# Why is the largest invariant factor of $A-xI$ the minimal polynomial of $A$?

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

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.