# minimal polynomials

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• Mar 5th 2009, 06:30 PM
robeuler
minimal polynomials
Let V be a finite dimensional vector space over the field F, with dim(V)=n. We know that End(V), the space of linear transformations from V to itself. dim(End(V)) is n^2.

The minimal polynomial of a linear transformation T is the unique monic polynomial which generates the ideal Ann(V) in F[x]. This polynomial divides the characteristic polynomial of T. Since the degree of the characteristic polynomial is n, the minimal polynomial is at most degree n.

I am asked to prove that the minimal polynomial of T has at most degree n^2, using that End(V) has dimension n^2. So I am asked to prove a weaker bound on the degree than I already know is true.
• Mar 5th 2009, 07:31 PM
ThePerfectHacker
Quote:

Originally Posted by robeuler
Let V be a finite dimensional vector space over the field F, with dim(V)=n. We know that End(V), the space of linear transformations from V to itself. dim(End(V)) is n^2.

The minimal polynomial of a linear transformation T is the unique monic polynomial which generates the ideal Ann(V) in F[x]. This polynomial divides the characteristic polynomial of T. Since the degree of the characteristic polynomial is n, the minimal polynomial is at most degree n.

I am asked to prove that the minimal polynomial of T has at most degree n^2, using that End(V) has dimension n^2. So I am asked to prove a weaker bound on the degree than I already know is true.

I think the idea here is to let $A$ be the matrix corresponding to $T$, then $\{ I,A,A^2,...,A^{n^2}\}$ is linearly dependent because it has $n^2+1$ elements. Therefore, there exists $a_j \in F, 0\leq j\leq n^2+1$ so that $\sum_{j=0}^{n^2+1}a_j A^j = \bold{0}$. Therefore, $f(X) = \sum_{j=0}^{n^2+1} a_j\cdot I ~X^j$ is a polynomial which has $A$ as a zero (I is the identity matrix).