inducing a linear map on quotient spaces Suppose $U$ is a subspace of $V$ invariant under a linear transformation $T :
V \to V$ Prove that $T$ induces a linear map $B:V/U \to V/U$ of quotients
given by  $B(v + U) = T(v) + U$. Prove that the minimal polynomial of $B$
divides the minimal polynomial of $T$.
I've been looking at quotient vector spaces recently and came across this problem. I don't really understand how to begin so could somebody please help me out with a solution.
 A: We begin by noting that for vector spaces $X,Y$, a subspace $S \subset X$ and a linear map $F \colon X \to Y$, we have an induced map
$$\tilde{F} \colon X/S \to Y$$
that satisfies $\tilde{F}(x+S) = F(x)$ for all $x\in X$ if and only if $S \subset \ker F$. [prove it, or cite a theorem]
Then we apply the above to the situation $S = U$, $X = V$, and $Y = V/U$, with $F = \pi\circ T$, where $\pi \colon V \to V/U$ is the canonical projection.
For the remaining part, note that $\widetilde{\pi\circ p(T)} = p(\widetilde{\pi\circ T}) = p(B)$ for all polynomials $p\in K[X]$. [prove it, or cite a theorem]
A: The induced map is obviously linear. We have to check that it is well-defined. Let $v+U\in V/U$ and $u\in U$. Then
$$B(v+U) = T(v) + U = T(v) + \underbrace{T(u)}_{\in U}+U = T(v+u)+U = B(v+u+U)$$
Now assuming the space $V$ is finite dimensional and over a field $k$, let $p_T,\ p_B\in k[x]$ be the minimal polynomials of $T$ and $B$ respectively, i.e. $p_T$ is the polynomial of least degree such that $p_T(T)=0$, similar for $B$. Then to show that $p_B$ divides $p_T$, it is enough to show that $p_T(B) = 0$. So let $v+U\in V/U$ be any element:
$$p_T(B)(v+U) \underbrace{=}_{\mathrm{you\ can\ check\ this}}p_T(T)(v)+U=0+U$$
This concludes the proof.
