I need to show:
How many possibilities for the Jordan normal form (up to order of the Jordan blocks on the diagonals) there are for matrices $B \in M_5 (\mathbb C) $ with $x_B(\lambda)=-(\lambda-2)^2(\lambda-3)^3$

What I already have:
The Eigenvalues are $2 $ and $3 $.
The Eigenvalue $2 $ has the algebraic multiplicity 2 and the geometric multiplicity $ 2 $ or $1 $.
The Eigenvalue $3 $ has the algebraic multiplicity 3 and the geometric multiplicity $ 3 $ or $ $2 or $1 $.

So I tried showing all the possibilites with:

1) $diag (J_1 (2), J_1 (2), J_1(3), J_1(3), J_1(3))$

2) $diag (J_2 (2), J_1(3), J_1(3), J_1(3))$

3) $diag (J_2 (2), J_3(3))$

4) $diag (J_1 (2), J_1 (2), J_3(3))$

5)$ diag (J_2 (2), J_2(3), J_1(3))$

6) $diag (J_1 (2), J_1 (2), J_2(3), J_1(3))$

I am not sure if this is right though..

  • $\begingroup$ I'm assuming you mean $\chi_B(\lambda)=-(\lambda-2)^2(\lambda-3)^3$? $\endgroup$ – Brian Fitzpatrick Jul 11 '17 at 19:51
  • $\begingroup$ Correct... Sry my mistake $\endgroup$ – PhysX Jul 11 '17 at 19:52

Yes, this is correct. Let $\DeclareMathOperator{am}{am}\am_B(\lambda)$ and $\DeclareMathOperator{gm}{gm}\gm_B(\lambda)$ denote the algebraic and geometric multiplicities of an eigenvalue $\lambda$ of $B$ respectively. Then the table of eigenvalues of $B$ along with their algebraic multiplicities and possible geometric multiplicities is $$ \begin{array}{c|c|c} \lambda & \am_B(\lambda) & \gm_B(\lambda) \\ \hline 2 & 2 & 1,2 \\ 3 & 3 & 1,2,3 \end{array} $$ There are thus six possible combinations of geometric multiplicities of the eigenvalues of $B$. These six possible combinations yield exactly the six Jordan forms you have listed.

Note, however, that in general a choice of geometric multiplicities can yield more than one possible Jordan form. For example, both \begin{align*} J_2(\lambda)\oplus J_2(\lambda) && J_3(\lambda)\oplus J_1(\lambda) \end{align*} have characteristic polynomial $\chi(t)=(t-\lambda)^4$ and hence have exactly one eigenvalue with geometric multiplicity two.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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