1
$\begingroup$

From what I see the definition of minimal polynomial is the polynomial of smallest degree that $A$ is a root of. Invariant factors are the diagonal entries of the Smith Normal Form of a matrix.

I understand that since determinant is unchanged by row and column operations, $\det(A-xI)=\det(SNF(A-xI))$, hence the characteristic polynomial is the product of the diagonal entries of $\det(SNF(A-xI))$. But how can I show that the minimal polynomial as defined above, is the largest invariant factor (which I know is a common multiple of all the other invariant factors).

Any comment or hint is greatly appreciated. Thank you

$\endgroup$
0
$\begingroup$

Because the largest invariant factor is an annihilator of $ F[x] $-module, and every annihilator is a multiple of the largest invariant factor since $ F[x] $ is a domain and have no zero divisor. And the minimal polynomial is defined by the annihilator polynomial that has the smallest degree, thus it is the largest invariant factor of $ F[x] $-module.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.