Characteristic and minimal polynomial in dual space

Let $V$ be a finite dimensional vector space over the field $\mathbb{F}$ and define a linear transformation $T:V\rightarrow V$.

We have that the dual space is $V'=\text{Hom}(V,\mathbb{F})$ and $T':f \mapsto f \ \circ \ T$, and know that if we have a basis $\mathcal{B}$ of $V$ then we can construct a dual basis $\mathcal{B}'$. I need to solve the question:

State a relationship between the characteristic polynomials of $T$ and $T'$, and the minimal polynomials of $T$ and $T'$. Explain your answer.

The matrix of $T'$ wrt $\mathcal{B}'$ is the transpose of the matrix of $T$ wrt $\mathcal{B}$. So my thinking is that because $\det A=\det A^{tr}$, we have that $\chi_{T'}(x)=\chi_T(x)$.

Is this correct? And how can I find a relationship between the minimal polynomials $m_{T'}(x)$ and $m_T(x)$?

Answer

Since, by definition $\chi_T(\lambda)=\det(T-\lambda\operatorname{Id})$, since a matrix and its transpose have the same determinants, and since

$$(T-\lambda\operatorname{Id})^{\mathrm{tr}}=T^{\mathrm{tr}}-\lambda\operatorname{Id},$$ $T$ and $T'$ have the same characteristic polynomials.

And if $P(T)=0$ for some polynomial $P(x)\in\mathbb F[x]$,then

$$P(T^{\mathrm{tr}})=P(T)^{\mathrm{tr}}=0^{\mathrm{tr}}=0.$$ So, $T$ and $T'$ have the same minimal polynomials.