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For $A in M_{ntimes n}(F)=V$, we define a linear operator $T$ such that $T(X)=AX$ for $Xin V$.

I need to show that the characteristic polynomial of $T$ is $(f_A(t))^n$, where $f_A(t)$ is the characteristic polynomial of $A$.

My latest approach has been to define $W$, the $T$-cyclic subspace of $V$ generated by some $X$. Then,

$beta = {X,,T(X),,T^2(X),,dots,,T^{k-1}(X)}$

is an ordered basis for $W$, where $k=dim(W)$. From here, I'm thinking I need to extend the basis to $n$ so that it's a basis for $V$ and then show somehow that, since $T^n(X)=A^nX$, my desired result falls out.

Am I on the right track? Any pushes in the right direction that could help me put the pieces together? Or is there a totally different way to solve this that has eluded me?

Extending the ordered basis so that $k=n$, we have

$beta = {X, AX, A^2X,...,A^{n-1}X}$

as a basis for $V$. Now we simply express $T$ as $[T]_{beta}$, the matrix form of the operator, and take the characteristic polynomial of the result. Note that:

$T(X)=AX$

$T(AX)=A^2X$

$dots$

$T(A^{n-1}X)=A^nX$

and thus

$[T]_{beta}=AI_n$

which has the characteristic polynomial given by $(f_A(t))^n$, as needed.