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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 $A$.

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    Welcome to MSE! This isn't a "We'll do your homework for you site." You're more likely to get an answer if you say, "Here's what I've tried, and here's where I got stuck, and I'd like some help moving past THIS step..." Of course, that entails more work for you, but it also gives you some hope of actually getting help. – 2017-01-06
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    Sorry ok I will do the next time .... – 2017-01-06

2 Answers 2

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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?

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    But we have in the hypothesis that the minimal and characteristic polynomials are different. – 2017-01-09
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    There must be a double root cause the matrix A isn't diagonalizable – 2017-01-09
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    How could we know the eigenvalues ? – 2017-01-09
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    @user401187 I added some further explanations in my answer. – 2017-01-09
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    What for the case r=_+2i – 2017-01-11
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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.

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    How can we know the eigenvalues? – 2017-01-09
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    Since $-1$, $2i$, and $-2i$ are eigenvalues, the minimal polynomial has at most $4$ roots _counted with multiplicity_ and in fact at least one of those roots is multiple, those three eigenvalues are all there is. (The first condition then also implies that $-1$ is the multiple, in fact double, root of the minimal polynomial.) – 2017-01-09