Similar matrix - Wikipedia, the free encyclopedia Jordan normal form - Wikipedia, the free encyclopedia
Matrices A and B are similar iff their Jordan normal forms are equal (this is because every matrix in an algebraically closed field has a unique Jordan form).

If you know the minimal polynomial of a matrix, then it's easier to calculate its jordan form:

Let $\displaystyle F$ be an algebraically-closed field, $\displaystyle V$ a vector space over $\displaystyle F$,

$\displaystyle A \in F^{n \times n}$ represent a linear operator $\displaystyle T:V \rightarrow V,$

$\displaystyle m_A = (x-c_1)^{r_1} \cdot (x-c_2)^{r_2} \cdot \ldots \cdot (x-c_k)^{r_k} $ be the minimal polynomial of A, where each $\displaystyle r_i$ is the geometrical multiplicity of $\displaystyle c_i$, and let $\displaystyle w_i = Ker(T-c_i I)^{r_i}.$

Then by the primary decomposition theorem, $\displaystyle V = w_1 \oplus w_2 \oplus \ldots \oplus w_k,$ and after a few more steps, we get that each $\displaystyle T_i = T|_{w_i}$ is represented by a Jordan block matrix:

$\displaystyle J_i = \begin{pmatrix}

c_i & 0 & 0 & 0 & \ldots & 0 \\

1 & c_i & 0 & 0 & \ldots & 0 \\

0 & 1 & c_i & 0 & \ldots & 0 \\

\vdots & & \ddots & \ddots \\

\vdots & & & \ddots & \ddots \\

0 & 0 & 0 & \ldots & 1 & c_i

\end{pmatrix}, J_i \in F^{r_i \times r_i}$

And the Jordan canonical form of A is the direct sum of its Jordan blocks.