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

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?

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  • $\begingroup$ Another route I have found to investigate is to start by using the fact that the characteristic polynomial of any $A\in M_{n\times n}(F)$ is equal to $(-1)^n(a^nt^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0$). $\endgroup$
    – notadoctor
    Nov 15, 2018 at 22:25
  • $\begingroup$ Nevermind, I've managed to solve it. See my answer below. $\endgroup$
    – notadoctor
    Nov 16, 2018 at 8:48

1 Answer 1

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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.

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