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Here is the exercise:

Let $A$ be a $5\times5$ complex matrix such that $(A-2)^3(A+2)^2=0$, where we define $A-\mu:=A-\mu I$ for scalar $\mu$. Assume that $\lambda=2$ is an eigenvalue of $A$ and its geometric multiplicity is at least $2$. What are the possibilities for the Jordan canonical form (JCF)?

What I know so far from the assumption is that

  • the minimal polynomial of $A$ is of the form $f(x)=(x-2)^i(x+2)^j$ where $1\leq i\leq 3$ and $0\leq j\leq 2$.
  • The number of blocks in the Jordan segment $J(2)$ is at least $2$.

One can write the possible minimal polynomial one by one, which gives the information of the size of the largest block in each Jordan segment, and use the possible geometric multiplicity of $\lambda=2$ to find JCF.

Here are my questions:

  • Is there an alternative approach?
  • Can we use the characteristic polynomial of $A$ here?
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Note that the annihilating polynomial that is given is not otherwise related to the characteristic polynomial, so you cannot reason based on knowledge of the characteristic polynomial. The information you do have is

  • No other (complex) eigenvalues than $2$ and $-2$ exist, and $\lambda=2$ is in fact an eigenvalue,
  • Size of Jordan blocks for $\lambda=2$ is at most $3$, and for Jordan blocks for $\lambda=-2$ it is at most $2$,
  • There are at least $2$ Jordan blocks for $\lambda=2$,
  • The sum of the sizes of all Jordan block (for $\lambda=2$ and maybe $\lambda=-2$) is $5$.

The sizes of the Jordan blocks for a given eigenvalues define a partition of an integer. So you are looking for pairs $(\mu,\nu)$ of partitions (for eigenvalues $2,-2$, respectively) satisfying

  1. $|\mu|+|\nu|=5$,
  2. $\mu_1\leq3$ and $\nu_1\leq2$
  3. $l(\mu)\geq2$.

Simple trial gives the following list of possible pairs

$$ \begin{matrix}\mu & \nu \\ (3,2) & (0) \\ (3,1,1) & (0) \\ (2,2,1) & (0) \\ (2,1,1,1) & (0) \\ (1,1,1,1,1) & (0) \\ (3,1) & (1) \\ (2,2) & (1) \\ (2,1,1) & (1) \\ (1,1,1,1) & (1) \\ (2,1) & (2) \\ (2,1) & (1,1) \\ (1,1,1) & (2) \\ (1,1,1) & (1,1) \\ (1,1) & (2,1) \\ (1,1) & (1,1,1) \end{matrix} $$

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I would work with Jordan blocks directly, without going through minimal polynomial. It's a packing problem: we need to pack a size 5 matrix with blocks, of which

  • at least two are labeled "2". The size of any such block is at most 3.
  • at least one is labeled "-2", and its size is at most 2.

Denoting a block for brevity as $(\lambda)$, $(\lambda\lambda)$, $(\lambda\lambda\lambda)$ etc, we can work out the combinations, sorting them by the number of -2 entries present.

  • (-2) and a partition of 4 elements into at least two blocks. This means:

    • (-2)(222)(2), or
    • (-2)(22)(22), or
    • (-2)(22)(2)(2), or
    • (-2)(2)(2)(2)(2)
  • (-2-2) and a partition of 3 elements into at least two blocks.

    • (-2-2)(22)(2), or
    • (-2-2)(2)(2)(2)
  • Same as the previous one, but with (-2)(-2) instead of (-2-2).

  • (2)(2) and a partition of 3 elements into at least two blocks:

    • (2)(2)(-2-2)(-2)
    • (2)(2)(-2)(-2)(-2)

I don't think it could be simpler: the answer takes about as long to write out as the solution itself.

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  • $\begingroup$ There need not be an eignevalue $\lambda=-2$ at all. $\endgroup$ – Marc van Leeuwen Jan 8 '14 at 9:20

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