Can we say that $m_{T_y} = m_y$?

Question

Let $L|K$ be a finite field extension with $[L:K]=n.$ Let $y \in L.$ Let $m_y$ be the minimal polynomial of $y$ over $K.$ Let $T_y$ be the $K$-linear map on $L$ defined by $$T_y(x) := yx,\ x \in L.$$ Then we know that the characteristic polynomial $\chi_y$ of $y$ over $K$ is defined by $$\chi_y := \chi_{T_y} = \det(XI_n - M)$$ where $M$ is the matrix representation of $T_y$ w.r.t. some fixed $K$-basis $B$ (say) of $L.$ Let $m_{T_y}$ denote the minimal polynomial of $T_y.$ Then I have a question regarding that. Can we say that $m_y=m_{T_y}$? If so how do I prove this equality?

Any help regarding this will be highly appreciated. Thank you very much.

EDIT $:$ I have proved that $\chi_{y} (y) = 0.$ Since all the zeros of characteristic polynomial of a linear operator are also zeros of it's minimal polynomial it follows that $m_{T_y} (y)=0.$ But that means $m_y\ |\ m_{T_y}.$

Answer 1

Hint: Prove that the set of polynomials that kill $T_y$ is the same as the set of polynomials that kill $y$. For that, prove that $P(T_y)(x)=P(y)x$.