# Jordan normal form 33 [closed]

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

## closed as off-topic by Jack, Leucippus, Claude Leibovici, José Carlos Santos, user91500Jun 20 '17 at 8:49

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• Isnt' this matrix already in Jordan normal form?? – Crostul Jun 19 '17 at 20:50
• The matrix is already in Jordan normal form. – Alberto Andrenucci Jun 19 '17 at 20:50
• Wolfram is writing that it is not – Student Jun 19 '17 at 20:52
• Wolfram is, again, high: that's the JNF of that matrix., with eigenvalues $\;3\;$ of algebraic mult. two and geometric mult. one, and the eigenvalue two, of alg. and geom. multiplicity one – DonAntonio Jun 19 '17 at 20:53
• Wolfram Alpha seems to want the eigenvalues in increasing order. There are various conventions, but ordinarily your matrix does qualify as a Jordan normal form. – Robert Israel Jun 19 '17 at 20:56

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

• so D and N are what? – Student Jun 19 '17 at 21:30
• The diagonal part and the nilpotent part of $A$, as shown in the first equation line. – Bernard Jun 19 '17 at 21:51
• but I need to find x′=Ax – Student Jun 19 '17 at 21:52
• As I explained, the solution is $x(t)=\exp(At)=\exp(Dt)\exp(Nt)$ since $At=(D+N)t=Dt+Nt$, and $D$ and $N$ commute. – Bernard Jun 19 '17 at 21:55
• is there a equation form of representation? – Student Jun 19 '17 at 21:57

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.