2
$\begingroup$

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

$\endgroup$

closed as off-topic by Jack, Leucippus, Claude Leibovici, José Carlos Santos, user91500 Jun 20 '17 at 8:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack, Leucippus, Claude Leibovici, José Carlos Santos, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 7
    $\begingroup$ Isnt' this matrix already in Jordan normal form?? $\endgroup$ – Crostul Jun 19 '17 at 20:50
  • 2
    $\begingroup$ The matrix is already in Jordan normal form. $\endgroup$ – Alberto Andrenucci Jun 19 '17 at 20:50
  • $\begingroup$ Wolfram is writing that it is not $\endgroup$ – Student Jun 19 '17 at 20:52
  • 1
    $\begingroup$ 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 $\endgroup$ – DonAntonio Jun 19 '17 at 20:53
  • 1
    $\begingroup$ 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. $\endgroup$ – Robert Israel Jun 19 '17 at 20:56
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ so D and N are what? $\endgroup$ – Student Jun 19 '17 at 21:30
  • $\begingroup$ The diagonal part and the nilpotent part of $A$, as shown in the first equation line. $\endgroup$ – Bernard Jun 19 '17 at 21:51
  • $\begingroup$ but I need to find x′=Ax $\endgroup$ – Student Jun 19 '17 at 21:52
  • $\begingroup$ 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. $\endgroup$ – Bernard Jun 19 '17 at 21:55
  • $\begingroup$ is there a equation form of representation? $\endgroup$ – Student Jun 19 '17 at 21:57
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.