Jordan form of a matrix Let 
$$A = \left( {\matrix{
   0 & 1 & 0 & 0  \cr 
   0 & 0 & 2 & 0  \cr 
   0 & 0 & 0 & 3  \cr 
   0 & 0 & 0 & 0  \cr 
 } } \right)$$
The characteristic polynomial is $f_A(x)=x^4$.
Questions:


*

*How do I conclude that $m_A=x^4$? Do I have to evaulate $A^3$ and figure that $A^3\ne 0$?

*The jordanian form is:
$A = \left( {\matrix{
   0 & 1 & 0 & 0  \cr 
   0 & 0 & 1 & 0  \cr 
   0 & 0 & 0 & 1  \cr 
   0 & 0 & 0 & 0  \cr 
 } } \right)$. Why?

 A: (2) As you observe, the characteristic polynomial of $A$ is $x^4$, so it has sole eigenvalue zero (of multiplicity $4$). Now, if we denote by $J_k$ the $k \times k$ Jordan block with eigenvalue $0$, the possible Jordan forms of such a matrix are
$$J_4, \quad J_3 \oplus J_1, \quad J_2 \oplus J_2, \quad J_2 \oplus J_1 \oplus J_1, \quad 0.$$
Now, $$\ker A = \langle e_4 \rangle,$$
and in particular $\dim \ker A = 1$, but the only Jordan-form matrix above with $1$-dimensional kernel is $J_4$, and the dimension of the kernel of two similar matrices agrees, so $J_4$ must be the Jordan form of $A$. In fact, as abel observes, the number of Jordan blocks in the Jordan form a matrix with all zero eigenvalues (a nilpotent matrix) is exactly the dimension of its kernel.
(1) Jordan blocks $J_k$ of eigenvalue zero have the property that $J_k^r \neq 0$ for $r < k$ and $J_k^r = 0$ for $r \geq k$. So, $J_4^3 \neq 0$ and hence $A^3 \neq 0$, that is, the minimal polynomial of $A$ must be $x^3$. More generally, the minimal polynomial of a matrix with all zero eigenvalues is $x^r$, where $r$ is the size of the largest Jordan block in its Jordan form.
A: i will use $e_1, e_2, e_3$ and $e_4$ to stand for the standard basis. with this convention, we have $$Ae_1 = 0, Ae_2 = e_1, Ae_3= 2e_2 \mbox{ and }Ae_4 = 3e_3.$$ so with respect to the basis $\{e_1, e_2, \frac{1}{2}e_3, \frac{1}{6}e_4 \} = \{f_1,f_2,f_3, f_4\}$ in the new $f$ basis, the
linear transformation is 
 $$Tf_1 = 0, Tf_2 = f_1, Tf_3 = f_2, Tf_4 = f_3$$ so $T$ has the representation $\pmatrix{0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr0&0&0&0},$ called the jordan form, similar to the matrix in question. 
if we permute the basis,e.g,$\{f_4,f_3,f_2,f_1\}$ the same transformation is now
represented by $$\pmatrix{0&0&0&0\cr1&0&0&0\cr0&1&0&0\cr0&0&1&0}$$ 
