Minimal and Characterisitic Polynomial

**aman_cc** · October 11th 2009, 12:25 AM

T be a linear operator on a finite dimensional vector space.
p(x) - Minimal Polynomial of T
q(x) - Characteristic Polynomial of T

I want to prove the p(x)|q(x)

I know that every root of q(x) is a root of p(x) and vice a versa. I guess we can use this fact to prove the above.

But I do not want to use this - because we might not be able to factorize the polynomials. Is there a better way please? Thanks

**NonCommAlg** · October 11th 2009, 12:48 AM

Originally Posted by aman_cc

T be a linear operator on a finite dimensional vector space.
p(x) - Minimal Polynomial of T
q(x) - Characteristic Polynomial of T

I want to prove the p(x)|q(x)

I know that every root of q(x) is a root of p(x) and vice a versa. I guess we can use this fact to prove the above.

But I do not want to use this - because we might not be able to factorize the polynomials. Is there a better way please? Thanks

by Cayley-Hamilton we have $q(T)=0.$ we also have $q(x)=s(x)p(x) + r(x),$ for some polynomials $s(x),r(x)$ with $r(x)=0$ or $0 < \deg r(x) < \deg p(x).$

but $r(T)=0$ because $q(T)=p(T)=0.$ thus either $r(x)=0$ or $\deg r(x) \geq \deg p(x),$ by minimality of $p(x).$ hence $r(x)=0.$

**aman_cc** · October 11th 2009, 01:15 AM

Originally Posted by NonCommAlg

by Cayley-Hamilton we have $q(T)=0.$ we also have $q(x)=s(x)p(x) + r(x),$ for some polynomials $s(x),r(x)$ with $r(x)=0$ or $0 < \deg r(x) < \deg p(x).$

but $r(T)=0$ because $q(T)=p(T)=0.$ thus either $r(x)=0$ or $\deg r(x) \geq \deg p(x),$ by minimality of $p(x).$ hence $r(x)=0.$

Thanks - I was trying to prove Cayley-Hamilton with this result

Stupid of me

If I told you that T is diagonizable - Then I guess I can use the root logic (every root of q(x) is a root of p(x) and vice a versa) result to prove it and not rely on Cayley-Hamilton Theorem. Correct?

**HallsofIvy** · October 11th 2009, 06:49 AM

What definition of "miminal polynomial" are you using? The one I would use is that the minimal polynomial of linear operator (matrix) A is the monic polynomial, P, of lowest degree satisfying P(A)= 0. Yes, the minimal polynomial and characteristic polynomial have the same roots which means they have almost the same factors. What is the difference between them? Why are they not exactly the same polynomial?

**aman_cc** · October 11th 2009, 07:03 AM

Originally Posted by HallsofIvy

What definition of "miminal polynomial" are you using? The one I would use is that the minimal polynomial of linear operator (matrix) A is the monic polynomial, P, of lowest degree satisfying P(A)= 0. Yes, the minimal polynomial and characteristic polynomial have the same roots which means they have almost the same factors. What is the difference between them? Why are they not exactly the same polynomial?

I do mean the same thing with I say minimal polynomial.

I guess (in the case when A is diagonalizable) chr polynomial will be of degree 'n' (equal to the dimension of the vector space) but have lesser num of distinct roots (as order of roots might be >1)

Thus chr poly might be different from min polynomial. Am I making sense?

**NonCommAlg** · October 11th 2009, 10:01 AM

Originally Posted by aman_cc

I do mean the same thing with I say minimal polynomial.

I guess (in the case when A is diagonalizable) chr polynomial will be of degree 'n' (equal to the dimension of the vector space) but have lesser num of distinct roots (as order of roots might be >1)

Thus chr poly might be different from min polynomial. Am I making sense?

if $T$ is diagonalizable and $\lambda_1, \cdots, \lambda_k$ are the distinct eigenvalues of $T,$ then the minimal polynomial of $T$ would be $\prod_{j=1}^k(x-\lambda_j).$ the characteristic polynomial of $T$ in this case would be in the
form $\prod_{j=1}^k (x-\lambda_j)^{n_j},$ where $n_j$ is the number of eigenvalues of $T$ which are equal to $\lambda_j.$

you can find standard facts like this (and much more) about minimal and characteristic polynomial of a linear transformation in any decent linear algebra textbook.

**HallsofIvy** · October 11th 2009, 10:45 AM

I wouldn't even worry about T being "diagonalizable". The characteristic polynomial of T is $(x- \lambda_1)^m(x- \lambda_2)^n\cdot\cdot\cdot(x- \lambda_n)^k$ . We can theoretically factor the polynomial into linear factors (possibly with complex eigenvalues) even if we can't actually find the factors. The minimal polynomial will have exactly the same factor but possibly to lower power. We can "factor" those lower powers out so it follows that the minimal polynomial divides the characteristic polynomial.

**aman_cc** · October 11th 2009, 07:03 PM

Originally Posted by NonCommAlg

if $T$ is diagonalizable and $\lambda_1, \cdots, \lambda_k$ are the distinct eigenvalues of $T,$ then the minimal polynomial of $T$ would be $\prod_{j=1}^k(x-\lambda_j).$ the characteristic polynomial of $T$ in this case would be in the
form $\prod_{j=1}^k (x-\lambda_j)^{n_j},$ where $n_j$ is the number of eigenvalues of $T$ which are equal to $\lambda_j.$

you can find standard facts like this (and much more) about minimal and characteristic polynomial of a linear transformation in any decent linear algebra textbook.

Thanks very much NonCommAlg

**aman_cc** · October 11th 2009, 07:05 PM

Originally Posted by HallsofIvy

I wouldn't even worry about T being "diagonalizable". The characteristic polynomial of T is $(x- \lambda_1)^m(x- \lambda_2)^n\cdot\cdot\cdot(x- \lambda_n)^k$ . We can theoretically factor the polynomial into linear factors (possibly with complex eigenvalues) even if we can't actually find the factors. The minimal polynomial will have exactly the same factor but possibly to lower power. We can "factor" those lower powers out so it follows that the minimal polynomial divides the characteristic polynomial.

Yes I get this. Thanks very much!

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