Define $l_a : F(a) → F(a) $ by $ l_a(x)=ax$, when $[F(a):F]=n$ .
show that the minimal polynomial of $a$ over $F$ is the same as the minimum polynomial of $l_a$ as defined in linear algebra.
this is what i've done to prove :
the associated matrix of $l_a$ is a $n*n$ matrix (because $F(a)$ is a $n$ dimensional space) with $"a"$ on all the main diagonal entries & zero everywhere else, so the characteristic polynomial of $l_a$ is $(a-\lambda)^n$,that is a monic polynomial with degree $n$ and with $a$ as a root.
and because we have $[F(a):F]=n$,so the minimal polynomial of $a$ over $F$ has degree $n$.
so the minimal polynomial of $a$ over $F$ is the same as the characteristic polynomial of $l_a$.
is there any mistake? is it a complete proof?
thank you.
Hint: if we take the basis $\{1,a,\dots,a^{n-1}\}$. Suppose furthermore that $a^n = \sum_{k=0}^{n-1} c_k a^k$. Then the matrix of this transformation with respect to the basis is $$ T_{\ell} = \pmatrix{&&&&c_0\\1&&&&c_1\\&1&&&\vdots \\&&\ddots\\&&&1&c_{n-1}} $$ Now, what is the characteristic/minimal polynomial of this linear transformation? Perhaps this matrix looks familiar?