Given a minimal polynomial, find a transformation

**Mimi89** · December 30th 2009, 05:49 AM

Hello!

Could you help me with the last part of this question?:

Let V be a vector space over C, and let T:V->V be a linear transforamtion with characteristic polynomial C(x)=(x^2+1)(x-1)^2\x^2 How many possibilities are there for the minimal polynomial? For each possibility, give an example of a suitable linear transformation.

Now, as the minimal polynomial (M(x)) divides the Characteristic polynomial, then M(x) is either x+i, x-i, x+1, 0, or any combination of these (sometimes up to powers of 2).

As every root (or irreducible polynomial) of C(x) is also a root of M(x), then we are left with four possibilities (where \ means we're no longer in the superscript):

M1=(x^2+1)(x-1)x , M2=(x^2+1)(x-1)^2\x, M3= (x^2+1)(x-1)x^2, or M4=(x^2+1)(x-1)^2\x^2

What I'm struggling with is finding functions whose minimal polynomials are M1, M2, M3, and M4, respectively. Is there an algorithm for this?

I know that the degree of the characteristic polynomial equals the dimention of the vector space. Hence, dim(V)=6. I'm thinking about representing the functions (F1-F4, respectively) as diagonal or upper triangular matrices (as otherwise the calculations would be enormous), but I'm not sure where to start from. (matrices wrt the standard basis) A friend suggested using the Jordan normal form, as well (almoust-diagonal matrices).

Could anyone offer any suggestions/do one as an example?

Your help is greatly appreciated!

M.

P.S. Sorry for the format, but I'm new to LaTex, and all formulas I tried ended up as "syntax errors"....

**tonio** · December 30th 2009, 07:39 AM

Originally Posted by Mimi89

Hello!

Could you help me with the last part of this question?:

Let V be a vector space over C, and let T:V->V be a linear transforamtion with characteristic polynomial C(x)=(x^2+1)(x-1)^2\x^2

What's written here? Is it $C(x)=(x^2+1)(x-1)^2x^2$ ?

How many possibilities are there for the minimal polynomial? For each possibility, give an example of a suitable linear transformation.

Now, as the minimal polynomial (M(x)) divides the Characteristic polynomial, then M(x) is either x+i, x-i, x+1, 0, or any combination of these (sometimes up to powers of 2).

No, this isn't correct: not only $M(x)\mid C(x)$ but ALSO they both have the same irreducible factors! Thus, $M(x)$ must be divided by each and all of the following factors: $x-i\,,\,x+i\,,\,x-1\,,\,x$ , and this already makes our work much easier.

As every root (or irreducible polynomial) of C(x) is also a root of M(x),

And this is exactly what I meant above...

then we are left with four possibilities (where \ means we're no longer in the superscript):

M1=(x^2+1)(x-1)x , M2=(x^2+1)(x-1)^2\x, M3= (x^2+1)(x-1)x^2, or M4=(x^2+1)(x-1)^2\x^2

What I'm struggling with is finding functions whose minimal polynomials are M1, M2, M3, and M4, respectively. Is there an algorithm for this?

Yes. Google "companion matrix". For example, let us build a matrix whose minimal pol. (and also characteristic, of course...these companion matrices kick ass!) is

$(x^2+1)(x-1)x^2=x^5-x^4+x^3-x^2$ :

$\begin{pmatrix}0&0&0&0&0\\1&0&0&0&0\\0&1&0&0&1\\0& 0&1&0&\!\!\!-1\\0&0&0&1&1\end{pmatrix}$

Tonio

I know that the degree of the characteristic polynomial equals the dimention of the vector space. Hence, dim(V)=6. I'm thinking about representing the functions (F1-F4, respectively) as diagonal or upper triangular matrices (as otherwise the calculations would be enormous), but I'm not sure where to start from. (matrices wrt the standard basis) A friend suggested using the Jordan normal form, as well (almoust-diagonal matrices).

Could anyone offer any suggestions/do one as an example?

Your help is greatly appreciated!

M.

P.S. Sorry for the format, but I'm new to LaTex, and all formulas I tried ended up as "syntax errors"....

.

