Jordan normal form 33 what is the Jordan normal form for matrix
$$\begin{bmatrix} 3&1& 0\\0& 3& 0\\0& 0& 2\\
\end{bmatrix}$$
can't figure out because some eigenvalues makes some rows 0
and how can I find solution for this equation x′=Ax
 A: THis matrix is in Jordan canonical form with a first Jordan block
$$J_1=\begin{bmatrix}
3&1\\0&3
\end{bmatrix}
$$
that means that the matrix has an eigenvalue $\lambda_1=3$ with algebraic multiplicity $2$ and geometric multiplicity $1$.
And a second Jordan block $J_2=2$ that means that the other eigenvalue is $\lambda_2=2$ ( with algebraic and geometric multiplicity $=1$).
Note that the Jordan normal form of a matrix is not unique because we can put the Jordan blocks in different order.
A: For the solutions  of $x'=Ax$ write 
$$A=\begin{bmatrix} 3&1& 0\\0& 3& 0\\0& 0& 2
\end{bmatrix}=\underbrace{\begin{bmatrix} 3&0& 0\\0& 3& 0\\0& 0& 2
\end{bmatrix}}_D+\underbrace{\begin{bmatrix} 0&1& 0\\0& 0& 0\\0& 0&
\end{bmatrix}}_N$$
$D$ and $N$ commute (I insist) so that $\;\exp(At)=\exp(Dt)\cdot\exp(Nt)$. As $N^2=0$, $\exp(Nt)=I+Nt$ and
$$\exp(At)=\begin{bmatrix} \mathrm e^{3t}&0& 0\\0&  \mathrm e^{3t}& 0\\0& 0& \mathrm e^{2t}
\end{bmatrix}\cdot\begin{bmatrix} 1&t& 0\\0&  1& 0\\0& 0& 1
\end{bmatrix} =\begin{bmatrix} \mathrm e^{3t}&t\mathrm e^{3t}& 0\\0&  \mathrm e^{3t}& 0\\0& 0& \mathrm e^{2t}
\end{bmatrix}$$
