Deduce minimum and characteristic polynomial of a $5\times5$ matrix from this list of properties

Question

Suppose $A$ is a $5\times5$ matrix with the following properties:

(i) $A$ has integer entries.

(ii) $A$ is not diagonalizable over $\Bbb C$ (the complex numbers)

(iii) the minimum and characteristic polynomial of $A$ are different.

(iv) $-1$ and $2i$ are two (but not necessarily all) eigenvalues of $A$.

Determine the minimum polynomial, the characteristic polynomial, and the Jordan form of $A$.

Answer 1

Hints:

The eigenvalues are $\;-1,\,2i,\,-2i,\,r,\,r\;$ , with $\;r\in\Bbb C\;$ (why there must a be a double (or more) root?).

Assume already that $\;r\neq-1,\,\pm2i\;$ for simplicity. You cover the other case:

Then, then minimal polynomial is $\;(x+1)(x-2i)(x+2i)(x-r)^2=\;$ the characteristic polynomial ( why can't the minimal one be $\;(x+1)(x-2i)(x+2i)(x-r)\;$ ? ).

Deduce now the Jordan form of $\;A\;$ .

Added on request:

(1) As the matrix is an integer one its characteristic and minimal polynomials are integer ones and, thus, real polynomials: if $\;z\in\Bbb C\;$ is a root then so is $\;\overline z\;$ .

(2) In case $\;r\neq-1,\,\pm2i\;$ , then we get the characteristic pol. = the minimal pol., so this can not be (did you read the "you cover the other case" part?).

Thus, either $\;r=-1\;$ and then we may have another real root $\;s\;$ , so the char. pol. is

$$\;(x-2i)(x+2i)(x+1)^2(x-s)\;$$

and since this is different to the min. pol. it must be that also $\;s=-1 $ , so char. pol. is $\;(x-2i)(x+2i)(x+1)^3\;$ and thus the minimal one must be...

The other case is when $\;r=\pm2i\;$, then the char. pol. is $\;(x-2i)^2(x+2i)^2(x+1)\;$, but then the min. one...what?

Answer 2

Hints:

1. Because the matrix has real entries, so does its minimal polynomial, and the set of it roots (eigenvalues) must be stable under complex conjugation, as must be their multiplicities.

2. Because the matrix is not diagonalisable over $\Bbb C$, the minimal polynomial has at least one multiple root over $\Bbb C$.

3. Because the minimal polynomial differs from the characteristic one, it can have degree $4$ at most.

From these conditions and the given eigenvalues, you can determine the multiset of roots of the minimal polynomial uniquely. From there you can get their multiplicities as roots of the characteristic polynomial as well, since these must be stable under complex conjugation too. Finally you will see that the values of the minimal and characteristic polynomials leave only one Jordan normal form as possibility.