**Mimi89** · December 31st 2009, 05:16 AM

Originally Posted by Tonio

Yes. Google "companion matrix". For example, let us build a matrix whose minimal pol. (and also characteristic, of course...these companion matrices kick ass!) is

:

Tonio

Thank you a lot for this! These companion matrices are really useful (especially in finding the matrix for
$M4=(x^2+1) (x-1)^2 x^2=x^6-2x^5+2x^4-2x^3+x^2$ , which would be

$\begin{pmatrix}0&0&0&0&0&0\\1&0&0&0&0&0\\0&1&0&0&0 &\!\!\!-1\\0&0&1&0&0&2\\0&0&0&1&0&\!\!\!-2\\0&0&0&0&1&2\end{pmatrix}$
)

But from what I read, these help find the matrix for a characteristic polynomial only when the characteristic polynomial and the minimal polynomial coincide. What about the other cases?

In the example you gave, the characteristic and minimal polynomials coincide. But what if I need a matrix with characteristic polynomial $C(x)=(x^2+1) (x-1)^2 x^2$ and minimal polynomial M3=

?

I would need a 6x6 matrices for this (and all of M1-M4), as the degree of the characteristic polynomial is 6.

Thank you again for suggesting companion matrices - these helped me in another problem I had been struggling with!

-M.

P.S. The code from the post makes a nice example for a beginner to copy from.

**tonio** · December 31st 2009, 06:02 AM

Originally Posted by Mimi89

Thank you a lot for this! These companion matrices are really useful (especially in finding the matrix for
$M4=(x^2+1) (x-1)^2 x^2=x^6-2x^5+2x^4-2x^3+x^2$ , which would be

$\begin{pmatrix}0&0&0&0&0&0\\1&0&0&0&0&0\\0&1&0&0&0 &\!\!\!-1\\0&0&1&0&0&2\\0&0&0&1&0&\!\!\!-2\\0&0&0&0&1&2\end{pmatrix}$
)

But from what I read, these help find the matrix for a characteristic polynomial only when the characteristic polynomial and the minimal polynomial coincide. What about the other cases?

In the example you gave, the characteristic and minimal polynomials coincide. But what if I need a matrix with characteristic polynomial $C(x)=(x^2+1) (x-1)^2 x^2$ and minimal polynomial M3=

?

I would need a 6x6 matrices for this (and all of M1-M4), as the degree of the characteristic polynomial is 6.

Thank you again for suggesting companion matrices - these helped me in another problem I had been struggling with!

-M.

P.S. The code from the post makes a nice example for a beginner to copy from.

Ok, then you may have to use Jordan's Canonical Form this time:

$\begin{pmatrix}i&0&0&0&0&0\\0&\!\!\!-i&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&1\\ 0&0&0&0&0&0\end{pmatrix}$

The above matrix has a char. pol $C(x)=(x^2+1)(x-1)^2x^2$ , but his minimal pol. is $m(x)=(x^2+1)(x-1)x^2$ ...note that all the irred. factors of the min. pol.

are linear BUT the factor $x^2$ , and that's why we've a Jordan Block of size 2 for the eigenvalue zero (the last right-down most block), and all the Jordan blocks for

the eigenvalues $\pm i\,,\,1$ are of size 1.

Tonio

**Mimi89** · December 31st 2009, 06:19 AM

Originally Posted by tonio

Ok, then you may have to use Jordan's Canonical Form this time:

$\begin{pmatrix}i&0&0&0&0&0\\0&\!\!\!-i&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&1\\ 0&0&0&0&0&0\end{pmatrix}$

The above matrix has a char. pol $C(x)=(x^2+1)(x-1)^2x^2$ , but his minimal pol. is $m(x)=(x^2+1)(x-1)x^2$ ...note that all the irred. factors of the min. pol.

are linear BUT the factor $x^2$ , and that's why we've a Jordan Block of size 2 for the eigenvalue zero (the last right-down most block), and all the Jordan blocks for

the eigenvalues $\pm i\,,\,1$ are of size 1.

Tonio

Thanks a lot for the example! It was very helpful (and I think I can do the rest of the question now).

And by the way, Happy New Year!

M.

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