Inducing quotients space and show the linear mapping

Question

Suppose U is a subspace of V invariant under a linear transformation T : V → V . Prove that T induces a linear map $\bar T : V /U → V /U$ of quotients given by $\bar T(v + U) = T(v) + U$. Prove that the minimal polynomial of $\bar T$divides the minimal polynomial of T.

I find similar question here but I am confused why it tries to show isomorphism since it is asked to show linear map. Do I miss anything? And for the quotient space linear mapping, $$ T(u + v) = T(u) + T(v) $$ for all $u, v ∈ V$ $$T(av) = aT(v)$$for all $a ∈ F$ and $v ∈ V$

How should it fully be shown?

Inducing a Linear Transformation on the Quotient Space

Answer 1

We have $v+U = (v+u)+U$ for each $u\in U$. The first step is to show that $\bar T(v+U) = \bar T(v+u+U)$ using the invariance of $T$. This gives well-definedness.

Addition is obvious: \begin{eqnarray*} \bar T((v+U)+(w+U)) &=& \bar T((v+w)+U) = T(v+w)+ U \\ &=& (T(v)+T(w))+U = (T(v)+U)+(T(w)+U) \\ &=& \bar T(v+U) + \bar T(w+U). \end{eqnarray*}

Scalar multiplication is similar.

Answer 2

The cosets $v+U$ and $(v+u)+U$ for $u\in U$ are the same. So they should have the same image: $\bar T((v+u)+U) =T(v+u)+U = (T(v)+T(u))+U =T(v) + (T(u)+U) = T(v)+ U$, since $T(u)\in U$ by assumption and so $T(u)+U=U$.