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Write down all possible Jordan normal forms for matrices with characteristic polynomial$ (x − λ)^5$.

In each case, calculate the minimal polynomial and the geometric multiplicity of the eigenvalue λ.

For the only eigenvalue $\lambda$, the possible JNF is just assigned 1 to every column above the diagonal since the min polynomial can be any degree from 1 to 5?

I figured out the possible JNF using the possible minimal polynomial

$(x-\lambda)$and $(x-\lambda)^2$ and $(x-\lambda)^3$...$(x-\lambda)^5$ In total, it is 7 possibility (corresponding to each minimal polynomial and $\lambda$ has to appear 5 times.) But I don't understand that why the number of blocks gives the geometric multiplicity since each represent one eigenspace.

Thank you so much!

Castalia520
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1 Answers1

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They basically correspond to partitions of $5$. Take the diagonal matrix $\lambda I_5$ and then insert some ones onto the superdiagonal. Let $J_n$ denote the matrix

$$ J_n := \begin{pmatrix} \lambda &&&& \\ 1 & \lambda && \\ & 1 & \lambda && \\ && & \ddots & \\ &&& 1 & \lambda \end{pmatrix} $$

with $n$ lambdas. Then the possible Jordan forms are

  • $J_1 \oplus J_1 \oplus J_1 \oplus J_1 \oplus J_1$

  • $J_1 \oplus J_1 \oplus J_1 \oplus J_2$

  • $J_1 \oplus J_2 \oplus J_2$

  • $J_1 \oplus J_1 \oplus J_3$

  • $J_2 \oplus J_3$

  • $J_1 \oplus J_4$

  • $J_5$

Try to figure out the minimal polynomial and geometric multiplicity on your own. If you need help:

If $\mu = \mu_1 \le \mu_2 \le \dots \le \mu_k$ is a partition of $5$ with $k$ parts then the Jordan matrix corresponding to $\mu$ has minimal polynomial $(x - \lambda)^{\mu_k}$ (i.e. the exponent is the largest part of $\mu$). The geometric multiplicity is $k$.

Sera Gunn
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