Finding minimal polynomial of a matrix

I have the following matrix:

$\displaystyle \begin{bmatrix}0 & 0 & c\\ 1 & 0 & b\\ 0 & 1 & a\end{bmatrix}$

its characteristic polynomial is $\displaystyle f(x)=x^3-ax^2-bx-c$

I need to prove that this is also its minimal polynomial.

How do I approach this problem?

I can't even factorize this polynomial.

Re: Finding minimal polynomial of a matrix

Hey Stormey.

You should be able to factorize a cubic polynomial exactly. Take a look at this:

Cubic function - Wikipedia, the free encyclopedia

Re: Finding minimal polynomial of a matrix

Hi chiro, and thanks for the help.

since I'm dealing with parameters a,b and c, the calculations became much too complicated and messy then I think they should be.

Isn't there another way to conclude that the polynomial of a lower degree will not satisfy the equation?

anyway, couple of things I thought that could work:

according to Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation, so I tried to solve this equation:

$\displaystyle A^3-aA^2-bA-c=0$

and tried to get some direction from there.

It yielded quite a nice, easy equation:

$\displaystyle \begin{bmatrix}0 & 0 & 0\\ 0 & ab-a & 0\\ 0 & 0 & 0\end{bmatrix}=0$

$\displaystyle a(b-1)=0$

not sure what now, though.

Re: Finding minimal polynomial of a matrix

my bad.

the result is:

$\displaystyle \begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\\ -a^2 & 0 & 0\end{bmatrix}=0$

meaning a=0.

what conclusion can I draw from that?

Re: Finding minimal polynomial of a matrix

It’s the minimum polynomial if it has distinct roots.

For general abc it has distinct roots because (x-g)^{2}(x-h) and (x-g)^{3} have only 2 and 1 constants respectiveley.

Re: Finding minimal polynomial of a matrix

Quote:

Originally Posted by

**Hartlw** It’s the minimum polynomial if it has distinct roots.

For general abc it has distinct roots because (x-g)^{2}(x-h) and (x-g)^{3} have only 2 and 1 constants respectiveley.

but how do you know that a, b and c are the roots without factorizing the polynomial?

(they can't be the roots)

Re: Finding minimal polynomial of a matrix

abc are not the roots. If x3-ax2-bx-c had multiple roots, it would be of the form (x-g)^{2}(x-h) or (x-g)^{3} which have less than 3 constants.

abc could be determined st f(x) had the form (x-h)^{3}, for example, from the identity:

x^{3}-ax^{2}-bx-c = (x-g)^{3} which would give a,b,c in terms of g.

EDIT: Fundamental theorm of algebra says every nth degree polynomial has n roots, some of which may be complex.

Re: Finding minimal polynomial of a matrix

so, to sum things up:

I know that this polynomial is the minimal because a, b and c are distinct, and therfore the polynomial has three distinct roots, and that's why it's also the minimal?

Re: Finding minimal polynomial of a matrix

Quote:

Originally Posted by

**Stormey** so, to sum things up:

I know that this polynomial is the minimal because a, b and c are distinct, and therfore the polynomial has three distinct roots, and that's why it's also the minimal?

No. abc can be distinct but related so that the polynomial does not have distinct roots.

Consider (x-l1)(x-l2)^{2}.

You are getting into the question, under what conditions on abc does f(x) have multiple roots. Their being distinct is not sufficient.