I have the following matrix:

its characteristic polynomial is

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

How do I approach this problem?

I can't even factorize this polynomial.

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- May 1st 2013, 12:14 PMStormeyFinding minimal polynomial of a matrix
I have the following matrix:

its characteristic polynomial is

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

How do I approach this problem?

I can't even factorize this polynomial. - May 1st 2013, 09:16 PMchiroRe: 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 - May 2nd 2013, 05:56 AMStormeyRe: 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:

and tried to get some direction from there.

It yielded quite a nice, easy equation:

not sure what now, though. - May 2nd 2013, 08:25 AMStormeyRe: Finding minimal polynomial of a matrix
my bad.

the result is:

meaning a=0.

what conclusion can I draw from that? - May 2nd 2013, 06:25 PMHartlwRe: 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. - May 3rd 2013, 02:24 AMStormeyRe: Finding minimal polynomial of a matrix
- May 3rd 2013, 06:34 AMHartlwRe: 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. - May 3rd 2013, 07:21 AMStormeyRe: 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? - May 3rd 2013, 07:51 AMHartlwRe: Finding minimal polynomial of a matrix