (using the canonical basis) is $\begin{pmatrix}1&0\\1&1\end{pmatrix}$; its minimal polynomial is $(x-1)^2$ and we are done.
In particular, the matrix
$\begin{pmatrix}1&0&0&0\\a&1&0&0\\b&c&2&0\\d&e&f&2\end{pmatrix}$ has $(x-1)^2(x-2)^2$ as minimal polynomial iff $af\not= 0$.
Generally knowing only the characteristic polynomial and the minimal polynomial is not enough to determine uniquely the Jordan normal form, like you showed in the question.
I think that the only times where just knowing these two polyomials gives you also the Jordan normal form is when the degree of the minimal polynomial is very low or very high.
For example, if you know that
- $f(x) = (x-\lambda)^n$ and $m(x)=(x-\lambda)$ you know that the Jordan normal form is the diagonal one
- $f(x) = (x-\lambda)^n$ and $m(x)=(x-\lambda)^n$ you know that the Jordan normal form is the one made up by only one Jordan block of dimension $n$
- $f(x) = (x-\lambda)^n$ and $m(x)=(x-\lambda)^{n-1}$ you know that the Jordan normal form is the one made up by one Jordan block of dimension $n-1$ and one Jordan block of dimension $1$.
I think that this cases, and the ones where every eigenvalue behave like one of these cases, are the only one where the two polynomials determines uniquely the Jordan normal form.
Best Answer
Hint: Assuming $A$ is supposed to be an $n \times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^\mbox{th}$ row and $j^\mbox{th}$ colmn of $A$ , which you know.