Assume V is finite dimensional,that $T:V \rightarrow V$ is a linear operator, and W is a T-invariant subspace. Define $\bar{T}: V/W \rightarrow V/W$ by $\bar{T}(v+W)=T(v)+W$ for any $v+W \in V/W$. Let f(t), g(t) and h(t) be the characteristic polynomials of $T,T_W$ and $\bar{T}$, respectively. Prove $f(t)=g(t) h(t)$
Choose an ordered basis $\gamma={v_1,...,v_k}$ for W, extend it to an ordered basis $\beta={v_1,...,v_k,v_{k+1},...,v_n}$ for V.
Let $\alpha={v_{k+1}+W,...,v_n+W}$ be an ordered basis for V/W.
Since W is a T-invariant subspace of V, the matrix representation of T is $[T]_\beta=\begin{pmatrix} B_1 \ B_2 \\ 0 \ B_3 \end{pmatrix}$, where $B_1=[T]_\gamma$ and $B_3=[\bar{T}]_\alpha$.
Since f(t) is the characteristic polynomial of T and g(t) is the characteristic polynomial of $T_W$, then$ f(t)=det([T]_\beta- tI) $and $g(t)=det(B_1-tI)$.
Since the set $\alpha={v_{k+1}+W,...,v_n+W}$ is a basis for V/W, then for each $j=k+1,k+2,...,n$.
$\bar{T}(v_j)=T(v_j)+W=[\sum_{l=1}^k(B_2)_{lj} v_l+\sum_{i=1}^k(B_3)_{ij} v_i]+W=\sum_{l=k+1}^k(B_3)_{ij}(v_i+W)$.
So $B_3=[\bar{T}]_\alpha$.
So $f(t)=g(t)h(t)$
Is it correct?