Jordan normal form 33

Question

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

Answer 1

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 $$\exp(At)=\begin{bmatrix} \mathrm e^{3t}&0& 0\\0& \mathrm e^{3t}& 0\\0& 0& \mathrm e^{2t} \end{bmatrix}\cdot\begin{bmatrix} 1&t& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix} =\begin{bmatrix} \mathrm e^{3t}&t\mathrm e^{3t}& 0\\0& \mathrm e^{3t}& 0\\0& 0& \mathrm e^{2t} \end{bmatrix}$$

Answer 2

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.